STRENGTH OF MATERIALS
By
Dr. Fathelarahamn Mohamed Adam
Course Content:
311 EngC Strength of Materials: (4: 3,3,0)
Review of basic principles of static. Plane stress and strain and
their relations. Normal force. Shear force and bending moment
diagrams for determinate members. Relation between load, shear
force and bending moment. Review of properties of plane sections.
Bending stresses in beams. Symmetric and unsymmetrical bending.
Shear flow and shearing stresses in beams. Torsion of circular
members. Compound stresses. Transformation of plane stress and
strain. Mohr’s circle of stress. Deflection of beams. Double
integration and moment-area methods. Buckling of columns.
Euler’s formula. Laboratory experiments.
Prerequisite:
221 EngC Engineering Mechanics (Static):
Refrences:
Text Books:
1. William A. Nash, Merle C. Potter, “Strength of Materials” 5th
Edition, Schaum’s Outlines Series, 2011
2. George F. Limbrunner and Leonard Spiegel “Applied Statics and
Strength of Materials” 5th Edition, Mar 9, 2008.
References:
1. Ferdinand P. Beer, Johnston, E.R., DeWof, J.T. and David F.
Mazurek "Mechanics of Materials", 6th edition, McGraw Hill,
2012.
2. Barry S. Onouye and Kevin Kane “Statics and Strength of
Materials for Architecture and Building Construction”, Jul 3,
2006.
About Strength of Material
Strength of material is the study of the behavior of structural
members under the action of external loads, taking into account the
internal forces created and the resulting deformations. Analysis is
directed towards determining the limiting loads which the member
can stand before the material fail or excessive deformation occurs.
Strength of Material – Dr. Fathelrahman Mohamed Adam
1
General review
System of Units (SI System):
The SI system (International System of Units) is the modern metric
system of measurement and the dominant system of international
commerce and trade. SI units are gradually replacing Imperial (old
English unit) and USCS units (United States Customary System Units are
the measuring units used in the U.S. consisting of the mile, inch, gallon,
second and pound)
The SI system includes:
• SI base and derived units described in terms of acceptable SI units.
• SI Prefixes.
Table contains some base and derived units and conversion factors
Quantity
Length
Mass
Time
Area
Volume
Velocity
Acceleration
Angular velocity
Force, Weight
Density
Specific weight
Pressure, stress
Work, Energy
Power
Symbol
L
m
t
A
V
V
a
w
F,W
r
g
P
W,E,U
W
SI Units
m
kg
s
m2
m3
m/s
m/s2
rad/s
rad/s
N
kg/m3
N/m3
kPa
J
W
W
Strength of Material – Dr. Fathelrahman Mohamed Adam
English Units
ft
lbm
sec
ft2
ft3
ft/sec
ft/sec2
rad/sec
rpm
lbf
lbm/ft3
lbf/ft3
psi
ft-lbf
ft-lbf/sec
hp
To Convert
from English
to SI units
Multiply by
0.3048
0.4536
1
0.0929
0.02832
0.3048
0.3048
1
9.55
4.448
16.02
157.1
6.895
1.356
1.356
746
2
SI prefixes:
Prefixes for SI Units
Prefix
Symbol
Abbreviation
1018
exa
E
1015
peta
P
1012
tera
T
109
giga
G
106
mega
M
103
kilo
k
102
hecto
h
101
deka
da
10-1
deci
d
10-2
centi
c
10-3
milli
m
10-6
micro
μ
10-9
nano
n
10-12
pico
p
10-15
femto
f
10-18
atto
a
Scalar quantities:
The scalar quantities are those quantities which have magnitude only such
as length, mass, time, distance, volume etc.
Vector quantities:
The vector quantities are those quantities which have both magnitude and
direction such as force, displacement, velocity, acceleration etc.
Analysis process:
Action
Resistance
Loads
Tension or Compression
Bending Action (uni-axial)
Bending Action (biaxial)
Strength of Material – Dr. Fathelrahman Mohamed Adam
Reaction
Axial Force
single bending moment, Shear Force
double bending moment , shear force
3
Coordinates System:
There are many types of system coordinates like: Cartesian coordinates
system, polar coordinate system, cylindrical coordinate system, natural
coordinate system, curvilinear coordinate system. The most general
coordinate system used in structural analysis is Cartesian coordinate
system, which classify to two axes the one is the local axis and the other
is global axis.
Local axis: is the axis defines the element locally and the terms
used for it is x, y and z. where x is the axial axis which is the axis
pass through the element between ends, y is perpendicular to x and
z is perpendicular to the plane xy.
Global axis: is the axis defines the structure, in other words all the
elements are corporate the same axis termed as X, Y and Z.
Supports:
The supports are special joints where located at the ends of elements and
where the applied loads are finally transmitted to. Paths at which the
applied loads are transferred to the system to its supports are called "load
paths". The resisting force forces against translation/rotation at the
support are called "support reactions"
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4
Loads:
The loads that act on common civil engineering structures can be grouped
according to their nature and source into three classes: (1) dead loads due
to the weight of the structural system itself and any other material
permanently attached to it; (2) live loads, which are movable or moving
loads due to the use of the structure; and (3) environmental loads, which
are caused by environmental effects, such as wind, snow, and
earthquakes. The loads are classified as its acting along the local axis to
axial loads (along the axial axis x) and transverse loads (along transverse
axis y or/and z). The loads are classified according to the location where it
acts to two types: 1. joint loads if act at place defined as joint and 2.
Strength of Material – Dr. Fathelrahman Mohamed Adam
5
member load if act between joints. The member loads may be
concentrated at point (concentrated load) or distributed along definite
length (distributed load) which it may be distributed uniformly i.e. has
the same value along the length or have variable value (varying linearly
or trapezoidal or any shape)
Forces:
Are the results of resistance of structures to loads accordance to its
behavior. The forces named as it acts along to direction of axis are:
1. Axial force: which it acts along the axial axis. Two types are
distinguished, according to the deformation they cause: tension
Strength of Material – Dr. Fathelrahman Mohamed Adam
6
causes elongation. Compression causes shortening. This force can
be found in bars and trusses and columns as well.
2. Shear force: which it acts along the transverse axis i.e. along the
x-section and at supports named as reaction. The shear force
generally accompanied by bending moment.
3. Bending moment: acts around the transverse axis. The shear force
and bending moments can be found in bending elements like beams
as an example.
4. Torsion moment: acts around the axial axis.
Forces on Supports:
Displacements:
Classify to two: Translation and Rotation.
Translation: mean the points translate from situation to situation
along the axes, and there are two types of translation: axial
translation (along axial axis) and transverse translation (along
transverse axes).
Rotation: is the bent of axial axis from its original situation and
equal to the tangent at the point considered, and it always about the
transverse axis.
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7
Relation between force and displacements:
At any points except supports:
axial force
–
axial translation
transverse force –
transverse translation
bending moment –
rotation
At support:
if there exist a force; the displacement is absent
Equilibrium Equation:
For a two dimensional body occupied a region on the X-Y plane,
equilibrium of this body implies that all forces and moments applied to
the body must satisfy the following three equations:
∑ FX = 0
∑ FY = 0
∑ MZ = 0
Strength of Material – Dr. Fathelrahman Mohamed Adam
(1)
(2)
(3)
8
Chapter 1
Review of properties of plane sections
Centroid and First Moment of Area
Definitions
Centroid is the geometric centre which represents a point in the plane
about which the area of the cross-section is equally distributed.
Centre of gravity (CG) for a body is a point which locates the gravity or
weight of the body. Centroid and CG are same for homogeneous material.
The First Moment of Area is analogous to a moment created by a Force
multiplied by a distance except this is a moment created by an area
multiplied by a distance and it is used to find out centroid.
Consider an area A located in the xy plane in the Figure shown below.
Denoting by x and y the coordinates of an element of area dA.
we define the first moment of the area A with respect to the x axis as the
integral:
=
(1)
Similarly, the first moment of the area A with respect to the y axis is
defined as the integral:
=
(2)
The centroid of the area A is defined as the point C of coordinates ̅ and
as shown in Figure which satisfy the following relations:
Strength of Material – Dr. Fathelrahman Mohamed Adam
9
=
,
=
̅ (3)
We note that the first moments of the area A can be expressed as the
products of the area and of the coordinates of its centroid:
Or
=
=
,
,
=
=
(4)
(5)
Example (1.1):
For the triangle shown in Figure below, find:
1. The first moment Qx of the area with respect to the x axis.
2. The ordinate of the centroid of the area.
Solution:
Strength of Material – Dr. Fathelrahman Mohamed Adam
10
From triangle similarity:
=
So:
ℎ−
ℎ
ℎ−
ℎ
=
The area of small element:
= ℎ−
ℎ
=
1. The first moment Qx of the area with respect to the x axis:
From equation (1):
=
=
ℎ
=
ℎ−
ℎ
!
ℎ
ℎ! ℎ!
− " =
− "
3
ℎ 2
3
2
ℎ!
ℎ
= # %=
ℎ 6
6
2. The ordinate of the centroid of the area:
From equation (5):
=
Where:
So:
=
ℎ
6
=
ℎ
2
2
1
= ℎ
ℎ
3
Note that: if an area A possesses an axis of symmetry, its centroid C is
located on that axis as shown in Figure below:
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11
From this note and from the above examples, the centriod of standard
shape can easily be found as tabulated in the Table below.
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12
Centriod of complex shape:
Where an area is of more complex shape, a simple method of determining
the location of the centroid can be used which divides the complex shape
into smaller simple geometric shapes for which the centroidal locations
may easily be determined such as (rectangles, right triangles, circles etc).
Then instead of using equations (4) and (5) to calculate the centriod, use
the following:
=&
' ' ,
and instead of A, use ∑
so by substitute in (5)
∑ ' '
=
,
∑ '
'
=&
=
∑
∑
' ̅' (6)
' ̅'
'
(7)
Example (1.2):
Locate the centroid C of the area A shown in Figure below.
Solution:
Divide the area to two areas and selecting the coordinate axes
shown as shown below:
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13
From symmetry of axis y, the ̅ is known and equal to zero.
From Figure:
A1 = 20(80) = 1600 mm2
A2 = 40(60) = 2400 mm2
from the x axis:
&
' '
*
= 70,,,
&
' '
=
= 30,,
* *
+
= 1600(70) + 2400(30) = 184,000,,!
&
&
'
=
'
=
*
+
= 1600 + 2400 = 4,000,,
∑
∑
' '
'
=
184,000
= 46,,
4,000
Example (1.3):
Locate the centroid of the shape shown in Figure below.
y
600 mm
1000 mm
All thicknesses are 10 mm
x
500 mm
Solution:
Divide the shape to three elements as shown in Figure with
specify the area and centriod of each area from the two axes x
and y.
Strength of Material – Dr. Fathelrahman Mohamed Adam
14
y
500
102.6 mm
3
300
550
C
365.8 mm
2
1
50
x
50
250
Element
b
h
1
500 100
2
100 400
3
1000 100
Summation
̅=
=
∑
∑
∑
∑
A (bxh)
50,000
40,000
100,000
x
250
50
50
190,000
=
=
y
50
300
550
Ax
Ay
2,500,000
12,000,000
55,000,000
19,500,000
2,500,000
12,000,000
55,000,000
69,500,000
19,500,000
= 102.6,,
190,000
69,500,000
= 365.8,,
190,000
Second Moment of Area or Moment of Inertia:
The second moment of area or second moment of inertia is a property of a
cross section that can be used to predict the resistance of beams to
bending and deflection, around an axis that lies in the cross-sectional
plane. The deflection of a beam under load depends not only on the load,
but also on the geometry of the beam's cross-section. This is why beams
with higher area moments of inertia, such as I-beams, are so often seen in
building construction as opposed to other beams with the same area.
The moment of inertia, of the area A shown in Figure with respect to the x
axis and y axis are defined, respectively, as:
Strength of Material – Dr. Fathelrahman Mohamed Adam
15
1 =
,
1 =
Polar moment of inertia:
The Polar moment of inertia of an area is a quantity used to predict an
object's ability to resist torsion. It is used to calculate the angular
displacement of an object subjected to a torque. It is analogous to the area
moment of inertia.
We now define the polar moment of inertia of the area A with respect to
point O shown in Figure as the integral:
23 =
4
where r is the distance from O to the element dA. While this integral is
again a double integral, it is possible in the case of a circular area to select
elements of area dA in the shape of thin circular rings, and thus reduce the
computation of JO to a single integration.
An important relation may be established between the polar moment of
inertia JO of a given area and the rectangular moments of inertia Ix and Iy
of the same area. Noting that:
r2 = x2 + y2
Strength of Material – Dr. Fathelrahman Mohamed Adam
16
so we can write:
23 =
4
=
(
+
)
=
+
Or:
JO = Ix + Iy
The radius of gyration:
The radius of gyration of an area A with respect to the x axis is defined as
the quantity rx, that satisfies the relation:
1 =4
where Ix is the moment of inertia of A with respect to the x axis.
Solving the Equation above for rx, we have:
1
4 =5
In a similar way, we define the radii of gyration with respect to the y axis
and the origin O. We write:
1
13
4 = 5 , 43 = 5
Example (1.4):
Find the moment of inertia and the radius of gyration about the x
axis for the rectangular section shown in Figure.
Solution:
select as an element of area a horizontal strip of length b and
thickness dy
Strength of Material – Dr. Fathelrahman Mohamed Adam
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Since the area at distance y from the x axis, the moment of
inertia of the strip with respect to that axis is:
1 =
where:
dA = b dy
the integration limited between –h/2 and +h/2
so:
1 =
1 =
!
3
"
6
6
=
3
ℎ
2
!
ℎ !
− −
"
2
ℎ!
ℎ! 1 1
1 =
7 + 8=
3 8 8
12
The radius of gyration about x axis is given by:
1
4 =5
where:
A=bh
so:
ℎ!
ℎ
ℎ
=5 =
= 3.464ℎ
4 =5
12( ℎ)
12
√12
Strength of Material – Dr. Fathelrahman Mohamed Adam
18
The moment of inertia about x axis (Ix), y axis (Iy) and polar
moment of inertia (IC), for standard shape can easily be found as
tabulated in the Table below:
Parallel axis theorem
The moment of area of an object about any axis parallel to the centroidal
axis is the sum of moment of inertia about its centroidal axis and the
product of area with the square of distance of from the reference axis.
1 =1 :+ Strength of Material – Dr. Fathelrahman Mohamed Adam
19
where:
1
:
is the moment of inertia about the centriod.
Example (1.5):
Find the second moment of inertia about the centriod C for the
section shown in Figure.
Solution:
Divide the area A into two areas A1 and A2, and compute the
moment of inertia of each area with respect to the x axis.
Note that the 1 : for the rectangular section is:
ℎ!
1 :=
12
for A1:
b1 = 80 mm, h1 = 20 mm
80(20)!
1 ;: =
= 53.33 10! ,,<
12
Strength of Material – Dr. Fathelrahman Mohamed Adam
20
for A2:
b2 = 40 mm, h2 = 60 mm
1
:
=
40(60)!
=
= 720 10! ,,<
12
Use the parallel axis theorem to find the 1 for the two areas.
Note that, the centriod is at distance 46 mm from bottom of
section as calculated from previous example.
1 =1 :+ Note that, d is the distance from the centriod of area to the axis x
for A1:
A1 = 80(20) = 1,600 mm2 , d1 = 10+14 = 24 mm
1
;
1
;
=1
:
;
+
* *
= 53.33 10! + 1,600(24) = 974.93 10! ,,<
for A2:
A2 = 40(60) = 2,400 mm2 , d1 = 16 mm
1
1
=
=1
:
=
+
= 720 10! + 2,400(16) = 1.334 10> ,,<
For the hole area:
1
=
1
=1 ; +1
=
= 974.93 10! + 1,334 10! = 2.31 10> ,,<
Strength of Material – Dr. Fathelrahman Mohamed Adam
21
Chapter 2
Simple Stress and Strain:
Stress:
Stress is defined as the force intensity or force per unit area. Here
we use a symbol σ to represent the stress.
@
? = (1)
where: A is the area of the x-section
P is the applied external load
Here we are using an assumption that the total load carried by the
rectangular bar is uniformly distributed over its cross section.
The basic units of stress in S.I units. (International system) are
N/m2 (or Pa)
KPa = 103 Pa
MPa = 106 Pa
GPa = 109 Pa
Some times N/mm2 for units are also used.
Types of Stresses:
Only two basic stresses exist:
1. Normal stress or Direct stress:
Is defined as force per unit area, If the stresses are normal to
the areas concerned, then these are termed as normal
Strength of Material – Dr. Fathelrahman Mohamed Adam
22
stresses. It must be uniaxial state of stress, biaxial state of
stress or triaxial state of stress as shown in Figures below:
Uniaxial state of stress
Biaxial state of stress
Triaxial state of stress
The normal stresses can be either:
tensile which the member is being stretched by the axial
force and is in tension and the deformation is characterized
by axial elongation.
or compressive which the member is being compressed by
the axial force and is in compression and the deformation is
characterized by axial shortening.
@
@
Tensile Stress
Strength of Material – Dr. Fathelrahman Mohamed Adam
23
@
@
Compressive Stress
2. Shear stress:
When a section is subjected to two equal and opposite
forces, acting tangentially across the resisting section, as a
result of which the body tends to shear off across the section,
the stress induced is called shear stress.
Note that the shear stress will always be tangential to the
area on which it acts.
The shear stresses can be computed from the following
equation:
@
A = (2)
where A is the shear area.
@
@
A
A
Complementary shear:
When a body influenced by a shear stress as shown in figure
below:
A
A
we observed that the body in unbalanced state and must be
balanced by the same shear but in opposite direction Á as
shown in figure below and the balanced shear stress called
the complementary shear stress.
A
Á
Á
A
Strength of Material – Dr. Fathelrahman Mohamed Adam
24
Strain:
If a bar is subjected to a direct load, and hence a stress, the bar will
change in length. If the bar has an original length L and changes in
length by an amount δ, the strain produced is defined as follows:
CD4EFG(H) =
IℎEGJKFGLKGJDℎ N
= (3)
M4FJFGELLKGJDℎ
O
Note that the strain is thus a measure of the deformation of the
material and is non-dimensional, i.e. it has no units; it is simply a
ratio of two quantities with the same unit.
Types of Strains:
The strain can be classified to two basic strains:
1. Normal strain or Direct strain:
It is as the direct stress due to the load and it must be tensile
or compressive according to the type of load as in the stress.
2. Shear strain:
Shear strain is defined as the strain accompanying a shear
stress. It is the angle in radian measure through which the
body gets distorted when subjected to an external shear
stress. It is denoted by γ and mathematically is equal to
change in right angle.
Strength of Material – Dr. Fathelrahman Mohamed Adam
25
Deformed shape
Original shape
S
A
A
Sign convention:
For the stress and strain as alike, tensile is positive and
compressive is negative.
The relation between stress and strain:
When a load is applied to an element of any material, there exist a
relationship between the load and small change in the element and
if the change is disappear (the element retain to the original shape)
when the load is removed, this state is called the elastic behavior
and the material be elastic under that load action. Experiments
have shown that the stress is directly proportional to the strain
within the straight line region. This relationship is expressed by
Robert Hook and defines by Hooke's Law which it stated by
“when a material is loaded within its elastic state, the stress is
directly proportional to the strain.” Mathematically:
? ∝ H(4)
The constant of the relation is stated by Thomas Young and known
by Young’s modulus or modulus of elasticity denoted by (E) and
its value is indicator to the material property. Its unit is the same as
stress N/mm2. Then the relation can be rewrite by equation:
? = QH(5)
In the case of shear stress and strain the constant relate the shear
stress and shear strain is called the modulus of rigidity or shear
modulus and is denoted by (G) and its unit is the same as stress
N/mm2. The relation can be writing as:
A = RS(6)
Young’s modulus and shear modulus for some materials are
tabulated later.
Strength of Material – Dr. Fathelrahman Mohamed Adam
26
Relation between load and deformation:
If a body of length L and x-section A subjected to an axial load P.
if the modulus of elasticity for the material of the body is E, find
an expression for the deformation in the body.
*Note that the deformation is the extension (δ) acted in the body by
the load
From equation (1), the stress can be calculated from:
@
? = (E)
From equation (3), the strain can be calculated from:
N
H = ( )
O
The relation between stress and strain from (5)
? = QH(I)
From (a) and (b), substitute in (c):
@
N
=Q
O
From that and for δ:
N=
O
Q
@( )
Equation (d) used to calculate the deformation by denotation of
load.
Also we can find the load by denotation of deformation as in
equation (e) below.
O
@=
N(K)
Q
Example (2.1):
A steel rod of 1 m length and square x-section 20 mm x 20 mm,
subjected to a tensile force of 40 kN. If modulus of elasticity E is 200
GPa (kN/mm2) , find:
1. The Stress in the bar σ.
2. The extension δ due to applied load.
3. The strain in the bar.
Strength of Material – Dr. Fathelrahman Mohamed Adam
27
Solution:
Given data:
L = 1 m = 1,000 1x103 mm
A = 20 x 20 = 400 mm2
P = 40 kN = 40,000 or 40x103 N
E = 200 kN/mm2 = 200,000 or 200x103 N/mm2
1. The stress σ:
?=
2. The extension δ:
@
=
N=
40,000
= 100T/,,
400
O
Q
@
1 10!
N =#
% 40 10! = 0.5,,
!
200 10 (400)
3. The strain ε:
H=
N
0.5
=
= 0.5 106!
!
O 1 10
Example (2.2)
A load of 5 kN is carried by steel wire, if the stress to be carried by
wire is 100 MPa, Dind the minimum diameter of the steel wire.
Solution:
Given data:
P = 5 kN = 5x103 N
σ = 100 MPa =100x106 N/m2 =100x106/1x106 = 100 N/mm2
If the diameter of the wire is d, then the area is:
V
=
4
W
W
from ? = so = , then:
X
V
5 10!
=
4
100
5 10! 4
=
= 63.66
100 V
= √63.66 = 7.98,,, CE 8,,
Strength of Material – Dr. Fathelrahman Mohamed Adam
28
Example (2.3)
Steel bar of cross section 500 mm2 is acted upon by the forces shown
in Figure. Determine the total elongation of the bar. For steel,
consider E = 200 GPa.
2
1
3
15 kN
50 kN
10 kN
0.6 m
1m
45 kN
1.25 m
Solution:
By superposition (element by element) and balance forces:
1
3
50
50
45
45
2
50
15
10
35
35
45
Given Data:
A = 500 mm2
E = 200 kN/mm2 = 200x103 N/mm2
L1 = 0.6 m = 600 mm, L2 = 1000 mm, L3 = 1250 mm
P1 = 50 kN = 50x103 N, P2 = 35x103 N, P3 = 45x103 N
O
@
Q
E A = 200x103 (500) = 100x106
N=
N* =
N =
600
50 10! = 0.3,,
>
100 10
1000
35 10! = 0.35,,
>
100 10
Strength of Material – Dr. Fathelrahman Mohamed Adam
29
N! =
1250
45 10! = 0.56,,
100 10>
Totalextension(δ)=δ1+δ2+δ3
δ=0.3+0.35+0.56=1.21mm
Tensile Test:
In order to compare the strengths of various materials it is necessary to
carry out some standard form of test to establish their relative properties.
One such test is the standard tensile test in which a circular bar of
uniform cross-section is subjected to a gradually increasing tensile load
until failure occurs by using Universal Testing Machine. Measurements
of the change in length of a selected gauge length of the bar are recorded
throughout the loading operation by means of extensometers and a graph
of load against extension or stress against strain is produced as shown in
Figure below; this shows a typical result for a test on a mild (low carbon)
steel bar; other materials will exhibit different graphs but of a similar
general form.
Stress
D
UltimateStress(σu)
FracturePoint
YieldStress(σy)
LimitofProportionality
E
B
A C
Strain
o
ElasticZone
PlasticZone
For the first part of the test it will be observed that Hooke’s law is
obeyed, i.e. the material behaves elastically and stress is proportional to
strain, giving the straight-line (O-A), where the deformations are
completely recovered when load is removed and point A is termed the
Strength of Material – Dr. Fathelrahman Mohamed Adam
30
limit of proportionality. For a short period beyond this point the
elongation increases more quickly and the diagram becomes curved. At B
a sudden elongation of the bar take place without an appreciable increase
in the tensile force. This phenomenon called yielding is shown in diagram
at points B and C where B, termed the upper yield point, and C, the lower
yield point, and the corresponding stress is called yield stress (σy ). Upon
further stretching, relatively rapid increases in strain occur without
correspondingly high increases in stress up to point D at which the stress
have a maximum value and called ultimate stress (σu ). Beyond point D
the strain increases with decreasing in stress and finally the fracture or
failure in the material occurs at point E.
During tensile testing, the zone between O and A corresponds to elastic
behavior and called elastic zone. Thus if the load were removed
within this zone, the material would return to its original state. But the
zone between A and E corresponds to plastic behavior and called
plastic zone whereby the material undergoes a permanent change
without recovering its original state.
Properties of the material:
There are two main types of material which described below:
Ductile Materials:
• Materials those are capable of undergoing large strains (at normal
temperature) before failure.
• An advantage of ductile materials is that visible distortions may
occur before failure.
• Have a high resistance to tensile and compression stress, but may
buckle under compression.
• Capable of absorbing large amounts of energy prior to failure.
• The failure in ductile materials not happens suddenly but takes
time and generally happens by yielding or buckling.
• As examples of ductile materials: mild steel, aluminum and some of
its alloys, copper, magnesium, nickel, brass, bronze and many
others.
Brittle Material:
• Materials that exhibit very little inelastic deformation.
• As an advantage have large amount of resistance to compression
stress, but have poor resistance to tensile stress.
• The failure in brittle materials occurs suddenly and generally
happens by crushing.
Strength of Material – Dr. Fathelrahman Mohamed Adam
31
• As examples of brittle materials: concrete, stone, cast iron, glass
and plaster etc.
The figure below is a picture for ductile and brittle material after failure.
Ductile material
Brittle material
In the following there are brief definitions to some properties of material:
Elasticity:
Is the ability of the material to return to its original dimensions
when the external applied load is removed.
Plasticity:
Is the property which permits materials to undergo permanent
change in shape without fracture, i.e the material does not return
to its original dimensions.
Ductility:
Is the ability of the material to stand large plastic deformation in
tension, i.e the property of the material which enables it to be
drawn out to a considerable extent before failure.
Strength:
Is the resistance of the material to any applied forces, and is
measured in the known stress units.
Stiffness:
Is the property of the material to resist any sort of deformation.
Strength of Material – Dr. Fathelrahman Mohamed Adam
32
A stiff material is that material which withstands high unit stress
with relatively small unit deformation. In axial tension and
compression tests, the stiffness is quantitatively measured by the
modulus of elasticity within the elastic limit.
Resilience:
It is the capacity of the material to store mechanical energy; given
in energy units, which is the maximum amount of mechanical
energy that may be stored in a unit volume and be completely
recovered upon the removal of load.
Hardness:
Is the ability of the material to resist scratching, abrasion, cutting,
or indentation.
Durability:
Is the ability of the material to resist the internal or external
destructive conditions over long period of time.
Elastic Constants:
The elastic constants of a material describe its response to an applied
stress and are derived from an experimental point of view, according to
the relation between the stress and strain.
Modulus of Elasticity or Young’s modulus (E):
Is the constant used to measure the stiffness of a material below the
proportional limit and mathematically is equal to the ratio between
the stress and strain. Its unit is N/mm2.
?
Q =
H
With the modulus of elasticity, we can describe the material by
isotropic, anisotropic or homogeneous.
Isotropic – Isotropic materials have elastic properties that
are independent of direction of the load applied. Most
common structural materials are isotropic.
Anisotropic – Materials whose properties depend upon
direction. An important class of anisotropic materials is
fiber-reinforced composites.
Homogeneous – A material is homogeneous if it has the
same composition at every point in the body. A homogeneous
material may or may not be isotropic.
Poisson’s ratio (ν):
Strength of Material – Dr. Fathelrahman Mohamed Adam
33
Any direct stress is accompanied by a strain in its own direction
and called longitudinal (linear) strain and an opposite kind strain
in every direction at right angles to it called lateral strain. The
ratio between the lateral strain and longitudinal strain is called the
Poisson’s ratio and is denoted by (ν) this constant is defined by
Simon-Denis Poisson. It has dimensionless unit.
Y = −
LEDK4ELCD4EFG
LMGJFD FGELCD4EFG
The minus sign is due to fact that longitudinal strains always
opposite to lateral strains (because in tensile stress, there exist
increase in length accompanied by decrease in x-section).
Modulus of Rigidity or Shear Modulus (G)
Is the constant obtained by finding the ratio between the shear
stress τ and the shear strain γ within elastic limit.
A
R =
S
Bulk Modulus (K):
When a body is subjected to three mutually perpendicular stresses
of equal intensity in the case of hydrostatic pressure, the ratio of
direct pressure to the corresponding volumetric strain is known as
bulk modulus. It is denoted by K
Z =
Where:
[F4KID\4KCC 4K
]ML ,KD4FI^D4EFG
Volumetric strain = εx + εy + εz
Relation between elastic constants:
1. Relation between Young’s modulus and shear modulus:
The relation gives by the equation:
Strength of Material – Dr. Fathelrahman Mohamed Adam
34
Q = 2R(1 + Y)
2. Relation between Young’s modulus and bulk modulus:
The relation gives by the equation:
Q = 3Z(1 − 2Y)
3. Relation between Young’s modulus, shear modulus and bulk
modulus:
The relation gives by the equation:
Q=
9RZ
(R + 3Z)
*For derivation of these equations refer to the references.
Values of Elastic constants for some materials
Material
Aluminum alloy
Brass
Concrete
Nickel
Steel
Timber
Glass
Young Modulus E
kN/mm2 (GPa)
70 – 79
96 – 110
17 – 31
210
195 – 210
11 – 14
48 – 83
Shear Modulus G
kN/mm2 (GPa)
26 – 30
36 – 41
80
80 – 100
1
19 – 35
Bulk Modulus K
kN/mm2 (GPa)
76
102
14
175
167
–
58
Poisson’s
ratio ν
0.33
0.34
0.1 – 0.2
0.31
0.27 – 0.3
0.5
0.17 – 0.27
Example (2.4)
A rod made of a homogenous and isotropic material, have 500 mm
length and 16 mm diameter. The rod is observed to increase in length
by 300 µm and to decrease in diameter by 2.4 µm when subjected to an
axial load of 12 kN. Determine the modulus of elasticity and Poisson’s
ratio of the material.
Solution:
Given data:
L = 500 mm, P = 12 kN = 12x103 N, d = 16 mm
Longitudinal extension = 300 µm = 300x10-6(1000) = 0.3 mm
Lateral contraction = -2.4 µm = 2.4x10-6(1000) = 2.4x10-3 mm
=
V
V(16 )
=
= 201.1,,
4
4
Strength of Material – Dr. Fathelrahman Mohamed Adam
35
?=
@
12 10!
=
= 59.7T/,,
201.1
H_`ab'cde'af_
H_fcghf_
N
O
0.3
=
= 0.6 106!
500
H=
−2.4 106!
=
= −0.15 106!
16
H_fcghf_
−0.15 106!
Y=
=
= 0.25
H_`ab'cde'af_
0.6 106!
*Note: ignore the minus sign
Q=
?
59.7
=
= 99.5 10! T/,, H 0.6 106!
Q = 99.5R@E
Strength of Material – Dr. Fathelrahman Mohamed Adam
36
Chapter 3
Simple Bending of Beams
When a beam is subjected to transverse loads it causes deflection in the
beam which produces bending. If the bending not accompanied by torsion
then the bending is said to be simple bending. If the shear force in the
beam is zero, then the beam is said to be subjected to pure bending.
The theory of simple bending was developed by Bernoulli. The following
assumptions are made in this theory:
1. The material of the beam is assumed to be homogeneous, perfectly
elastic and isotropic.
2. All transverse sections of the beam, which are planes before
bending, remain plane after bending.
3. The radius of curvature of the beam before bending is very large in
comparison to its transverse dimensions. This implies that the
beam is initially straight.
4. The resultant pull or push across any transverse section of the
beam is zero. This implies that total tensile force is equal to the
compressive force in the beam cross- section.
5. The Young’s modulus of elasticity is same in tension and
compression.
6. The stresses are within the proportional limit.
If we now consider a beam subjected to a constant bending moment (M)
along its length, i.e. pure bending, it will bend to a radius R as shown in
Figure below. As a result of this bending the top fibers of the beam will
be subjected to compression and the bottom to tension. It is reasonable to
suppose, therefore, that somewhere between the two there are plane at
which the stress is zero called the neutral plane and the axis passed
through it called neutral axis (N.A). The radius of curvature R is then
measured to this axis. For symmetrical sections the N.A is the axis of
symmetry, but whatever the section the N.A. will always pass through the
centre of area or centroid.
Strength of Material – Dr. Fathelrahman Mohamed Adam
37
O
θ
R
Y
Compression
M
M
y
N
S
N
S
δA
X
σ
N.A
y
Tension
Consider that the beam is bent into an arc of a circle through angle θ
radians. NN is on the neutral axis and is the same length before and after
bending. The radius of the neutral axis is R.
There is a layer of material at distance y from the neutral axis and this is
stretched because it must become longer (SS).
The original length (L) before and after bending is NN:
L = NN = R θ
The length at distance y is SS:
SS = (R + y) θ = R θ + y θ
The extension (δ):
δ = SS – NN
δ = R θ + y θ –R θ = y θ
The strain (ε) at level SS:
H=
N
O
i
=
ji j
According to the relation between stress and strain:
?
Q =
H
So:
H=
Strength of Material – Dr. Fathelrahman Mohamed Adam
38
H =
?
Q
?
=
Q j
Q
?= j
Therefore, the bending stress at any point in the cross-section is
proportional to its distance from the neutral axis.
? Q
= (1)
j
In other form:
Now consider an elementary area dA at a distance y from the neutral axis,
as shown in Figure. And the stress at this level is σ.
Force acting on dA is:
The total F is:
Substitute for σ, so:
k=&
dF = σ dA
F = ∑ σ dA
Q
j
=
Q
&( j
)
The total moment of all the forces acting on various elements composing
the cross-section forms a couple which is equal to the bending moment M
and is called the moment of resistance (MR).
The moment of resistance can be obtained by taking moment about the
N.A, so:
MR = M = F y
Substitute for F, so:
Q
l = &( ) j
Strength of Material – Dr. Fathelrahman Mohamed Adam
39
Know that:
So:
In other form:
l=
Q
&(
j
&(
l=
Q
&(
j
l=
) = 1
Q1
j
)
)
l Q
= (2)
1
j
Combine equation (1) and (2), we get:
l ? Q
= = (3)
1
j
And this equation called Bending Equation or Engineering Equation
• The engineering equation indicates that the stress in a beam
depends on the bending moment and so the maximum stress will
occur where the bending moment is a maximum along the length
of the beam.
• It also indicates that stress is related to distance y from the
neutral axis so it varies from zero to a maximum at the top or
bottom of the section. One edge of the beam will be in maximum
tension and the other in maximum compression.
• Note that the stress σ is called bending stress, it results from
bending moment and it differ from the direct stress which results
from direct load.
• Note that the neutral axis must always pass through the centroid
The section Modulus:
From equation (3), the bending stress can be calculated by using the
relation:
?=
l
1
Strength of Material – Dr. Fathelrahman Mohamed Adam
40
Or
?=
Where:
l
l
=
1/
m
m=
1
Z is called the section modulus and used to find the maximum bending
stress according to the maximum value of y which it clearly is the far
distance from N.A at outer fibers at top and bottom.
Bending stress for symmetrical sections:
1. Rectangular section:
if a beam have a rectangular section of width b and depth h as
shown in Figure below:
b
h/2
y
h
N.A
y
h/2
ymax = h/2 , top or bottom
mnf =
ℎ!
1=
12
1
nf
?nf = ℎ! 2
ℎ
=
=
12 ℎ
6
l
mnf
=
2. For circular section of diameter d:
6l
ℎ
V <
1=
64
mnf =
1
nf
ymax = d/2
V < 2
V !
=
=
64
32
Strength of Material – Dr. Fathelrahman Mohamed Adam
41
?nf = l
mnf
=
32l
V !
Example (3.1)
Abeamhasarectangularcrosssection80mmwideand120mm
deep.Itissubjectedtoabendingmomentof15kNmatacertain
point along its length. Calculate the maximum stress on the
section.
Solution:
Given data:
b=80mm,h=120mm
M=15kN.m15x106N.mm
ℎ!
1=
12
80(120)!
1=
= 11.52 10> ,,
12
mnf =
=
nf
1
nf
?nf = l
ℎ 120
=
= 60,,
2
2
11.52 10>
=
= 192 10! ,,! 60
mnf
15 10>
=
= 72.1T/,, 192 10!
Example (3.2)
Abeamhasahollowcircularcrosssection40mmouterdiameter
and30mminnerdiameter.Itismadefrommetalwithamodulusof
elasticityof205GPa.Themaximumtensilestressinthebeammust
notexceed350MPa.Calculatethefollowing.
1. Themaximumallowablebendingmoment.
2. Theradiusofcurvature.
Solution:
Given data:
douter=40mm,dinner=30mm
σmax=350MPa=350N/mm2
E=205GPa=205x103 N/mm2
1=
Vo(
`dcgh )
−(
64
<
<
'aagh ) p
Strength of Material – Dr. Fathelrahman Mohamed Adam
Vo(40)< − (30)< p
=
64
42
I=85x103mm4
nf
mnf =
1
=
nf
`dcgh
2
=
40
= 20,,
2
85 10!
=
= 4.25 10! ,,! 20
1. Themaximumallowablebendingmoment.
From that:
?nf = l
mnf
l = ?nf . mnf
l = 350(4.25 10! ) = 1.5 10> T. ,, = 1.5qT. ,
2. Theradiusofcurvature.
From Equation (2)
j=
Q1
l
205 10! (85 10! )
j=
= 11.6 10! ,, = 11.6,
>
1.5 10
Bending stress for unsymmetrical sections:
The only difference between the symmetrical and
unsymmetrical sections in calculating the bending stress is
that:
The bending stress for symmetrical sections is the same at
tension and compression, which means the bending stress at
bottom fiber, is equal to the bending stress at top fiber, i.e y
measured upward and downward is equals. This is result from
thecentriodliesatequidistancebetweentopandbottom.
But for unsymmetrical sections and because of the centriod not
liesatequidistancetopandbottom,sothevaluesofymeasured
fromneutralaxistotopandbottomfibernotequals.
Strength of Material – Dr. Fathelrahman Mohamed Adam
43
b
σt
yt
t2
N.A
d
yb
t1
σb
mc =
?c = 1
c
, mr =
1
r
l
l
, ?r = mc
mr
Example (3.3)
Two wooden planks 150 mm x 50 mm each, connected to form a
T-section of a beam. If a moment of 6.4 kN.m is applied around
the horizontal neutral which inducing tension below the neutral
axis, find the bending stresses at both the extreme fibers of the
cross-section.
150
50
50
150
Solution:
Firstly determine the location of centriod by divide the section
to two areas as shown:
Strength of Material – Dr. Fathelrahman Mohamed Adam
44
150
1
50
175
2
= 125
75
Element
b
1
150
2
50
Summation
=
c=
h
50
150
A (bxh)
7,500
7,500
y
175
75
15,000
∑
∑
=
Ay
1,312,500
262,500
1,875,000
1,875,000
= 125,,
15,000
1
75
50
50
r=
125
2
Secondly calculate the moment of inertia:
Element
b
h
A
d
Ad2
Ix'
6
1
150 50 7,500 50 18.75x10 1.56x106
2
50 150 7,500 50 18.75x106 14.1x106
Summation
Ad2+ Ix'
20.3x106
32.8x106
53.1x106
I = 53.1x106 mm4
Strength of Material – Dr. Fathelrahman Mohamed Adam
45
1
53.1 10> mc = = = 708.2 10! 75
c
mr =
1
r
=
53.1 10 = 424.8 10!
125
>
M=6.4kN.m=6.4x106N.mm
l
6.4 10>
?c = =
= 9T/,, mc 708.2 10!
l
6.4 10>
?r = =
= 15.1T/,,
mr 424.8 10!
Strength of Material – Dr. Fathelrahman Mohamed Adam
46
Chapter 4
Shear stress in beams
When a beam is subjected to non-uniform bending, both bending
moments (M) and shear forces (V), act on the cross section. We will now
consider the distribution of shear stresses (τ), associated with the shear
force (V).
The shear stress in a beam at any transverse cross-section in its length,
and at a point a vertical distance y from the neutral axis, resulting from
bending is given by:
A=
]
1C
Where:
V is the applied vertical shear force at that section.
A is the area of cross-section “above” y, i.e. the area between y
and the outside of the section, which may be above or below the
neutral axis (N.A.).
is the distance of the centroid of area A from the N.A.
I is the second moment of area of the complete cross-section.
y is the position where shear stress τ is being calculated
s is the actual width of the section at position y.
Rectangular section:
b
A
y
h
=
=
ℎ
−
2
ℎ
ℎ
s2 − t 2 + −
2
+
=
2
2
Strength of Material – Dr. Fathelrahman Mohamed Adam
h/2 – y
τ
N.A
=
2
+
ℎ
4
47
1=
s=b
ℎ!
12
substitute in Equation of shear above:
ℎ
ℎ
] s − ts + t
2
2 4
A=
!
ℎ
12
A=
6] ℎ
# −
ℎ! 4
%
From the above Equation τ denotes the shearing stress on a fiber at a
distance y from the neutral axis. The distribution of vertical shearing
stress over the rectangular cross section is thus parabolic, varying
from zero at the outer fibers where y = h/2 to a maximum at the
neutral axis where y = 0.
Anf =
6] ℎ
]
=
1.5
#
%
ℎ! 4
ℎ
Example (4.1)
A timber beam 10 cm wide by 15 cm deep, carries a uniformly
distributed load over a span of 2 m. if the permissible bending
stress is 28 N/mm2 and shear stress is 2 N/mm2, calculate the
maximum load which can be carried by beam.
Solution:
Given Data:
b = 10 cm = 100 mm, d = 15 cm = 150 mm
L = 2 m = 2000 mm
σ = 28 N/mm2, τ = 2 N/mm2
Assume the uniformly distributed load = w N
The maximum bending stress can be calculated from:
l=
?nf =
l
m
uL
u(2000)
=
= 500 10! u
8
8
Strength of Material – Dr. Fathelrahman Mohamed Adam
48
1=
!
12
m=
1
=
100(150)!
= 28.125 10> ,,<
12
=
2
=
150
= 75,,
2
28.125 10>
=
= 375 10! ,,! 75
l = ?nf m
500 10! u = 28(375 10! )
28(375 10! )
u =
= 21T/,,
500 10!
The shear stress can be calculated from:
Anf = 1.5
]=
u=
]=
]
Anf
1.5
uL u(2000)
=
= 1000u
2
2
100(150)(2)
= 20T/,,
1.5(1000)
For maximum load select w=20N/mm
I-Section:
B
v−ℎ
2
2
H
ℎ
2
1
N.A
y
*
τ
h
b
Strength of Material – Dr. Fathelrahman Mohamed Adam
49
*
=
=w
ℎ
−
2
,
*
v−ℎ
,
2
=
ℎ
s2 − t
=
2
+
=
ℎ
+
4 2
ℎ
v−ℎ
v ℎ
v+ℎ
+
= + =
2
4
4 4
4
The shear stress calculated from the same Equation:
=
s=b
Where:
* *
* *
=
=w
So:
A=
+
ℎ
−
2
v−ℎ
2
ℎ
= # −
2 4
A=
]
1C
ℎ
ℎ
+
= # −
4 2
2 4
%
v+ℎ
w
=
v −ℎ
4
8
%+
w
v −ℎ
8
]
ℎ
# −
1C 2 4
%+
w
v −ℎ "
8
Here again the maximum shear stress be at N.A, i.e at y = 0
Note:
Anf =
]
o ℎ +w v −ℎ p
81C
More than 95% of the shear stress was found to be carried by the
web, so the flange carried the rest of shear stress and almost be
neglected by compare.
Example (4.2)
A 12 cm by 5 cm I-beam is subjected to a shearing force of 10 kN.
Calculate the value of the transverse shear stress at the neutral
axis and at the top of the web. Take I = 220 cm4, Area = 9.4 cm2,
web thickness = 0.35 cm, Dlange thickness = 0.55 cm.
Strength of Material – Dr. Fathelrahman Mohamed Adam
50
Given data:
b = 0.35 cm, B = 5 cm
h = 12 – 0.55(2) = 10.9 cm, H = 12 cm
I = 220 cm4, V = 10 kN = 10,000 N
A=
]
ℎ
# −
1C 2 4
%+
w
(v − ℎ )"
8
s = 0.35 cm
A=
= 0.175(29.7 −
10,000 0.35 10.9
−
#
220(0.35) 2
4
A = 129.87o0.175(29.7 −
A = 129.87(20.94 − 0.175
) + 15.74
5
% + (12 − 10.9 )"
8
) + 15.74p
)T/I,
At neutral axis y = 0, gives max shear stress:
A = 129.87(20.94) = 2719.15T/I,
A = 27.2T/,,
At the top of the web y = (12/2-0.55) = 5.45 cm:
A = 129.87(20.94 − 0.175(5.45) ) = 2,044T/I, A = 20.44T/,,
Strength of Material – Dr. Fathelrahman Mohamed Adam
51
Chapter 5
Torsion
When a uniform circular shaft of radius (r) and length (L) is subjected to
a torque (T) it can be shown that every section of the shaft is subjected to
a state of pure shear (τ) as shown in Figure below, the moment of
resistance developed by the shear stresses being everywhere equal to the
magnitude, and opposite in sense, to the applied torque. For the purposes
of deriving a simple theory to describe the behavior of shafts subjected to
torque it is necessary to make the following basic assumptions:
1. The material is homogeneous, i.e. of uniform elastic properties
throughout.
2. The material is elastic, following Hooke's law with shear stress
proportional to shear strain.
3. The stress does not exceed the elastic limit or limit of
proportionality.
4. Circular Sections remain circular.
5. Cross-sections remain plane.
6. Cross-sections rotate as if rigid, i.e. every diameter rotates
through the same angle (θ) through the length L.
T
L
δr
τ
A τ
θ
r
A
B
θ
τ
τ
B
γ
T
d
From the Figure:
θ
is the angle of twist suspended by an arc AB and it
constant through the length (L).
γ
is the shear strain and it constant with constant torque (T)
in other way it is the angle suspended by an arc AB.
The length AB = rθ = Lγ
Previously you know that γ can be given by:
γ =
A
R
Strength of Material – Dr. Fathelrahman Mohamed Adam
52
So:
4i = O
A
R
A Ri
=
(1)
4
O
The torque can be equated to the sum of the moments of the tangential
stresses on the elements:
T = Force x r
kM4IK =
A=
kM4IK =
x=
Ri
O
x=
Ri
4
O
Ri
O
x=
Aδ
Ri
4δ O
(4δ )4
42 δ
Ri
2
O
x Ri
=
(2)
2
O
Or
Where J is polar modulus, computed from the following formula:
V <
2=
, yM4CMLF CKIDFMG
32
V([< −
2=
32
<)
, yM4ℎMLLMuCKIDFMG
Combining (1) and (2):
x A Ri
= =
(3)
2 4
O
Equation (3) called the Equation of simple theory of torsion.
Strength of Material – Dr. Fathelrahman Mohamed Adam
53
Torsion stiffness (k) is defined as torque per radian twist
q=
x
i
=
R2
O
Example (5.1)
If a twisting moment of 1100 N m is impressed upon a 4.4 cm
diameter shaft, what is the maximum shearing stress developed?
Also, what is the angle of twist in a 150 cm length of the shaft?
The material is steel for which G = 85 GPa.
Solution:
Given data:
T = 1100 N m = 1.1x106 N mm
D = 4.4 cm = 44 mm
L = 150 cm = 1500 mm
G = 85 GPa = 85x103 N/mm2
For the shear stress and from the general Equation of torsion:
x A
=
2 4
A=
x
4
2
V[4 V(44)4
3
2=
=
= 368 10 ,,<
32
32
A=
4=
[ 44
=
= 22,,
2
2
1.1 10> 368 10
3
(22) = 65.8T/,,
A = 65.8l@E
For angle of twist and from the general Equation of torsion:
x Ri
=
2
O
i=
xO
2R
Strength of Material – Dr. Fathelrahman Mohamed Adam
54
i=
1.1 106 1,500
368 10! 85 103 )
i = 0.0527
180
= 0.05274E
= 3°
V
Example (5.2)
Ahollow3mlongsteelshaftmusttransmitatorqueof25kNm.
Thetotalangleoftwistinthislengthisnottoexceed2.5°andthe
allowable shearing stress is 90 MPa. Determine the inside and
outsidediametersoftheshaftifG=85GPa.
Solution:
Givendata:
T=25kNm=25x106Nmm
L=3m=3000mm
2.5(V)
θ = 2.5° = = 0.04364E 180
τ=90MPa=90N/mm2
G=85GPa=85x103N/mm2
From:
x Ri
=
2
O
2 =
xO
Ri
25 10> (3000)
2=
= 20.24 10> ,,<
!
85 10 (0.0436)
Assume the outer diameter is D and inner diameter is d, then:
V s [4 −
2=
32
V ([ < −
32
From:
[< −
<
<)
4
t
= 20.24 106
= 206.2 106 (1)
x A
=
2 4
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55
x4
A
[
4=
2
2=
V([ < −
32
[< −
(1)-(2),gives:
Fromthis:
2(90)
= 1.415 106 [(2)
206.2 10>
[=
= 145.7,,
1.415 10>
From(1):
<
=
25 106 [
0 = 206.2 10> − 1.415 10> [
<
<)
<
= [ < − 206.2 106 = (145.7)< − 206.2 106 = 244.4 106 = {244.4 10> = 125,,
|
Example (5.3)
Theworkingconditionstobesatisfiedbyashaftofdiameter165
mmtransmittingpowerare:
1.
2.
Thattheshaftmustnottwistmorethan1degreeona
lengthof2.5meters.
Theshearstressmustnotexceed55N/mm2
If G = 80,000 N/mm2, calculate the torque which can be
transmitted for a given diameter according to the two
conditions.
Solution:
Givendata:
θ = 1° = 1(V)
4E = 0.01744E 180
L=2.5m=2500mm,D=165mm
τ=55N/mm2
G=80,000N/mm2
Condition1:
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56
Byusingequation2:
x=
x=
80,000(0.0174)
2,500
Ri
2
O
V[4 V(165)4
6
2=
=
= 72.8 10 ,,< 32
32
Condition2:
FromEquation3:
x Ri
=
2
O
x=
4=
s72.8 10 t = 40.5 10 T. ,,
6
6
x = 40.5qT. ,
x A
= 2 4
A
x = 2
4
[ 165
=
= 82.5,,
2
2
55
6
6
s72.8 10 t = 48.5 10 T. ,,
82.5
x = 48.5qT. ,
Tomeetthetwoconditionsselect:
T=40.5kN.m
Strength of Material – Dr. Fathelrahman Mohamed Adam
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Chapter 6
Complex (Compound) Stress
Complex Stress
Materials in a stressed component often have direct and shear stresses
acting in two or more directions at the same time. This is a complex stress
situation. The engineer must then find the maximum stress in the
material. We will only consider stresses in two dimensions, x and y.
Derivation of equations:
Consider a rectangular part of the component material. The stresses σx
and τxy are act on the AB plane and the stresses σy and τyx are act on the
BC plane. (note that τxy = τyx = τ).
Frequently it is desirable to investigate the state of stress on a plane
inclined at an angle θ, measured counterclockwise to the vertical, as
shown in Figure above.
σy
τ
C
B
τ
τθ
σx
σθ
θ
A
Find the forces in all planes by multiplying the stress by the area
of the plane exerted on it.
Let t denote the thickness of the element.
Area of Planes:
plane AB: AAB = AB(t)
plane BC: ABC = BC(t)
plane AC: AAC = AC(t)
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58
Then the forces are explained at Figure below:
σy (ABC)
τ (ABC)
C
B
τ (AAB)
τθ (AAC)
σx (AAB)
θ
σθ (AAC)
A
From the Figure we can write the lengths AB and BC by using the
length AC:
AB = AC cos(θ)
BC = AC sin(θ)
Then the area can be rewrite by:
AAB = AC cos(θ) (t)
ABC = AC sin(θ) (t)
Resolve in the direction of σθ and τθ
C
[σy (ABC) +τ (AAB)] cos θ
[σy (ABC) +τ (AAB)]
θ
[σy (ABC) +τ (AAB)] sin θ
τθ (AAC)
[σx (AAB) +τ (ABC)] cos θ
θ
σθ (AAC)
θ
[σx (AAB) +τ (ABC)]
[σx (AAB) +τ (ABC)] sin θ
A
Strength of Material – Dr. Fathelrahman Mohamed Adam
59
Find the summation of forces at direction σθ:
σθ (AAC) = σy (ABC) sin θ + σx (AAB) cos θ + τ (AAB) sin θ + τ (ABC) cos θ
Substitute for AAB , AAC and ABC:
σθ AC(t) = σy AC sin θ (t) sin θ + σx AC cos θ (t)cos θ +
τ AC cos θ (t) sin θ + τ AC sin θ (t) cos θ
Reduce for AC and t and rearrange:
σθ = σy sin2 θ + σx cos2 θ + τ (2 sin θ cos θ)
Note:
From the trigonometry:
sin 2i = 2 sin i cos i
1 − cos 2i
2
1 + cos 2i
IMC i =
2
CFG i =
so:
?‚ = ?‚ = ? (1 − cos 2i) ? (1 + cos 2i)
+
+ τ^FG2i
2
2
ƒ? + ? „ ƒ? − ? „
+
cos 2i + τCFG2i(1)
2
2
Find the summation of forces at direction τθ:
τθ (AAC) = σx (AAB) sin θ – σy (ABC) cos θ + τ (ABC) sin θ - τ (AAB) cos θ
Substitute for AAB , AAC and ABC:
τθ AC(t) = σx AC cos θ (t) sin θ – σy AC sin θ (t) cos θ
+τ AC sin θ (t) sin θ– τ AC cos θ (t) cos θ
Reduce for AC and t and rearrange:
τθ = (σx – σy ) sin θ cos θ – τ (cos2 θ – sin2 θ)
A‚ = ƒ? − ? „
sin 2i − τIMC2i(2)
2
Example (6.1):
If the stresses on two perpendicular planes through a point are
60 N/mm2 tension applied horizontally, 40 N/mm2 compression
applied vertically and 30 N/mm2 shear, find the stress
components on a plane inclined by 60° to the vertical.
Strength of Material – Dr. Fathelrahman Mohamed Adam
60
Solution:
The Figure below explain the stress as described
40
30
τθ
θ = 60°
σθ
30
60
All data in N/mm2
Given Data:
σx = 60 N/mm2, σy = -40 N/mm2, τ = 30 N/mm2,
Calculate the stresses from the following equations:
?‚ = ?‚ = ƒ? + ? „ ƒ? − ? „
+
cos 2i + τCFG2i
2
2
A‚ = ƒ? − ? „
sin 2i − τIMC2i
2
(60 + (−40)) (60 − (−40))
+
cos(120) + 30CFG(120)
2
2
?‚ = 10 + 50(−0.5) + 30(0.866) = 11T/,,
A‚ = (60 − (−40))
sin 2(60) − 30IMC2(60)
2
A‚ = 43.3 + 15 = 58.3T/,,
The maximum and minimum stresses:
The place of maximum and minimum stresses can be located by
angle θ by finding the first derivative of stress with relative to θ.
Strength of Material – Dr. Fathelrahman Mohamed Adam
61
put
ƒ? − ? „
…?‚
= −2
sin 2i + 2τIMC2i
…i
2
†X‡
†‚
= 0 to find θ
−2
ƒ? − ? „
sin 2i + 2τIMC2i = 0
2
sin 2i
2τ
=
IMC2i
ƒ? − ? „
tan 2i =
By using the right angle theorem:
2τ
2τ
ƒ? − ? „
sin 2i = ±
cos 2i = ±
2θ
‹4τ + ƒ? − ? „
ƒ? − ? „
From triangle:
2τ
‹4τ + ƒ? − ? „
ƒ? − ? „
‹4τ + ƒ? − ? „
The above equations gives two values for angle θ, which are
defined two planes called the principal planes. These two planes
differ by angle 90°. The normal stresses that exist on these planes
are designated as principal stresses. They are the maximum
and minimum values of the normal stress in the element under
consideration.
Substitute for sin 2θ and cos 2θ, to Dind the values of stresses:
Strength of Material – Dr. Fathelrahman Mohamed Adam
62
?‚nf
Ž
‘
Ž
‘
ƒ? + ? „ ƒ? − ? „
ƒ? − ? „
2τ
•
•
•
•
=
±
± τ
2
2
•
•
•
•
‹
‹
Œ 4τ + ƒ? − ? „ •
Œ 4τ + ƒ? − ? „ •
?‚nf
Ž
ƒ? + ? „ 1 4τ + ƒ? − ? „
=
± •
2
2•
‹
Œ 4τ + ƒ? − ? „
?‚nf = ƒ? + ? „ 1
± ‹4τ + ƒ? − ? „
2
2
‘
•
•
•
This gives the maximum value says (σ1) and minimum value says
(σ2), so:
ƒ? + ? „ 1
?* = + ‹4τ + ƒ? − ? „ (3)
2
2
? =
ƒ? + ? „ 1
− ‹4τ + ƒ? − ? „ (4)
2
2
By repeating the above process for equation of shear stress, by
†’
using ‡ = 0 we get:
†‚
In other form:
1
Anf = ‹4τ + ƒ? − ? „
2
1
Anf = (?* − ? )(5)
2
Example (6.2):
For the stresses shown in Figure, find the following:
1. The principal stresses.
2. The position of the principal planes.
3. The maximum shear stresses.
200
All data in MPa
150
150
100
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63
Solution:
Given data:
σx = 100 N/mm2, σy = 200 N/mm2, τ = 150 N/mm2
(Note that Pa = 1 N/m2, MPa = 1 N/mm2)
1. The principal stresses:
?* = ? =
?* = ƒ? + ? „ 1
+ ‹4τ + ƒ? − ? „
2
2
ƒ? + ? „ 1
− ‹4τ + ƒ? − ? „
2
2
(100 + 200) 1
+ {4(150) + (100 − 200)
2
2
?* = 150 + 158.1 = 308.1T/,,
? = 150 − 158.1 = −8.1T/,,
2. The position of the principal planes:
tan 2i =
tan 2i =
2τ
ƒ? − ? „
2(150)
= −3
(100 − 200)
2i = tan6* (−3)
2i = −71.6°
or:
i = −71.6/2 = −35.8°
i = −35.8 + 90 = 54.2°
3. The maximum shear stress:
1
Anf = (?* − ? )
2
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64
1
Anf = (308.1 − (−8.1)) = 158.1T/,,
2
Example (6.3):
For the state of plane stress shown in Figure below, determine:
(a) The principal planes.
(b) The principal stresses
(c) The maximum shearing stress.
(a)
(b)
(c)
Ans.
26.6°, 116.6°
σ1 = 70 MPa, σ2 = -30 MPa
τmax = 20 MPa
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65
Mohr’s Stress Circle:
Known also as Mohr’s circle, refers to German Civil Engineer Otto
Mohr. Mohr’s circle for stress it is the graphical representation for
stress used to locate the principal planes and to calculate the
principal stresses and maximum shear stresses instead of using the
equations. In this representation normal stresses are plotted along
the horizontal axis and shearing stresses along the vertical axis.
The stresses σx, σy, and τxy or (τ) are plotted and a circle is drawn
through these points having its center on the horizontal axis.
General steps to draw the Mohr’s circle:
1. Draw point “S” at a suitable position.
2. By using suitable scale, measure σx to right if positive, start
from S and end at x
S
x
σ
σx
3. From S and to right if positive measure σy to y.
S
x
y
σ
σy
σx
4. Measure τ vertically from y upward to A and from x
downward to B if positive.
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66
τ
A
τ
σ
x
y
S
τ
σy
B
σx
5. Draw the diagonal AB.
τ
A
τ
x
O
σ
y
S
τ
½(σx – σy)
B
σx – σy
σy
σx
Note:
• The line AB divide the distance xy (xy = σx – σy) at O to
two distance ox and oy, where ox = oy = ½(σx – σy).
• The point O is the centre of the circle of radius OA or
OB, where:
“ = “w = 5A + s
? −?
1
t = ‹4τ + ƒ? − ? „
2
2
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67
6. Draw the circle of radius OA and centre O, along the line SO
mark point 1 at intersection between circle and line and mark
2 at intersection between circle and line.
σ1
A
σ2
τmax
τxy
x
1
y
S
O
2
2θ1
τxy
½(σx – σy)
B
σx – σy
σy
σx
7. From the drawing you can measure S1 and S2
Note:
Proof:
• The length of S1 equal to σ2
• The length of S2 equal to σ1.
^“ = ? +
? −?
? +?
=
2
2
1
“1 = “ = ‹4τ + ƒ? − ? „
2
S1 = SO – O1
^1 =
? +?
1
− ‹4τ + ƒ? − ? „
2
2
Which is equivalent to equation (4), so S1 = σ2
S2 = S1 + 2(OA)
^2 =
? +?
1
− ‹4τ + ƒ? − ? „ + ‹4τ + ƒ? − ? „
2
2
Strength of Material – Dr. Fathelrahman Mohamed Adam
68
^2 =
? +?
1
+ ‹4τ + ƒ? − ? „
2
2
Which is equivalent to equation (3), so S2 = σ1
8. forτmax:
τmax = radius (OA) or O1or O2
from diagram:
?* − ?
“1 = s
t
2
This mean:
?* − ?
Anf = s
t
2
which is equivalent to equation (4)
9. for the position of principal planes measure the angle
between line S2 and line AB to give 2θ1
10.for the stresses at any angle θ, draw angle 2θ anticlockwise
start from line AB:
τ
C
A
τθ
S
2 σ
2θ
1
O
D
σθ
B
By measuring the length CD vertically, this gives τθ
• By measuring the length SD horizontally, this gives σθ
•
Example (3):
With referring to example 2, recalculate by using the Mohr's
circle.
Solution:
Given data:
σx = 100 N/mm2, σy = 200 N/mm2, τ = 150 N/mm2
Strength of Material – Dr. Fathelrahman Mohamed Adam
69
Required:
σ1, σ2, τmax, θ1
Follow the steps of drawing circle:
take the scale:
1 cm ≡ 25 N/mm2
For σx = 100, draw Sx = 4 cm horizontally
For σy = 200, draw Sy = 8 cm horizontally
For τ = 150, draw yA = 6 cm vertically upward, and xB= 6
cm downward
τ
A
τmax
O
S
2
2θ1
x
1
y
σ
B
Connect AB and determine O (intersection between AB and Sy.
Measure the length OA, which is the radius of circle, you find
OA = 6.325 cm, use it to draw the circle.
For σ1, measure S2 = 12.3 cm, so σ1 = 12.3x25 = 307.5 N/mm2
σ1 = 307.5 N/mm2
Strength of Material – Dr. Fathelrahman Mohamed Adam
70
For σ2, measure S1 = -0.325 cm, so σ2 = -0.325x25 = -8.125
N/mm2
σ2 = -8.1 N/mm2
For τmax, OA = 6.325 cm, so τmax = 6.325x25 = 158.1 N/mm2
τmax = 158.1 N/mm2
For θ1, you can measure it directly or measured Oy and calculate
tan-1 (Ay/Oy). Oy = 2.05 and Ay = 6, so
2θ1 = tan-1 (6/2.05) = 71.1° , so
θ1 = 71.1/2 = 35.6° anticlockwise
Example (4):
If the stresses on two perpendicular planes through a point are
60 N/mm2 tension applied horizontally, 40 N/mm2 compression
applied vertically and 30 N/mm2 shear, find the following:
1. The principal stresses.
2. The location of principal stresses.
3. The maximum shear stress.
4. The stress components on a plane inclined by 60° to the
vertical.
Solution:
Given data:
σx = 60 N/mm2, σy = -40 N/mm2, τ = 30 N/mm2
Required:
σ1, σ2,θ, τmax, θ1, σθ , τθ
Follow the steps of drawing circle:
take the scale:
1 cm ≡ 10 N/mm2
For σx = 60, draw Sx = 6 cm horizontally
For σy = -40, draw Sy = 4 cm horizontally to left S
For τ = 30, draw yA = 3 cm vertically upward, and xB= 3
cm downward
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71
τ
C
A
τθ
120°
1
y
S
2
x
O D 2θ1
σ
σθ
B
Connect AB and determine O (intersection between AB and
Sy.
Measure the length OA, which is the radius of circle, you
find OA = 3.16 cm, use it to draw the circle.
For σ1, measure S2 = 6.8 cm, so σ1 = 6.8x10 = 68 N/mm2
σ1 = 68 N/mm2
For σ2, measure S1 = -4.84 cm, so σ2 = -4.84x10 = -48.4 N/mm2
σ2 = -48.4 N/mm2
For τmax, OA = 5.83 cm, so τmax = 5.83x10 = 58.3 N/mm2
τmax = 58.3 N/mm2
For θ1, you can measure it directly or measured Oy and calculate
tan-1 (Ay/Oy). Oy = 5.03 cm and Ay = 3 cm, so
Strength of Material – Dr. Fathelrahman Mohamed Adam
72
2θ1 = tan-1 (3/5.03) = 30.8° , so
θ1 = 30.8/2 = 15.4° clockwise
For θ = 60° , draw line OC by rotating anticlockwise 2θ = 120° from AB
then measure CD to find τθ and measure SD to find σθ
SD = 1.1 cm, so σθ = 1.1 x 10 = 11 N/mm2
CD = 5.8 cm, so τθ = 58 N/mm2
Strength of Material – Dr. Fathelrahman Mohamed Adam
73
Chapter 7
SHEAR FORCE AND BENDING MOMENT
Introduction:
The internal forces in any point or section can be determined by making
cut at specified point or section and there are three forces appear which
are:
1. Axial forces (acting along the axial axis x).
2. Shear force (acting at transverse axes y and/or z).
3. Bending moment (acting about transverse axes y and/or z).
A shear force is the force tending to produce a shear failure at a given
point in a beam or frame, the value of it at any point in a beam or
frame = the algebraic sum of all upward and downward forces to the
left of the point. (The term "algebraic sum" means that upward forces
are regarded as being positive and downward forces are considered to
be negative). The unit of shear force is kN or N.
The bending moment is the magnitude of the bending effect at any
point in a beam or frame which is a force multiplied by a
perpendicular distance, it's either clockwise or anticlockwise and is
measured in kN.m or N.mm. The value of bending moment at any
point on a beam or frame = the sum of all bending moments to the left
of the point.
The sign convention:
The sign convention is the determination of the signs of forces (positive
or negative) according to the rule set from starting
Take the following rule:
• For the axial force N, if the force tends to elongate the segment
consider this force is positive.
Strength of Material – Dr. Fathelrahman Mohamed Adam
74
• For the shear force V, if the force tends to rotate the segment
clockwise, consider this force is positive.
• For the bending moment if the moment tends to make a sag
(concave or downwards curvature) in the segment, consider this
moment is positive.
Sag
Note that:
the opposite to sag is hog which is the curvature upwards
hog
the moment create sag called sagging moment (positive) and the
moment create hog called hogging moment (negative).
in the sag the tension at bottom and in the hog the tension at top.
Example (7.1):
Determine whether is the applied load(s) create sag or hog in the
following beams by drawing the deflected shape:
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75
Solution:
Figures explains the deflected shape under loads
Strength of Material – Dr. Fathelrahman Mohamed Adam
76
Distributed load:
• Uniformly distributed full span:
x
w = 10 kN/m
A
5m
VA
B
x
VB
x
For the beam shown above, find the shear force and bending
moment at section x-x when x = 0, x = 1 m, x = 2 m, x = 3 m, x = 4 m
and x = 5 m.
Solution:
For symmetry the vertical reactions at A and B are equal
VA = VB = total load/2
Total load = 10x5 = 50 kN
VA = VB = 50/2 = 25 kN
Sx = VA – w(x) = 25 – 10(x)
Mx = VA (x) – w (x)(x/2) = 25 (x) – 10(x2/2)
x (m)
Sx (kN)
Mx (kN.m)
0
1
2
3
4
5
25 – 10(0) = 25
25 – 10(1) = 15
25 – 10(2) = 5
25 – 10(3) = -5
25 – 10(4) = -15
25 – 10(5) = -25
25 (0) – 10(02/2) = 0
25 (1) – 10(12/2) = 20
25 (2) – 10(22/2) = 30
25 (3) – 10(32/2) = 30
25 (4) – 10(42/2) = 20
25 (5) – 10(52/2) = 0
Strength of Material – Dr. Fathelrahman Mohamed Adam
77
• Uniformly distributed short than span:
x
A
VA
w = 10 kN/m
B
3m
x
x
VB
5m
For the beam shown above, find the shear force and bending
moment at section x-x when x = 0, x = 1 m, x = 2 m, x = 3 m, x = 4 m
and x = 5 m.
Find reactions:
By taking moment about B:
∑MB = 0; VA (5) – 10(3)(3/2+2) = 0
VA = 105/5 = 21 kN
from ∑FY = 0; VA + VB = 10(3)
VB = 30 – 21 = 9 kN
for0≤x≤3
Sx = VA – 10(x)
Mx = VA (x) – 10(x2/2)
for3≤x≤5
Sx = VA – 10(3)
Mx = VA (x) – 10(3)(x-1.5)
x(m)
Sx(kN)
Mx(kN.m)
0
1
2
3
4
5
21 – 10(0) = 21
21 – 10(1) = 11
21 – 10(2) = 1
21 – 10(3) = -9
21 – 10(3) = -9
21 – 10(3) = -9
21 (0) – 10(02/2) = 0
21 (1) – 10(12/2) = 16
21 (2) – 10(22/2) = 22
21 (3) – 10(32/2) = 18
21 (4) – 30(4 – 1.5) = 9
21 (5) – 30(5 – 1.5) = 0
Strength of Material – Dr. Fathelrahman Mohamed Adam
78
• distributed load varying linearly
x
y
A
B
x/3
VA
x
x
F
w=10kN/m
VB
5m
y/x=10/5;y=2x
F=½(0+y)(x)
Sx=VA–F
Mx=VA(x)–F(x/3)
for reactions:
taking moment about B:
x=5m
y=2(5)=10kN/m
F=½(0+10)(5)=25kN
∑MB=0;VA(5)–F(5/3)=0
5VA = 25(5/3); VA = 8.33 kN
∑FY=0;VA+VB = F; VB=F–VA
VB=25–8.33=16.67kN
x (m)
Y (kN/m)
0
1
2
3
4
2(0)=0
2(1)=2
2(2)=4
2(3)=6
2(4)=8
F (kN)
½(0+0)(0)=0
½(0+2)(1)=1
½(0+4)(2)=4
½(0+6)(3)=9
½(0+8)(4)=16
Sx (kN)
8.33–0=8.33
8.33–1=7.33
8.33–4=4.33
8.33–9=-0.67
8.33–16=-7.67
Strength of Material – Dr. Fathelrahman Mohamed Adam
Mx (kN.m)
8.33(0)–0(0/3)=0
8.33(1)–1(1/3)=8
8.33(2)–4(2/3)=14
8.33(3)–9(3/3)=16
8.33(4)–16(4/3)=12
79
5
2(5) = 10
½(0 + 10)(5) = 25
8.33 – 25 = -16.67
8.33 (5) – 25(5/3) = 0
Notes about shear force:
You can observe that shear force is constant between concentrated
loads, but it varies linearly along uniformly distributed load.
The maximum values of shear force are at ends and equal in values
to the reactions.
Notes about bending moment:
The bending moment is varies linearly between concentrated loads,
but it draw curve throw the distributed loads.
Maximum bending moments obtained at points of concentrated
loads and points of reactions.
The points of zero moment fall in-between ends called the points of
contraflexure.
Relation between Shear Force and Bending Moment
If the bending moment at any section x-x (Mx), is the function of x, where
x is the distance measured from left side to the section considered, then
the slope at any point is equal to the shear force.
^ =
l
When the applied load is distributed load, notices that:
• the maximum bending moment corresponds zero shear force.
• The first derivative of shear force gives the distributed load
u =
^
Shear force and bending moment diagram:
The shear force and bending moment diagrams is the drawing represents
the variation of shear force and bending moment along the beam or
frames by draw the values using graphical representations.
To draw the shear force diagram follow the following steps:
1. Draw the datum (Zero Shear Force Line).
2. Marks any points of:
a. Concentrated loads.
b. Start and end of distributed load.
c. Reactions.
Strength of Material – Dr. Fathelrahman Mohamed Adam
80
3. Calculate the values of shear force at the marking points, and
notice that at any point of concentrated load the shear force is
calculated twice one exclude the load and the other include the
load.
4. Select suitable scale and put a dot at any marked point. The dot
must be upward or downward the datum according to the sign of
shear force value if positive put the dot upward if negative put the
dot downward the datum.
5. Connect between the dots start from the zero point at the left and
continue in series until reach the zero point at right. The final shape
is shear force diagram.
To draw the bending moment diagram follow the following steps:
1. Draw the datum (Zero Bending Moment Line).
2. Marks any points of:
a. Concentrated loads.
b. Start and end of distributed load.
c. Reactions.
d. Zero shear force.
3. Calculate the values of bending moment at the marking points.
4. Calculate the values of maximum moment at the points that are
coincide with the points of zero shear force.
5. Select suitable scale and put a dot at any marked point. The dot
must be upward or downward the datum according to the sign of
shear force value if positive put the dot upward if negative put the
dot downward the datum.
6. Connect between the dots start from the zero point at the left and
continue in series until reach the zero point at right with notice that
the line drawn under distributed load is not drawn as straight line
but it drawn as curve. The final shape is shear force diagram.
Examples (7.2):
Draw the shear force and bending moment diagram for the beams
shown in Figures below and find the points of contraflexure if exist.
Strength of Material – Dr. Fathelrahman Mohamed Adam
81
50 kN
A
B
C
3m
7m
VA
VB
Fig. (1)
50 kN
30 kN
A
B
C
D
3m
3m
4m
VA
A
Fig. (2)
VB
10 kN/m
B
10 m
VA
25 kN
A
VB
Fig. (3)
50 kN
10 kN/m
B
D
C
3m
5m
10 m
VA
Fig. (4)
Strength of Material – Dr. Fathelrahman Mohamed Adam
VB
82
30 kN
20 kN
A
10 kN/m
B
4m
C
D
E
2m
2m
VA
2m
VB
Fig. (5)
Solutions:
1.
50 kN
A
B
C
3m
7m
VA
VB
Fig. (1)
First find the reactions:
By taking moment about B ∑ l” = 0 :
] 10 − 50(3) = 0
150
] =
= 15qT
10
By taking the vertical balance: (∑ k = 0):
] + ]” = 50
]” = 50 − 15 = 35qT
50 kN
A
B
C
7m
15 kN
x
Strength of Material – Dr. Fathelrahman Mohamed Adam
3m
35 kN
83
Shear Force:
Point
A
C
B
Sx Before (kN)
0
15
-35
Sx After (kN)
15
15 - 50 = -35
-35 + 35 = 0
S.F.D
15
15
(+)
0
0
(-)
all numbers are in kN
35
35
Bending Moment:
Point
A
C
B
Mx (kN.m)
0
15(7) = 105 (sag)
0
B.M.D
0
0
(+)
all numbers are in kN.m
Strength of Material – Dr. Fathelrahman Mohamed Adam
105
84
2.
50 kN
30 kN
A
B
C
D
3m
3m
4m
VA
VB
Fig. (2)
First find the reactions:
By taking moment about B ∑ l” = 0 :
] 10 − 30(7) − 50(3) = 0
] (10) = 210 + 150
360
= 36qT
] =
10
By taking the vertical balance: (∑ k = 0):
] + ]” = 80
]” = 80 − 36 = 44qT
50 kN
30 kN
A
B
C
D
3m
36 kN
3m
4m
44 kN
x
Shear Force:
Point
A
Sx Before (kN)
0
Sx After (kN)
36
Strength of Material – Dr. Fathelrahman Mohamed Adam
85
D
C
B
36
3
-44
36 - 30 = 3
36 -30 -50 = -44
-44 + 44 = 0
S.F.D
36
36
(+)
3
0
0
(-)
all numbers are in kN
44
44
Bending Moment:
Point
A
C
D
B
Mx (kN.m)
0
36(3) = 108 (sag)
36(7) – 30(4) = 132 (sag)
0
B.M.D
0
0
(+)
108
132
all numbers are in kN.m
Strength of Material – Dr. Fathelrahman Mohamed Adam
86
3.
A
10 kN/m
B
10 m
VA
VB
Fig. (3)
First find the reactions:
By taking moment about B ∑ l” = 0 :
] 10 − 10(10)(
10
)=0
2
] (10) = 500
500
= 50qT
] =
10
By taking the vertical balance: (∑ k = 0):
] + ]” = 100
]” = 100 − 50 = 50qT
A
10 kN/m
B
10 m
50 kN
50 kN
Shear Force:
Strength of Material – Dr. Fathelrahman Mohamed Adam
87
Point
A
B
Sx Before (kN)
Sx After (kN)
0
50 – 100 = -50
50
-50 – 50 = 0
S.F.D
50
(+)
0
0
(-)
all numbers are in kN
50
From shear force diagram the maximum at centre (x = l/2)
Bending Moment:
l = ] − u
Point
A
B
max
x (m)
0
10
5
2
Mx (kN.m)
0
50(10) – 10(10)(10/2) = 0
50(5) – 10(5)(5/2) = 125 (sag)
B.M.D
0
0
(+)
125
all numbers are in kN.m
Strength of Material – Dr. Fathelrahman Mohamed Adam
88
4.
25 kN
A
50 kN
10 kN/m
B
D
C
5m
3m
10 m
VA
VB
Fig. (4)
First find the reactions:
By taking moment about B ∑ l” = 0 :
5
] 10 − 10(5) + 5 − 25(5) − 50(3) = 0
2
] (10) = 375 + 125 + 150
650
] =
= 65qT
10
By taking the vertical balance: (∑ k = 0):
] + ]” = 10(5) + 25 + 50
]” = 125 − 65 = 60qT
25 kN
A
50 kN
10 kN/m
B
D
5m
C
3m
10 m
65 kN
60 kN
Shear Force:
Strength of Material – Dr. Fathelrahman Mohamed Adam
89
Point
A
D
C
B
Sx Before (kN)
Sx After (kN)
0
65 – 10(5) = 15
-10
-60
65
15 – 25 = -10
-10 – 50 = -60
-60 + 60 = 0
50
S.F.D
(+)
0
15
0
10
(-)
all numbers are in kN
50
Between A & D no maximum moment
Bending Moment:
Point
A
D
C
B
x (m)
0
5
7
10
Mx (kN.m)
0
65(5) – 10(5)(5/2) = 200 (Sag)
65(7) – 10(5)(5/2+2) – 25(2) = 180 (Sag)
65(10) – 10(5)(5/2+5) – 25(5) – 50(3) = 0
B.M.D
0
0
(+)
200
180
all numbers are in kN.m
Strength of Material – Dr. Fathelrahman Mohamed Adam
90
5.
30 kN
20 kN
10 kN/m
A
B
D
E
4m
C
2m
2m
VA
2m
VB
Fig. (5)
First find the reactions:
By taking moment about A ∑ l = 0 :
4
10 4
+ 20 6 − ]” (8) + 30(10) = 0
2
]” (8) = 80 + 120 + 300
500
]” = = 62.5qT
8
By taking the vertical balance: (∑ k = 0):
] + ]” = 10(4) + 20 + 30
] = 90 − 62.5 = 27.5qT
30 kN
20 kN
10 kN/m
A
B
D
E
4m
C
2m
2m
27.5 kN
2m
62.5 kN
Shear Force:
Point
A
Sx Before (kN)
0
Sx After (kN)
27.5
Strength of Material – Dr. Fathelrahman Mohamed Adam
91
E
D
B
C
27.5 – 10(4) = -12.5
-12.5
-32.5
30
-12.5
-12.5 – 20 = -32.5
-32.5 + 62.5 = 30
30 – 30 = 0
30
27.5
4-x
S.F.D
(+)
(+)
0
0
x
(-)
12.5
all numbers are in kN
32.5
32.5
Between A & E there is a maximum bending moment coincide with
zero shear force at distance x from A
Shear force between A & E can be calculated from the equation:
^ = ] − u( )
When ^ = 0 ⇒ u( ) = ]
From that
= ] /u
Bending Moment:
Point
A
E
D
B
C
max
x (m)
0
4
6
8
10
2.75
=
27.5
= 2.75,
10
Mx (kN.m)
0
27.5(4) – 10(4)(4/2) = 30 (Sag)
27.5(6) – 10(4)(4/2+2) = 5 (Sag)
27.5(8) – 10(4)(4/2+4) – 20(2) = -60 (hog)
27.5(10) – 10(4)(4/2+6) – 20(4) + 62.5(2) = 0
27.5(2.75) – 10(2.75)(2.75/2) = 37.8 (Sag)
Strength of Material – Dr. Fathelrahman Mohamed Adam
92
60
Point of contraflexure
(-)
0
0
5
a
(+)
2-a
30
37.8
all numbers are in kN.m
B.M.D
to find the location of point of contraflexure:
Assuming the point is at distance a from D, then:
E
2−E
=
5
60
60(E)
=2−E
5
12E + E = 2
2
= 0.154,
13
This means the point of contraflexure is at distance of 0.154 m from D.
E=
Strength of Material – Dr. Fathelrahman Mohamed Adam
93
Chapter 8
Buckling in Columns
Structural members which carry compressive loads may be divided into
two broad categories depending on their relative lengths and crosssectional dimensions.
1. Short, thick members are generally termed short columns and these
usually fail by crushing when the yield stress of the material in
compression is exceeded
2. Slender columns or struts, however, fail by buckling some time
before the yield stress in compression is reached.
In general the buckling is a tendency of slender compression
members to bow out, which causes lateral bending. Buckling effects all
compression members, such as columns, truss bars, bracing, etc.
Strength of Material – Dr. Fathelrahman Mohamed Adam
94
Classification of slender columns:
The slender column is classified according to the ratio of length (L) and
radius of gyration (r) which it called the slenderness ratio.
CLKG K4GKCC4EDFM = O
4
The radius of gyration (r) describes the way in which the area of a
cross-section is distributed around its centroidal axis. If the area is
concentrated far from the centroidal axis it will have a greater value of
r and a greater resistance to buckling. The section tends to buckle
around the axis with the smallest value. The radius of gyration, r,
mathematically is calculated as:
4=5
1
where I is the moment of inertia and A is the area of the cross-section.
The Critical Load:
When a load P of a value below the buckling load is applied to a slender
column, the column will be in stable equilibrium condition and the lateral
displacement caused by load will be totally recovered when the is
removed.
If the load exceed a value of Pcr which called the critical load it is
possible for slender column to achieve a condition of unstable
equilibrium and the lateral displacement happened by this load is not
recovered and this is case of failure called by buckling.
Strength of Material – Dr. Fathelrahman Mohamed Adam
95
The critical load is determined through the study of stability of columns
by using Euler's theorem with considering different end conditions.
Euler's Theorem:
The following assumptions are made in this theorem:
1. The column is initially straight and the applied load is truly axial.
2. The material of the column is homogeneous, linear and isotropic.
3. The length of the column is very large as compared to the crosssectional dimensions of the column.
4. The cross-section of the column is uniform throughout.
5. The shortening of the column due to axial compression is
negligible.
6. The self weight of the column is neglected.
7. The ends of the column are frictionless.
There are three conditions at the end of the column.
1. Column Pinned at both ends:
Consider an elastic column of length L, pin-ended so free to rotate at
its ends, subjected to an axial load P, as shown in Figure below.
Assume that it undergoes a lateral deflection denoted by y. Moment
equilibrium of a section of the deflected column cut at a typical point
at distance x from A, and using the relation:
l Q
= , M4
1
j
1
1
l = Q1 EG =
j
j
l = Q1
(1)
Strength of Material – Dr. Fathelrahman Mohamed Adam
, CM
96
P
A
y
x
y
C
L
B
P
x
From Figure, moment at point C is:
Mx = - P y
(2)
From (1):
Q1
Let:
Q1
+
= −@
+ @ = 0, M4
@
= 0(3)
Q1
@
=–
Q1
Rewrite (3):
+ – = 0(4)
Equation (4) is the differential equation has a general solution
of:
y = A sin α x + B cos α x
(5)
where A and B are constants, which can be determined using the
column’s kinematic boundary conditions
Strength of Material – Dr. Fathelrahman Mohamed Adam
97
by applying end condition at equation (5):
at x = 0, y = 0:
0 = A (0) + B (1), B =0
at x = L, y = 0:
0 = A sin αL
Either A =0, in which case y = 0 for all values of x and column in
this case not buckle, so the preponderant
sin αL = 0
from that αL = π, corresponding to:
α2 L2 = π2
From that:
α2 = π2/ L2
@
V
=
Q1
O
@ =
V Q1
O
The least value of the above Equation which causes buckling is
called the Euler crippling load (Pcr), so:
Where:
Pcr
E
I
L
@—h = V Q1
(6)
O
is the critical or maximum axial load on the column
just before it begins to buckle
is the Young’s modulus of elasticity
is the least second moment of area for the column’s
cross sectional area
is the unsupported length of the column, whose ends
are pinned.
Example (8.1)
A 2-cm-diameter, 120-cm-long steel column with pinned ends supports
a load P. Calculate the slenderness ratio and Euler’s buckling load.
Take E = 200,000 N/mm2.
Strength of Material – Dr. Fathelrahman Mohamed Adam
98
Solution:
Given data:
L = 120 cm = 1200 mm, d = 2 cm = 20 mm
E = 200,000 N/mm2
Required:
Slenderness ratio, Pcr
CLKG K4GKCC4EDFM = 4=5
@—h = Calculate least value of I:
1
O
4
V Q1
O
V < V(20)<
1 =
=
= 7,854,,<
64
64
=
V
V(20)
=
= 314.2,,
4
4
7,854
4=5
= 5,,
314.2
CLKG K4GKCC4EDFM = @—h = 1,200
= 240
5
V (200,000)(7,854)
= 10,766T = 10.8qT
(1200)
Example (8.2)
Steel bar of rectangular cross section 40 mm x 50 mm and pinned at
each end is subject to axial compression. If the proportional limit of the
material is 230 MPa and E = 200 GPa, determine the minimum length
for which Euler’s equation may be used to determine the buckling load.
Strength of Material – Dr. Fathelrahman Mohamed Adam
99
Solution:
Given data:
b = 50 mm, h = 40 mm
σy = 230 MPa = 230 N/mm2, E = 200 GPa = 200x103 N/mm2
Required:
L=?
Since the column is pinned at tow ends then the Euler load can be
calculated by using the following Equation:
@—h =
Q1V
O
ℎ! 50(40)!
1=
=
= 266.67 10! ,,<
12
12
The critical load can be calculated by using the yield stress:
? =
@—h
, @—h = ? A = bxh = 40(50) = 2,000 mm2
@—h = 230(2,000) = 460,000T
Use this to calculate length from:
Q1V
O=5
@—h
200 10! (266.67 10! )V
O=5
= 1070,, = 1.07,
460,000
Example (8.3)
A straight column of alloy, 1 m length and 12.5 mm by 4.8 mm in
section, is mounted in a buckling testing machine and loaded axially
until it buckles. Assuming the Euler theorem to apply, estimate the
Strength of Material – Dr. Fathelrahman Mohamed Adam
100
maximum central deflection before the materials attains its yield point
of 280 N/mm2. Use E = 72,000 N/mm2.
Solution:
Given data:
L = 1 m = 1000 mm, b = 12.5 mm, h = 4.8 mm
σy = 280 N/mm2, E = 72,000 N/mm2
Required:
Max central deflection (δmax)
Calculate least value of I:
ℎ! 12.5(4.8)!
1 =
=
= 115.2,,<
12
12
Find critical load by using Equation (6):
@—h = V (72,000)(115.2)
= 81.9T
(1000)
The maximum bending moment at centre:
Mmax = Pcr (δmax) = 81.9 δmax
The stress in section is compound stress compose of direct stress
(σd) and bending stress (σb), where:
?e =
@
=
81.9
= 1.4T/,,
12.5(4.8)
?r =
lnf
m
1
115.2
m = =
= 48,,!
(4.8/2)
?r =
81.9(Nnf )
= 1.7Nnf
48
σy = σd + σb
280 = 1.4 + 1.7 δmax
Nnf =
280 − 1.4
= 163.9,,
1.7
2. Column fixed at both ends:
Strength of Material – Dr. Fathelrahman Mohamed Adam
101
P
M
y
A
x
y
C
L
M
B
P
x
Mx = - P y + M
(7)
From (1):
Q1
Q1
+
Let:
Rewrite (8):
= −@ + l
+ @ = l, M4
@
l
= (8)
Q1
Q1
@
=–
Q1
+– =
The complete solution is:
l
(9)
Q1
= CFG– + wIMC– +
l
(10)
Q1–
by applying end condition at equation (10):
at x = 0, y = 0:
Strength of Material – Dr. Fathelrahman Mohamed Adam
102
l
Q1–
l
l
w = −
=−
Q1–
@
0 = 0 + w(1) +
@ = Q1–
Note that:
at x = 0,
†
†
=0:
= –IMC– − w–CFG–
0 = –(1) − 0
=0
Substitute A & B in (10):
=
l l
l
− IMC– = (1 − IMC– )
@
@
@
l
= (1 − IMC– )(11)
@
at x = L, y = 0 (substitute in (11)):
cos αL = 1
the least solution is:
αL=2π
(12)
By quadrate the two ends of (12)
– O = 4V
Since:
– =
Then:
@
Q1
@
O = 4V
Q1
From that:
@=
4Q1V
(13)
O
Strength of Material – Dr. Fathelrahman Mohamed Adam
103
Equation (13) is equivalent to Equation (6) with taking L/2
insteadofL
IfEquationrewrittenbythegeneralformas:
@—h =
Q1V
(14)
(qO)
Wherek=1forpinnedends,andk=0.5forDixedends.
Byfollowingthesamepreviousprocedure,wecanderivethelength
factorforotherconditionsbasedonEquation(14).
ThelengthfactorfortheendconditionsaresummarizedinTable
below:
Euler Load
Effective Length
End Conditions
›œ•ž
Factor (k)
s˜™š = (Ÿ )ž t
BothEndsPinned
BothEndsFixed
OneEndFixedand
otherPinned
OneEndFixedand
otherFree
1
0.5
0.7
2
Q1V
O
4Q1V
@—h =
O
2Q1V
@—h =
O
Q1V
@—h =
4O
@—h =
Chapter 9
DEFLECTION OF STRUCTURES
Strength of Material – Dr. Fathelrahman Mohamed Adam
104
1. MOMENT AREA METHOD:
The moment-area method, developed by Otto Mohr in 1868, is a
powerful tool for finding the deflections of structures primarily
subjected to bending.
Consider the beam shown in Figure below.
A
N.A
B
A
A´
Original N.A
B
B´
Bended N.A
θ
R
R
dθ
R
A´
B´
θA
θB
P
dS
Q
dθ
dΔ
Δ
θ
M
̅
B.M.D
Q
P
dx
x
From diagram:
AB - is the original length of beam before loading.
Strength of Material – Dr. Fathelrahman Mohamed Adam
105
A´B´ - is the deflected length along the curve after loading.
ds - is the very short curve length between P & Q.
dx - is the very short straight length between P & Q.
θ - is the angle subtended at the centre of arc A´B´.
dθ - is the angle subtended at the centre of arc ds.
M - is the average bending moment over the portion x between P & Q.
∆ is the vertical intercept, measured from point B to point meet the
extension of tangent at point A.
with referring to the diagram:
1. Change of slope:
C = j. i, y4M,DℎED:
1
i
=
(1)
j
C
from the Euler-Bernoulli theory of bending:
Q l
1 l
= M4 = (2)
j
1
j Q1
From (1):
i l
l
= M4 i =
C(3)
C Q1
Q1
noting that for small deflections, we can take ds ≈ dx
from (3) find the total change of rotation between A & B , by taking
the bound integration between A & B then:
”
i=
”
l
Q1
l
i” − i = 7 8 (4)
Q1 ”
where the right side is the area of
2. Change of slope:
from Fig.:
from (3) substitute i =
Let
=
£
¤¥
, then:
£
¤¥
diagram between A & B.
∆= . i
£
¤¥
∆=
∆=
l
Q1
Strength of Material – Dr. Fathelrahman Mohamed Adam
106
Integrate each side, then:
”
”
∆=
”
where:
∆ = ∆” , uℎFIℎFCDℎK§K4DFIELFGDK4IK\DECK \LEFGFGkFJ.
”
=
£
̅ ,
which is the first moment of area of diagram about B
¤¥
therefore:
∆” =
̅ (5)
Example (9.1):
Find the rotation and deflection at the free end for the cantilever shown
in Fig.
P kN
B
A
EI = const
Lm
θ = (θB – θA) = θB
θA = 0
θB
M = PL
ΔBA
(-)
B.M.D
=
(-)
̅=
1
@O
@O
−
O=−
2
Q1
2Q1
2O
3
£
¤¥
Diagram
calculate the rotation by using equation (4):
Strength of Material – Dr. Fathelrahman Mohamed Adam
107
l
i” − i = 7 8
Q1 ”
note that θA = 0, then:
θB = A =
6W¨=
¤¥
calculate the deflection by using equation (5):
∆” =
̅
note that the slope at A is straight, then the deflection is
directly equal to the intercept:
KyLKIDFMG = ∆”
−@O
=
2Q1
2O
−@O!
=
3
3Q1
Example (9.2):
Find the rotation and deflection at the free end for the cantilever
shown in Fig.
10 kN
2EI
A
B
EI
2m
2m
C
θ = (θB – θA) = θB
θA = 0
M = 10(4) = 40
ΔBA
θB
20
(-)
B.M.D
40
20
=
2Q1 Q1
10
Q1
10
Q1
A3
A2
20
Q1
10
Q1
A1
x1
£
¤¥
Diagram
x2
x3
1 20
20
(2)
=
,
*
2 Q1
Q1
Strength of Material – Dr. Fathelrahman Mohamed Adam
=
*
2
4
= (2) =
3
3
108
=
!
=
10
Q1
1 10
2 Q1
2 =
2 =
20
,
Q1
=
10
,
Q1
!
=
1
2 +2=3
2
10
2
2 +2=
3
3
calculate the rotation by using the equation:
l
i” − i = 7 8
Q1 ”
note that θA = 0, then:
20 20 10 50
i” = * +
+ !=
+
+
=
Q1 Q1 Q1 Q1
calculate the deflection by using the equation:
∆” =
̅
note that the slope at A is straight, then the deflection is
directly equal to the intercept:
KyLKIDFMG = ∆” =
[KyLKIDFMG =
Example (9.3):
* *
+
+
! !
20 4
20
10 10
120
+ (3) +
=
Q1 3
Q1
Q1 3
Q1
32 kN
A
B
3m
C
2m
EI = 36x103 kN/m2
For the cantilever shown in Fig. find the rotations and
displacements at joint B and C.
Solution:
Draw the bending moment diagram:
Strength of Material – Dr. Fathelrahman Mohamed Adam
109
96
l FEJ4E,
(-)
0
B
A
0
C
96
= 2.67 106! !
36 10
(-)
A
0
xB
From diagram:
”
l
FEJ4E,
Q1
0
xC
1
3 −2.67 106! = −4 106!
2
2
2
= 3 = 2,, © = (3) + 2 = 4,
3
3
=
Rotation:
Apply in the equation:
l
i” − i = 7 8
Q1 ”
i = 0:
KDuKKGªMFGD &w:
i” = 4KE KDuKKG &w = −4 106! 4E KDuKKGªMFGD &¬:
i© = 4KE KDuKKG &¬ = −4 106! 4E Deflection:
As before for cantilever the deflection is directly equal to vertical
intercept (∆) by using the equation:
∆” =
̅
Between A & B:
δB = moment area about B:
6!
6!
N” = ∆” =
” = −4 10 (2) = −8 10 ,
Between A & C:
δC = moment area about C:
6!
6!
N© = ∆© =
© = −4 10 (4) = −16 10 ,
Strength of Material – Dr. Fathelrahman Mohamed Adam
110
2. DEFLECTION BY CALCULUS:
From the general equation of bending (Engineering Equation), the
curvature can be expressed as:
1 l
= (1)
j Q1
where:
M: is the bending moment.
I: is moment of inertia.
E: is the modulus of elasticity.
*
-
from calculus equal to
†=
† =
When a beam is subjected to a transverse loading, this gives a bending
moment through the length of beam and can be calculated at any section
at distance x from one end of the beam and can be denoted as Mx. so
equation (1) can be rewritten by equation:
†
Q1
…
…
= l (2)
The rotation can be found by differentiate each side of equation (2),
†
as:
…
Q1
= l
+ ¬* (3)
…
The deflection y can be found by differentiate each side of equation (3),
as:
Q1 = ® l
+ ¬* + ¬ (4)
where C1 and C2 are constant.
*note that:
for double integrating shown in equation (4), this method
sometimes called (Method of Double Integration)
The two constants C1 and C2 can be obtained by substituting known
values of slope or deflection at particular points. This called applying
boundary conditions.
Strength of Material – Dr. Fathelrahman Mohamed Adam
111
Example (9.4)
Find the rotation and deflection at the free end for the cantilever
shown in Fig. by using method of calculus.
P kN
B
A
EI = const
Lm
x
Solution:
By taking the origin of x at B, then the moment can be calculated
by:
l =@
Substitute in equation (2):
…
Q1
=@
…
Integrate each side:
…
@
Q1
=
+ ¬* (E)
…
2
Integrate each side of equation (a):
@ !
Q1 =
+ ¬* + ¬ ( )
6
from figure:
at x = L:
†
= 0 and y=0
†
Substitute in (a) and (b):
@O
Q1(0) =
+ ¬* 2
from that:
@O
¬* = −
2
Substitute in (b):
@O!
@O
Q1(0) =
+ #−
%O + ¬
6
2
from that:
1 1
@O!
!
¬ = @O
−
=−
6 2
3
Strength of Material – Dr. Fathelrahman Mohamed Adam
112
Rewrite equations (a) & (b):
…
@
@O
Q1
=
−
(1)
…
2
2
@ ! @O
Q1 =
−
6
2
@O!
−
(2)
3
Equations (1) & (2) are general equations to calculate rotation
and deflection anywhere through the length.
So at free end this means at x =0:
from (1):
so the rotation is:
Q1
…
@(0) @O
=
−
2
…
2
…
@O
=−
…
2Q1
@(0) @O
@O!
Q1 =
−
(0) −
6
2
3
so the deflection is:
@O!
=−
3Q1
from (2):
Example (9.5):
A
10 kN/m
C
6m
EI = const
For the simply supported beam shown in figure, find:
1. Rotation at support A & B.
2. Deflection at centre.
Strength of Material – Dr. Fathelrahman Mohamed Adam
113
Solution:
10kN/m
A
C
x
¯°(±)
= ²°
ž
²°
Calculate the reaction and take origin of x at A.
Then:
l = 30 − 10 #
Q1
…
…
2
= 30 − 10 #
%
2
%
Integrate each side:
!
…
Q1
= 30 # % − 10 # % + ¬* (1)
…
2
6
Integrate each side of equation (1):
Q1 = 30 #
!
6
<
% − 10 # % + ¬* + ¬ (2)
24
Apply boundary condition:
y =0, at x= 0 and x = 6:
substitute in (2):
0
0
Q1(0) = 30
− 10
+ ¬* (0) + ¬ ⇒ ¬ = 0
6
24
6!
6<
Q1(0) = 30 # % − 10 # % + ¬* (6) + 0
6
24
¬* = −
540
= −90
6
Substitute for C1 & C2 in (1) and (2):
!
…
Q1
= 30 # % − 10 # % − 90(1)
…
2
6
Strength of Material – Dr. Fathelrahman Mohamed Adam
114
Q1 = 30 #
!
<
% − 10 # % − 90 (2)
6
24
1. Rotation at A, x =0:
Substitute in (1):
…
0
0
Q1
= 30
− 10
− 90
…
2
6
…
90
=−
…
Q1
Rotation at B, x = 6:
Substitute in (1):
…
6
6!
Q1
= 30 # % − 10 # % − 90
… ”
2
6
…
…
”
=
90
Q1
2. Deflection at centre: x = 3:
Substitute in (2):
3<
3!
Q1 —gachg = 30 # % − 10 # % − 90(3)
6
24
—gachg
=−
168.75
Q1
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115