strength of materials

1. TCBE1201 STRENGTH OF MATERIALS Chapter II
CHAPTER II. TENSION AND COMPRESSION
SECTION I. STRESS AND STRAIN IN TENSION (OR COMPRESSION)
1. Direct stress (σ )
It has been noted above that external force applied to a body in equilibrium is reacted by internal forces set up
within the material. If, therefore, a bar is subjected to a uniform tension or compression, i.e. a direct force, which is
uniformly or equally applied across the cross-section, then the internal forces set up are also distributed uniformly and
the bar is said to be subjected to a uniform direct stress, the stress being defined as
stress (σ ) =
A
P
=
area
load
Stress σ may thus be compressive or tensile depending on the nature of the load and will be measured in units of
Newton’s per square metre (N/m2
) or multiples of this.
In some cases the loading situation is such that the stress will vary across any given section and in such cases the
stress at any point is given by the limiting value of AP δδ / as Aδ tends zero.
2. Direct strain (ε )
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 Lδ , the strain produced is defined as follows:
strain ( ε )=
L
Lδ
=
lengthoriginal
lengthinstrain
Strain is thus a measure of the deformation of the material and is non-dimensional, i.e. it has units; it is simply a
ratio of two quantities with the same unit (Fig. 2.1.1).
Since, in practice, the extensions of materials under load are very small, it is often
convenient to measure the strains in the form of strain x 10-6
, i.e. microstrain, when the
symbol used becomes µε .
3. Sign convention for direct stress and strain
Tensile stresses and strains are considered POSITIVE in sense.
Compressive stresses and strains are considered NEGATIVE in sense.
Thus a negative strain produces a decrease in length:
4. Definitions
Elastic materials — Hooke's law
A material is said to be elastic if it returns to its original, unloaded dimensions when
load is removed. A particular form of elasticity which applies to a large range of engineering
materials, at least over part of their load range, produces deformations which are
proportional to the loads producing them. Since loads are proportional to the stresses they
produce and deformations are proportional to the strains, this also implies that, whilst materials are elastic, stress is
proportional to strain. Hooke's law, in its simplest form, therefore states that
Stress ∈)(σ strain )(ε
=
strain
stress
constant
It will be seen in later sections that this law is obeyed within certain limits by most ferrous alloys and it can even
be assumed to apply to other engineering materials such as concrete, timber and non-ferrous alloys with reasonable
accuracy.
Whilst a material is elastic the deformation produced by any load will be completely recovered when the load is
removed; there is no permanent deformation.
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2. TCBE1201 STRENGTH OF MATERIALS Chapter II
Modulus of elasticity — Young's modulus
Within the elastic limits of materials, i.e. within the limits in which Hooke's law applies, it has been shown that
=
strain
stress
constant
This constant is given the symbol E and termed the modulus of elasticity or Young's modulus.
Thus
)2.2(
)1.2(
LA
PL
L
L
A
P
strain
stress
E
δ
δ
ε
σ
=÷=
==
Young's modulus E is generally assumed to be the same in tension or compression and for most engineering
materials has a high numerical value. Typically, E = 200 x 109
N/m2
for steel, so that it will be observed from (2.1)
that strains are normally very small since
E
σ
ε = (2.3)
In most common engineering applications strains do not often exceed 0.001 or 0.1%.
The actual value of Young’s modulus for any materials is normally determined by carrying out a standard
tensile on a specimen of the material.
Poisson's ratio.
When a material is subjected to longitudinal deformation then the lateral dimensions also change. The
ratio of the lateral strain to longitudinal strain is a constant quantity called the Poisson's ratio and is
designated by ν or 1/m.
strainalLongitudin
strainLateral
=ν (2.4)
Modulus of rigidity
It is defined as the ratio of shearing stress to shearing strain, i.e.
γ
τ
=G (2.5)
Factor of safety
Because of uncertainties of loading conditions, we introduce a factor of safety, defined as the ratio of
the maximum stress to the allowable or working stress. The maximum stress is generally taken as the
yield stress. This is also called the ‘ factor of ignorance’
Free body Diagram
The free body diagram of an element of a member in equilibrium is the diagram of only that member or
element, as if made free from the rest, with all the internal and external forces acting on it.
Coefficient of Linear Thermal Expansion
Linear thermal strain ( Tε ) due to change in temperature ( T∆ ) is obtained by using this coefficient (α )
TT ∆=αε (2.6)
α has units of per degrees Centigrade (or Fahrenheit)
5. Bar of Varying Cross-section
Consider a bar of varying circular cross-section as shown in Fig. 2.1.2
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3. TCBE1201 STRENGTH OF MATERIALS Chapter II
1 2
3
d1
d2 d3
L1 L2 L3
P P
Fig. 2.1.2 Bar of varying cross-section.
and subjected to axial load P throughout. The area of different cross -sections is:
2
33
2
22
2
11
4
,
4
,
4
dAdAdA
πππ
===
Let 21 , σσ and 3σ be the corresponding stresses, then,
3
3
2
2
1
1 ,,
A
P
A
P
A
P
=== σσσ
The strains become,
3
3
3
2
2
2
1
1
1 ,,
EEE
σ
ε
σ
ε
σ
ε ===
The changes in lengths become,
,,, 333222111 llllll εεε =∆=∆=∆
Total change in length,






++=∆+∆+∆=∆
33
3
22
2
11
1
321
EA
l
EA
l
EA
l
Pllll
or in general, we have
∑=
=∆
n
i ii
i
EA
l
Pl
1
d1P1
1
d2
d3
2
3
P4
P3P2
L1 L2 L3
d1P1
1
d2
d3
2
3
P4P1
P1 - P2 P1 - P2
(P4- P )3
P4
(a)
(b) Free body diagrams
Fig.2.1.3. Bar of varying cross-section.
If the loads in different sections of the bar are different as shown in Fig. 2.1.3 (a), then free body diagrams
may be drawn for each section as shown in Fig. 2.1.3 (b), and the net forces acting in each section may be
determined. Thus the stresses, strains and total elongation may be determined.
Page 3 of 6

4. TCBE1201 STRENGTH OF MATERIALS Chapter II
∑=
=
∆+∆+∆=∆
=∆=∆=∆
===
===
n
i ii
ii
EA
lP
llll
llllll
EEE
A
P
A
P
A
P
1
321
333222111
3
3
3
2
2
2
1
1
1
3
3
3
2
2
2
1
1
1
,,
,,
,,
εεε
σ
ε
σ
ε
σ
ε
σσσ
Example 2.1 A mild steel rod 20mm diameter is subjected to an axial pull of 50KN. Determine the tensile stress
induced in the rod and the elongation if the unloaded length is 5m. 2
/210 mGNE = .
Solution. Given
d=20mm ; P=50KN ; l=5m
Area of cross-section of the rod
26622
1031410)20(
44
mdA −−
×=×==
ππ
Stress 2
6
33
/155.159
10314
1051050
mMN
A
P
=
×
×××
== −
σ
Elongation mm
AE
Pl
789.3
1021010314
1051050
96
33
=
×××
×××
== −
δ
Example 2.2 A short hollow cast iron cylinder of wall thickness 10mm is to carry a compressive load of 600KN.
Determine the outside diameter of the cylinder if the ultimate crushing stress for material is 2
/540 mKN . Use a
factor of safety of 6.
Solution. Let 0d be the outside diameter of the cylinder in mm. Then area of cross-section of the cylinder is,
{ } 25
0
62
0
2
0 10)10(10)20(
4
mdddA −−
×−=×−−= π
π
Safe load
25
0 10)10( md −
×−=π
3
0 10600)10(900 ×=−×= dπ
mmd 2.2220 =∴
Example 2.3 A round bar as shown in Fig.2.1.4 is subjected to an axial tensile load of 100KN. What must be the
diameter ‘d’ if the stress there is to be 2
/100 mMN ? Find also the total elongation. 2
/210 mGNE = .
Solution.
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5. TCBE1201 STRENGTH OF MATERIALS Chapter II
100KN100KN d 10cm 8cm
15cm10cm 15cm
Fig.2.1.4.
Stress 2
4
d
P
π
σ =
2
3
6
4
10100
10100
d
π
×
=×
∴ Diameter, mmmd 68.3503568.0
10
4
3
==
×
=
π
Total elongation,
mm
A
l
A
l
A
l
E
P
l
0745.0
10
1
)80(
4
15.0
)100(
4
15.0
10
10.0
10200
10100
6
22
39
3
2
2
2
2
1
1
=
×












×
+
×
+
×
×
=






++=∆
−
ππ
Example 2.4 A steel bar 25mm diameter is loaded as shown in Fig.2.1.5. Determine the stresses in each part and
the total elongation. 2
/210 mGNE = .
Solution.
50cm 40cm 20cm
40KN 30KN20KN
10KN
A B C D
Fig.2.1.5
40KN 30KN40KN
20KN
A B C D
20KN
B
30KN
C
Fig.2.1.6 Free body diagrams.
Area of cross-section,
26622
1087.49010)25(
44
mdA −−
×=××==
ππ
The free body diagrams for each portion have been shown in Fig.2.1.6. Stress in various parts are:
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6. TCBE1201 STRENGTH OF MATERIALS Chapter II
2
6
3
2
6
3
2
6
3
/166.61
1087.490
1030
/744.40
1087.490
1020
/488.81
1087.490
1040
mMN
mMN
mMN
CD
BC
AB
=
×
×
=
=
×
×
=
=
×
×
=
−
−
−
σ
σ
σ
Total elongation ∑=∆ iilP
AE
l
1
[ ] mm3298.02.0304.0205.040
102101087.490
10
96
3
=×+×+×
×××
= −
Page 6 of 6

7. TCBE1201 STRENGTH OF MATERIALS Chapter II
2
6
3
2
6
3
2
6
3
/166.61
1087.490
1030
/744.40
1087.490
1020
/488.81
1087.490
1040
mMN
mMN
mMN
CD
BC
AB
=
×
×
=
=
×
×
=
=
×
×
=
−
−
−
σ
σ
σ
Total elongation ∑=∆ iilP
AE
l
1
[ ] mm3298.02.0304.0205.040
102101087.490
10
96
3
=×+×+×
×××
= −
Page 6 of 6