If
E
=
200 kN/mm
2
,
v
=
0.3.
Tapered Bar
Consider a circular bar uniformly tapered from diameter
d
1
at one end and gradually increasing to diam
eter
d
2
at the other end over an axial length
L
as shown in Fig. 1.18. Take an elementary disc of diameter
d
x
and length
dx
at a distance of
x
from end
A
.
Diameter
d
x
of elementary disc
=
d
d
d
d
L
d
kx
x
=
+

=
+
1
2
1
1
Figure 1.17
20
30
500 mm
P
P
Simple Stresses and Strains
13
where constant
k
d
d
L
=

2
1
Crosssectional area,
A
d
d
kx
x
x
=
=
+
π
π
4
4
2
1
2
(
)
(
)
Axial stress,
σ
π
x
x
P
A
P
d
kx
=
=
+
4
1
2
(
)
Axial strain,
ε
σ
π
x
x
E
P
E d
kx
=
=
+
4
1
2
(
)
Change in length over
dx
,
δ
π
d
Pdx
E d
kx
x
=
+
4
1
2
(
)
Total change in length,
δ
π
L
Pdx
E d
kx
O
L
=
+
∫
4
1
2
(
)
=
+


4
1
1
P
E
d
kx
k
O
L
π
(
)
= 
+

4
1
1
1
1
P
Ek
d
kL
d
π
= 

4
1
1
2
1
P
Ek
d
d
π
as
d
kL
d
d
d
L
L
d
1
1
2
1
2
+
=
+

×
=
Therefore,
δ
π
π
L
P
Ek
d
d
P d
d
Ekd d
=

=

4
1
1
4
1
2
2
1
1
2
(
)
Putting the value of
k
d
d
L
=

2
1
,
we get
Figure 1.18
Circular tapered bar
d
2
d
1
P
L
B
d
dx
x
x
A
P
14
Chapter 1
δ
π
L
P d
d
E d d
L
d
d
=

×

4
2
1
1
2
2
1
(
)
(
)
(
)
Total change in length
=
δ
π
L
PL
Ed d
=
4
1
2
Example 1.7
A brass bar uniformly tapered from diameter 20 mm at one end to diameter 10 mm
at the other end over an axial length 300 mm is subjected to an axial compressive load of 7.5
kN.
If
E
=
100 kN/mm
2
for brass, determine (Fig.1.19):
(a) the maximum and minimum axial stresses in bar and
(b) the total change in length of the bar.
Solution
Axial load,
P
=
7.5 kN. Maximum stress occurs at end
A
with minimum crosssectional area:
(a)
σ
π
π
max
(
)
.
,
.
= 
= 
×
×
×
= 
4
10
4
7 5
1 000
100
95 94
2
P
N/mm
2
at end
A
(b)
σ
π
π
min
(
)
.
,
.
= 
= 
×
×
×
= 
4
20
4
7 5
1 000
20
23 87
2
P
N/mm
2
at end
B
δ
π
L
PL
Ed d
= 
4
1
2
, as the load is compressive
= 
×
×
×
×
×
×
×
= 
4
7 5
1 000
300
100
1 000
10
20
0 143
.
,
,
.
π
mm
Figure 1.19
B
P
A
P
300 mm
10 mm
f
20 mm
f
Exercise 1.7
A 400mmlong aluminium bar uniformly tapers from a diameter of 25 mm to a
diameter of 15 mm. It is subjected to an axial tensile load such that stress at middle section is 60
MPa.
(a) What is the load applied?
(b) What is the total change in the length of the bar if
E
=
67,000 N/mm
2
?
[Hint: At middle, the diameter is
(
)
25
15
2
20
+
=
/
mm]
Simple Stresses and Strains
15
Tapered Flat
Consider a flat of same thickness
t
throughout its
length, tapering uniformly from a breadth
B
at one end
to a breadth
b
at the other end over an axial length
L
.
The flat is subjected to an axial force
P
as shown in
Fig. 1.20.
Thickness of elementary strip
abc
is
t
at a distance
of
x
from end
A.
Breadth,
b
b
x =
x
B
b
L
b
kx
= +

+
(
),
where
k
B
b
L
=

Crosssectional area of elementary strip,
A
x
=
b
x
t
=
(
b
+
kx
)
t
Stress at elementary strip,
σ
x
x
P
A
P
b
kx t
=
=
+
(
)
Change in length over
d
x
,
δ
d
Pdx
Et b
kx
x
=
+
(
)
Total change in the length of flat,
δ
L
Pdx
Et b
kx
L
=
+
∫
(
)
0
δ
L
P
Etk
B
b
=

[
]
ln(
)
ln( )
=
×
P
Etk
B
b
ln
Putting the value of
k
,
δ
L
PL
Et B
b
B
b
=

(
)
ln
Example 1.8
A steel flat of an axial length of 600 mm and a uniform thickness of 10 mm has a uni
formly tapered width of 40 mm at one end to 20 mm at the other end. If the axial force is
P
=
18 kN and
E
=
208 kN/mm
2
, then what is the change in axial length of flat?
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 Force, Shear Stress, Tensile strength, Professor Hooke