STRENGTH OF MATERIALS
| Site: | LEARNING FOR ALL |
| Course: | LEARNING FOR ALL |
| Book: | STRENGTH OF MATERIALS |
| Printed by: | |
| Date: | Tuesday, 10 June 2025, 1:37 AM |
1. STRAIN AND STRESS
1.1 Introduction. When a load is applied to a member of a machine or
structure, the material distorts. The stress intensity (usually abbreviated to
stress) is the load transmitted per unit area of cross-section and the strain
is a measure of the resulting distortion.
Assuming that the load is insufficient to cause rupture, it is resisted by
the force of attraction between the molecules of the material and the
deformation is the result of the slight re-orientation of the molecules.
If the material returns to its former shape when the load is removed, it
is said to be elastic; if the strain is permanent, it is said to be plastic. Most
engineering materials are elastic up to a certain stress (referred to as the
elastic limit), after which they are partly elastic and partly plastic. The
transition is not always abrupt, but for the purposes of calculation it is
usually assumed to be so, an assumption which is reasonably justified for
common mild steel.
In the simple theory of Strength of Materials, it is assumed that the
material is isotropic (i.e. displays the same properties in all directions) and
that it is equally rigid in tension and compression. It is further assumed
that the stress is uniformly distributed over the area resisting the load;
this is approximately true, except in the near vicinity of the point of
application of the load or a sudden change of section (St Venant's Prin
ciple).
1.2 Tensile and compressive stress and strain. If a piece of material
of cross-sectional area a is subjected to equal and opposite forces P, either
tensile, as in Fig. 1.1(a) or compressive, as in Fig. 1.1(6), then
force
stress = cross-sectional area
i.e. σ= - (1.1)
a
If the original length of the bar is I and under the effect of the force Ρ it
extends or compresses a distance x, then
change in length
strain = original length
i.e. e= | (1.2)
1 2
STRENGTH OF MATERIALS
The deformed shapes of the bars are as shown dotted in Fig. 1.2; the
strain in directions perpendicular to that of the load is proportional to that
in the direction of the load and is of the opposite sign.
The ratio ^
e r
^
s
^
r a m
j
s ca
j j
e (
j p
0
isson's Ratio and is denoted by v.
axial stram
Thus if the axial strain is ε, the lateral strain is —νε.
(a)
FIG. 1.1
(b)
(a)
Τ
(b)
FIG. 1.2
1.3 Shear stress and strain. If a piece of material of cross-sectional
area a is subjected to equal and opposite forces Ρ which produce a state of
shear, as shown in Fig. 1.3, then
force
shear stress :
i.e.
=
cross-sectional area
Ρ
(1.3)
If the deformation in the direction of Ρ is χ and the perpendicular dis
tance between the applied forces is Z, then
deformation
shear strain :
couple arm
i.e.
(1.4)
is the angular displacement in radians, since - is very small.
V
i
/
/
/
1/ p
FIG. 1.3
τ
FIG. 1.4 SIMPLE STRESS AND STRAIN
3
When a shear stress r is applied to the faces AB and CD of an element of
the material, Fig. 1.4, a clockwise couple ( X AB χ t) χ BC is applied to
the element, t being the thickness of the material. Since it does not rotate,
however, an equal anticlockwise couple must be applied by means of shear
stresses induced on faces AD and BC.
If the magnitude of these stresses is τ', then for equilibrium,
(r χ AB χ t) X BC = (τ' X BC χ t) X AB
τ
' = τ
Thus a shear stress in one plane is always accompanied by an equal shear
stress (called the complementary shear stress) in the perpendicular plane.
1.4 Hooke's Law. Hooke's Law states that when a load is applied to
an elastic material, the deformation is directly proportional to the load
producing it. Since the stress is proportional to the load and the strain is
proportional to the deformation, it follows that the stress is proportional to
the strain, i.e. the ratio stress/strain is a constant for any given material.
For tensile or compressive stresses, this constant is known as the Modulus
of Elasticity (or Young's Modulus) and is denoted by E.
Thus Ε = - = -L- = — . . . . (1.5)
x/l αχ
For shear stress, this constant is known as the Modulus of Rigidity and
is denoted by G.
Thus 0 * ...
. (1.6)
x/l ax
Instead of basing this factor on the stress at failure, it is sometimes based
on the stress at the yield point (where the material suddenly becomes
plastic) or, for materials which have no well-defined yield point, on the
stress at which the extension is a certain percentage (e.g. 0-1 per cent) of
the original length.
1.5 Factor of safety. The maximum stress used in the design of a
machine or structure is considerably less than the ultimate stress (i.e. the
stress at failure), to allow for .possible overloading, non-uniformity of stress
distribution, shock loading, faults in material and workmanship, corrosion,
1.1. STRENGTH OF MATERIAL
1.8 Stress in thin rotating rims. Let a thin rim (one in which the
radial depth is small in comparison with the mean radius) of cross-sectional
area a, mean radius r and density m rotate at a tangential speed v
y
Fig.
1.10. Then the centrifugal force, F, on an element subtending an angle
do is
v
2
m Χ a X r άθ —
r
This is resisted by the radial components of the forces on the ends of the
element. If the stress induced is σ, then
dö
mav
2
dd
= 2 χ σα χ —
from which
mir
(1.13)
α
FIG. 1.1 0
θ
2
α
α
FIG. 1.1 1
α
Alternatively, the centre of gravity of the half of the rim above the
2/*
diametral plane XX , Fig. 1.11, is at a distance — from XX . The centrifugal
71
force on this part is then resisted by the stress a acting on the section XX,
i.e.
from which
2r
«iXaXjirxcü 2 X - = 2σα
π
2 r 2 .
mir
1.9 Stresse s in composite bars. A composite bar is a load-resisting
member which is made up of two different materials.
Let the cross-sectional areas of the two materials be a
x
and a
2
, the
moduli of elasticity be E
1
and E
2
and the coefficients of expansion be a
x
and a
2
.
Stresse s due to external load. If the ends through which the load is
applied are rigid, Fig. 1.12, the change in length of each part is the same,
i.e. χ
= x
2
i.e.
(1.14) SIMPLE STRESS AND STRAIN 7
FIG. 1.1 2 FIG. 1.1 3
Stresse s due to change in temperature. Let XX , Fig. 1.13, be the
initial level of the top of the composite bar and let Y Y be its final level after
a temperature rise t. If both parts were free to expand, the extension of
material (1 ) would be l^t and that of material (2) would be l
2
ac
2
t; if, how
ever, the two materials are rigidly connected at the top, material (1) is
forced to extend a distance x
x
and material (2) is forced to compress a
distance x
2
. It is only these forced changes in length, x
x
and x
2
, which pro
duce stresses in the materials.
From Fig. 1.13, it will be seen that
+
χ
2
=
h&é ~~ h0 *-^
ti1 Ά
2
(1.16)
Also, since no external force is applied to the bar,
tensile force in material (1 ) = compressive force in material (2),
i.e
Λ
ji* ! —
(T2
a
2 ...
.
and a
2
can then be obtained from equations (1.16) and (1.17)
If the bar is subjected to an external load P, Fig. 1.14,
as well as to a temperature rise t, then, from the
equilibrium of the end plate,
P + P
1
= P
i
or σ
1
1
— σ
2
2
= Ρ . . . (1.18)
assuming Ρ to be compressive and a
2
> cc
v
FIG. 1.14
(1.17)
Also the sum of the loads carried by each part is equal to the applied load,
i.e. P
1
+ P
2
= Ρ
or σ
+ σ
2
2
= Ρ . . . (1.15)
and σ
2
can then be obtained from equations (1.14) and (1.15). 8
STRENGTH OF MATERIALS
Equation (1.16) will still be applicable and hence
σ
1
and
σ
2
may be calcu
lated. In many instances, however, it may be simpler to determine the
stresses due to the external load and temperature change separately and
then combine these values to obtain the resultant stresses, particular care
being taken over the nature of the separate stresses.
1.10 Strain energy. When a body is stressed, it distorts and work is
done on it. This energy is stored in the material and is recoverable when the
stress is relieved, provided that the material remains elastic. The energy is
termed strain energy or resilience and that stored when the material is
stressed to the elastic limit is termed the proof resilience.
Gradually applied load. If an axial load Ρ is gradually applied to a bar
and produces an extension x, then the work done, or strain energy, is
represented by the area under the load/extension diagram, Fig. 1.15,
U=\Px
But
= a X a and χ = — σΐ
Ε
= — X volume
2E
(1.19)
FIG. 1.1 5
The expression — χ volume represents the strain energy in the material
when the stress is a and however this stress is caused, the strain energy will
always be given by the same expression.
Suddenly applied load. Let a mass M be dropped from a height h on
to a collar at the lower end of a bar, Fig. 1.16, producing an instantaneous
extension χ and an instantaneous stress σ. Then the loss of potential
energy of the weight is Mg(h + x)
t
so that, ignoring any loss of energy at
impact,
T2
Mg(h χ) = — χ volume
i.e.
Mg
2E X volume
(1.20)
This is a quadratic from which σ can be found; the positive
solution will represent the tensile stress at the point of
maximum extension and the negative solution will repre
sent the compressive stress at the end of the rebound.
The oscillation of the weight will die away due to internal
IM
Fia. 1.16 SIMPLE STRESS AND STRAIN
9
friction in the material and its final position will be the same as when
gradually applied.
ol o
2
When h = 0, Μα χ — == — χ volume
from which
Ε
σ=2
2Ε
Mg
a
i.e. the maximum stress is twice that due to a gradually applied load of the
same magnitude.
1.11 Shear strain energy. Referring to Fig. 1.3, the work done by the
shearing force Ρ is \Px, assuming it to be gradually applied.
rl
G
But
= τ
X
a and χ •
.*. U:
2G
2G
X al
X volume
(1.21)
1. A tension specimen of circular cross-section tapers uniformly from 20 mm
to 16 mm diameter over a gauge length of 200 mm. When an axial load of
40 JcN is applied, the extension measured on this gauge length is 0-4 mm. Find
the modulus of elasticity of the material.
Compare the strain energy in this specimen with that in a specimen of the
same material of uniform diameter 18 mm carrying the same load. (U. Lond.)
20mm D
Referring to Fig. 1.17,
0-016
L
t =
χ
Stress on element =
FIG. 1.1 7
0-02 - 0-016
0-2
0-016
40 χ 10
3
- X (0-02x)
2
4
/. L = 0-8 m
/. d = 0-02z m
127-4 χ 10·
N/m
2
extension of element = — ol =
Ε
127-4
χ
10
6
ax
"Ζ
Χ
~
ΖΓ
"
2
Ε 10 STRENGTH OF MATERIALS
1274 χ 10
6
f
1
'
0
çb _ 31-85 χ 10
6
J
os x
2
Ε
= 0-000 4 m
=
31-85 χ 10*
=
79-6GN/m*
0-000 4
7
—
For a bar of uniform diameter 0-018 m, extension under a load of 40 kN
=
«
=
40 χ 103 χ 0-2
=Q-000395 m
- χ 0-018
2
x 79-6 χ 10
9
4
.'. ratio of strain energies = ratio of extensions for the same load
° -
m 4
=1-012
0-000 395
nrE
3. A straight steel bar of uniform cross-section, 1 m long, rotates at 2 500
revJ min about an axis at mid-length perpendicular to the length of tine bar. Find
the maximum stress and total extension if Ε = 200 GN/m
2
and steel den
sity = 7-8 Mg/m
z
. (U. Lond.)
Tensile force on element
= centrifugal force on part of bar to right of element, Fig. 1.19
= Mœ
2
r
.". total extension of bar =
2. A steel ball, radius r, has equal and parallel flats machined on opposite
sides so that the thickness across the faces is l
#
6r. Calculate the decrease in
thickness when an axial load W is applied to these faces.
Keferring to Fig. 1.18, ι \y
area of element
stress on element =
decrease in thickness of element
total decrease in thickness SIMPLE STRESS AND STRAIN 11
= 7-8 χ ΙΟ
3
χ
a χ (0-5 — sc) X (2 500 χ χ ^_!1^ N
\ 60/ 2
where a is the cross-sectional area
= 267·5α(0·25 - χ
2
) ΜΝ
The maximum stress is at the axis of rotation (i.e. where χ = 0)
267·5
χ
0-25
β
OÛ
_ _ .
9
i.e. maximum stress = = 66-8
8 MN/m
2
1m
-
*
Extension of element =
FIG. 1.1 9
PI
ÔÈ
_ 267-5 χ 10
6
a (0-25 - χ
2
) ax
X 200 χ 10
9
= 1-338 X 10-
3
(0-25 - x
2
) dx
r
0-5
total extension = 2 χ 1-338 X 10~
3
(0-25 — x
2
) dx
J ο
= 0-229 Χ ΙΟ"
3
m or 0-229 mm
4. Explain the meaning of the term 'efficiency of a riveted joint
9
.
A cylindrical vessel having a diameter of 2 m is subjected to an internal
pressure of 1-25 MN/m
2
. The vessel is made of steel plates 15 mm thick which
have an ultimate tensile strength of 450 MN/m
2
. If the efficiencies of the
longitudinal and circumferential joints are 80 and 60 per cent respectively
t
what is the factor of safety? (U. Lond.)
m
, „ . „ . , , . .
L
.
xl
^ strength of riveted joint
The efficiency of a riveted joint is the ratio ——
J
— ·
strength oi undrilled plate
pd
1-25 χ 2
From equation (1.8), a
c
=
pd
From equation (1.10) a
%
= -f—
=
104 MN/m
2
2 χ 0-015 χ 0-80
1
pd
4tyc
1-2 5 X 2
= 69-5 MN/m
2
450
Therefore factor of safety = = 4-32
4 χ 0-015 X 0-60
450
Ϊ04 12
STRENGTH OF MATERIALS
5. Derive an expression for the tensile strength of a thin rotating ring.
Find the greatest speed in rev/min for a rotor so that the stress due to rotation
does not exceed 120 MN/m
2
. The rotor may be treated as equivalent to a ring
oflm mean diameter, and the material has a density of 7-S Mg/m
3
.
(U. Lond.)
From equation (1.13), a
i.e. 120 χ ΙΟ
6
.'. ν
Ν
124-2 60
ΛΟ
_
= Χ — = 2 370 rev/mm
0-5 2π
6. A brass rod 6 mm diameter and 1 m long is joined at one end to a rod of
steel 6 mm diameter and 1-3 m long. The compound rod is placed in a vertical
position with the steel rod at the top and connected top and bottom to rigid
fixings in such a way that it is carrying a tensile load of 3-5 kN.
An attachment is fixed at the junction of the two rods and to this a vertical
axial load of 1-3 kN is applied downwards. Calculate the stresses in the steel
and brass.
The temperature is then raised 30 deg C. What are the final stresses in the
steel and brass?
E
s
= 200 GN/m
2
;
a
8
= 12 χ 10-
6
/deg C;
E
h
= 85 GN/m
2
;
a
b
= 19 X 10-
6
/deg C. (U. Lond.)
Let the forces in the steel and brass be P
s
and
P
h
respectively, Fig. 1.20. Then, equating upward
and downward forces at the junction,
Ps=Ph
+ 1 300 . . (1)
The tension in the steel increases by (P
s
—
3 500)
Ν
and that in the brass decreases by
(3 500 - P
h
) N. Since the overall length of the
rod remains unaltered, the increase in length of
the steel is equal to the decrease in length of
the brass,
— mv
2
= 7-8 χ 10 V
= 124-2 m/s
_ ν 60
~ r 2π
Steel
π
1-3 kN
Brass-
77777,
FIG. 1.2 0
1-3m
1m
i.e.
(P
s
- 3 500) χ Z
8
_ (3 500 - P
h
) χ l
b
(2) SIMPLE STRESS AND STRAIN
13
Substituting for P
s
from equation (1), equation (2) becomes
1
85
P
b
= 3 040 Ν and P
s
= 4 340 Ν
3 040
(P
b
- 2 200) X
from which
20 — = 0 (3 500
π
N/m
2
= 107-5 MN/m
2
and
0-006
2
4 340
- χ 0-006
2
4
N/m
2
= 153-5 MN/m
2
Let the compressive stress in the rod due to the increase in temperature
be a (which will be the same in the steel and brass since they both have the
same cross-sectional area). The reduction in overall length due to a must
be the same as the free expansion due to the temperature increase, since the
ends remain fixed,
σϊ
8
ol
h
i.e
i.e
Eh
1-3
(1-3x12x10-· + 1X1 9 ΙΟ"
6
) 30
-m
50mm
30mm
\200xl0
9
* 85xl0
9
y
from which a = 56-9 MN/m
2
resultant stress in steel = 153-5 — 56-9 '= 96-6 MN/m
2
and the resultant stress in brass = 107-5 — 56-9 = 50-6 MN/m
2
7. Fig. 1.21 shows a round steel rod supported in a recess and surrounded
by a coaxial brass tube. The upper |
end of the rod is 0-1 mm below that 0-1 mm
of the tube and an axial load is
applied to a rigid plate resting on the
top of the tube.
(a) Determine the magnitude of the
maximum permissible load if the
compressive stress in the rod is not to 30 0 mm
exceed 110 MN/m
2
and that in the
tube is not to exceed 80 MN/m
2
.
(b) Find the amount by which the
tube will be shortened by the load if ,
the compressive stress in the tube is
1
the same as that in the rod. 100 mm
£
8teel
= 200GN/m
2
;
J
^braes = 100 GN/m
2
. (U. Lond.) F i g . 1.21
45mrrJ
Steel
-Brass 14
STRENGTH OF MATERIALS
The sum of the loads carried by the brass and steel is equal to the total
load,
i.e. Ph + P
s
= P
i.e. <j
b
Χ
-(0-05
2
- 0-045
2
) + σ
3
X - χ 0-03
2
= Ρ
4 4
i.e. 0-000 373 o
h
+ 0-000 707 a
s
= Ρ (1)
After the plate has made contact with the top of the rod, the compression
of the brass exceeds that of the rod by 0-000 1 m,
i.e. x
h
= x
s
-I- 0-000 1
a
h
Χ
0-3 _ σ
8
X 0-4*
100 χ 10
9
200 χ 10
9
or a
h
= 0-667 σ
3
+ 33-3 Χ 10
6
. . . (2)
The maximum stresses of 110 MN/m
2
and 80 MN/m
2
in the steel and
brass respectively will not occur simultaneously. Equation (2) shows that
if a
h
= 80 MN/m
2
, σ
8
= 70 MN/m
2
and if σ
= 110 MN/m
2
, a
h
= 106-7
MN/m
2
. Hence the maximum permissible load is that which will produce
stresses of 80 MN/m
2
and 70 MN/m
2
in the brass and steel respectively.
Therefore, from equation (1),
= 0-000 373 χ 80 χ 10
6
+ 0-000 707 χ 70 χ ΙΟ
6
Ν
= 79-33 kN
AVhen a
s
= cr
b
, equation (2) becomes
a
b
= 0-667 a
h
+ 33-3 χ 10
6
N/m
2
from which a
h
= 100 MN/m
2
_ 100 χ ΙΟ
6
X 0-3 _ q.qqq ^
or 0
.
3 m m
b
100 χ 10
9
8. A mass of 150 kg is suspended by three vertical wires. The two outer wires
are of steel and the middle one of aluminium, each of area 8 mm
2
. The lengths
are adjusted so that each wire carries an equal share of the load. If the tempera
ture is raised by 50 deg C, find the stresses in the wires.
Find also what rise of temperature would just cause the aluminium wire to
become slack.
Steel Aluminium
E(GN/m
2
) 210 70
Coefficient of expansion per deg C 12 X 10~
6
24 X 10~
6
(U. Lond.)
Initial stress in steel = initial stress in aluminium
50
ν
Q.ftl
=
—^— — N/m
2
= 61-3 MN/m
2
(tensile)
8 χ 10-
6
1
/ ν
/
* Allowance for the 0Ό0 0 1 m in the original length of the rod will have a negligible
effect on the answers. SIMPLE STRESS AND STRAIN
15
Let σ
8
and σ
&
be the stresses in the steel and aluminium respectively due
to the temperature rise only. Then
: σ
. from equation (1.17)
. . . . (1)
i.e.
Also
σΛ
+
2a
&
a
s
σΛ
210 χ ΙΟ
9
7
0
χ ΙΟ
9
I (24 - 12)
χ
ΙΟ-
6
χ 50
from equation (1.16)
126 MN/m
2
£4-0 MN/m
2
(compressive)
MN/m
2
(tensile)
N/m
2
- 12 MN/m
2
49 χ - χ 0-001 6
2
4
. stress due to falling load = 90— 12 = 78 MN/m
2
If M is the mass of the falling load, then, from equation (1.20),
78χ10β^5
\
=
(78Χ10
, 0-001 6
2
χ
25
75xl0
9
/ 2x75xl0
9
4
9-81M( 0-1 +
from which
ί^0·1
M= 80-8 kg
from which
Substituting for σ
8
from equation (1),
and
resultant stress in aluminium =
and resultant stress in steel
When aluminium wire becomes slack,
σ
&
= 61-3 MN/m
2
(compressive).
Therefore, since o*
a
is proportional to the change in temperature,
temperature rise =
9. It is estimated that the loads to be carried by a lift may be dropped through
a distance of 100 mm on to the floor. The cage itself has a mass of 100 kg and it
is supported by 25 m of wire rope of mass 0-8 kg/m, consisting of 49 wires
each 1-6 mm diameter. The maximum stress in the wire is not to exceed 90
MN/m
2
and Ε for the rope may be taken as 75 GN/m
2
. Find the maximum
safe load that can be carried, neglecting loss of energy at impact. (U. Lond.)
Stress in wire due to dead weight of lift and rope 16
STRENGTH OF MATERIALS
10. A solid steel bar, 200 mm long, is 10 mm square over part of its length.
The remainder is circular, and alone has a volume of li: χ 10 ~ m
3
. The bar
is subjected to a pull of 20 kN. Find the dimensions of the circular portion so
that the total strain energy may be a minimum, and find its value. Ε = 200
GN/m
2
. (U. Lond.)
Let suffices 1 and 2 refer to the square and circular portions respectively,
Fig. 1.22.
Then
and
a
x
=
Λ
_ N/m
2
= 200 MN/m
2
2
=
0-01
2
20 χ 10
3
N/m
2
d
2
0-08 MN/m
2
10 mm sq]
X
200m 200mm m
FIG. 1.2 2
I
u
=ïè
x r
-
+
â *
r
<
(
0-0
8
*
10
6
\
2
]
X 24 X 10-
nd
2
6
y
X 24
χ
10-
e
j
But
π
-d
2
(0-2-x) = 24xl0-
e
4
nd2 . 96xl0-
6
0-2-x
1_
2Ë
1 0
1 2
Ï2Ê
2
J
2
x24xlO-« |
4sxl0
1 2
+ | ^(0·2-ζ)χ10
1 2
| X24xl0
-
{lOOx
2
- 16x + 4}
άϋ
For the strain energy to be a minimum, — = 0,
ax
i.e.
or
200x - 16 = 0
= 0-08 m
length of circular portion = 0-12 m SIMPLE STRESS AND STRAIN
17
0-2-0-12
.·. d = 0-025 5 ra or 25-5 mm
ΙΟ
12
U = ———————{100χ0·08
2
— 16 0Ό8 -j- 4}
12χ200χ10
1
'
= 1-4 J
11. The maximum safe compressive stress in a hardened steel punch is limited to
1 GN/m
2
and the punch is used to pierce circular holes in mild steel plate 20 mm
thick.
(a) If the ultimate shearing stress of the plate is 300 MN/m
2
, calculate the
smallest diameter of hole that can be pierced.
(6) If the effective length of the punch is 75 mm, calculate the maximum strain
energy stored in the punch during the piercing operation. Assume the modulus of
elasticity for the material of the punch to be 200 GN/m
2
.
( Ans. : 24 mm ; 84·8 J )
d2
FIG. 1.2 3
12. Fig. 1.23 shows a knuckle joint in a tie bar. Allowing stresses of 105, 75
and 150 MN/m
2
for tension, shearing and bearing respectively, obtain suitable
dimensions for Z), dv t and d2 if the load on the rod is 125 kN.
(Ans.: 39 mm; 32-5 mm; 12-8 mm; 79 mm)
13. A steel bar 40 mm diameter and 4 m long is raised in temperature through
60 deg C, after which its ends are firmly secured. After cooling to normal tem
perature again, the length of the bar is found to be 1-2 mm less than when at its
highest temperature. Determine the total pull exerted by the cold bar and the
intensity of stress in it. Ε = 200 GN/m
2
and α = 000 0 011/deg C.
(Ana.: 90-5 kN; 72 MN/m
2
)
14. A mild steel rod, 600 mm long, is 25 mm in diameter for 150 mm of its
length and 50 mm for the rest of its length. It carries an axial tensile pull of
18 kN. With the axial pull applied, the ends of the rod are secured by rigid fixings.
Find the temperature through which the rod must be raised to reduce the axial
pull by two thirds. a8 te e l = 11 X 10"
e
/deg C;
Ε
steel = 200 GN/m
2
. (U.
Lond.)
(4ws.;4/degC )
15. A uniform rectangular slab of concrete, 2-5 m by 1-5 m, of mass 1 200 kg
rests on vertical columns a t the four corners. One of these columns may be
regarded as rigid and the others as three identical columns of the same length,
cross-section and elasticity. Assuming that the slab may be treated as rigid and 18
STRENGTH OF MATERIALS
that it remains sensibly horizontal when the elastic columns are slightly coin
pressed under load, determine the reaction at each support. (U. Lond.)
(Ans.: 3 924 Ν; 3 924 Ν; 1 962 Ν; 1 962 Ν)
16. A flat steel bar, 10 m long and 10 mm thick tapers from 60 mm at one end
to 20 mm at the other. Determine the change in length of the bar when a tensile
force Ρ = 12 kN is acting along its axis. Ε = 200 GN/m
2
. (U.
Lond.)
(Ans.: 1-648 mm)
17. A boiler shell, 2 m mean diameter, is constructed of steel plate having an
ultimate tensile strength of 450 MN/m
2
. If the thickness of the shell plates is
20 mm, calculate the maximum internal gauge pressure to which the boiler may
be subjected, assuming a factor of safety of 6 and a longitudinal joint efficiency of
80 per cent. (Ans.: 1-2 MN/m
2
)
18. Derive formulae to give the longitudinal and circumferential tensions in a
thin boiler shell, stating the assumptions made in your argument.
A cylindrical compressed air drum is 2 m in diameter with plates 12-5 mm thick.
The efficiencies of the longitudinal and circumferential joints are respectively 85
and 45%. If the tensile stress in the plating is to be limited to 100 MN/m
2
, find
the maximum safe air pressure, (U. Lond.) (Ans.: 1-063 MN/m
2
)
19. Derive an expression for the tensile stress in a thin spherical shell of thick
ness t and internal diameter d when subjected to an internal pressure p.
A thin spherical pressure vessel is required to contain 18 000 1 of water at a
gauge pressure of 700 kN/m
2
. Assuming the efficiency of all riveted joints to be
75 per cent, determine the diameter of the vessel and the thickness of the plate.
The stress in the material must not exceed 140 MN/m
2
. (Ans. 3·248 m ; 5-41 mm)
20. Derive a formula for the hoop stress in a thin cylinder having a mean radius
R and made of material of density m when rotating at ω rad/s about its axis.
What is the most important assumption you make?
Apply this theory to find the maximum allowable speed in rev/min for a fly
wheel 1 -25 m external diameter and 50 mm thick. The material has a density of
7-3 Mg/m
3
and the hoop stress is limited to 20 MN/m
2
. (U.
Lond.)
(Ans.: 834 rev/min)
21. Deduce an expression for the centrifugal stress induced in a thin rotating
rim.
A thin rim 1-5 m mean diameter rotates at 600 rev/min. The cross-section of
the rim is rectangular, 125 mm χ 12-5 mm thick, and the density of the material
of which it is constructed is 7-8 Mg/m
3
. Calculate the stress and force produced
in the rim. (Ans.: 17-34 MN/m
2
; 27-15 kN)
22. A thin rim, 1-5 m diameter and 150 mm wide, is made of steel plate 12-5 mm
thick. The rim is made in two halves with the joints parallel to the axis. Determine
the centrifugal stress produced in the material at a speed of 420 rev/min. Density
of steel = 7-8 Mg/m
3
.
The two halves are fastened by three bolts on each side of each joint. The bolts
are 20 mm diameter and are in single shear. Determine the shear stress in each
bolt when the rim is rotating at 420 rev/min. (Ans.: 8-5 MN/m
2
; 16-9 MN/m
2
)
23. State Hooke's Law. Comment briefly on its limitations.
A straight rod of steel, 1 m long, of constant section, rotates at 1 200 rev/min
about an axis at one end, perpendicular to its length. Calculate (a) the maximum
stress in the rod, (b) the extension. Density of steel = 7-8 Mg/m
3
;
Ε
= 200 GN/m
2
.
(U. Lond.) (Ans.: 61-65 MN/m
2
; 0-205 5 mm)
24. Two elastic rods, A and B, of equal free length hang vertically 0-6 m apart
and support a rigid bar horizontally. The bar remains horizontal when a vertical
load of 60 kN is applied to the bar 0-2 m from A. If the stress in A is 100 MN/m
2
, SIMPLE STRESS AND STRAIN
19
find the stress in Β and the cross-sectional areas of the two rods ; Ε A = 200 GN/m
2
;
Β
= 130 GN/m
2
.
(I.CE.) (Ans.: 65 MN/m
2
; 400 mm
2
; 307-5 mm
2
)
25. A wire strand consists of a steel wire 2-7 mm diameter, covered by six bronze
wires each of 2-5 mm diameter. The tensile modulus for the steel is 200 GN/m
2
and for the bronze 85 GN/m
2
.
If the working stress for the bronze is 60 MN/m
2
, calculate the strength of the
strand, also the equivalent tensile modulus for the complete strand. (U. Lond.)
(Ans.: 2 575 Ν; 104 GN/m
2
)
26. A round steel bar, 28 mm diameter and 400 mm long, is placed concentric
ally within a brass tube which has an outside diameter of 40 mm and an inside
diameter of 30 mm; the length of the tube exceeds that of the bar by 0-12 mm.
Rigid plates are placed on the ends of the tube through which an axial compressive
force is applied to the compound bar. Determine the compressive stresses in the
bar and tube due to a force of 60 kN. Esteei = 200 GN/m
2
; ^brasa =
100 GN/m
2
.
(U. Lond.) (Ans.: 48-2 MN/m
2
; 54-1 MN/m
2
)
27. A steel bolt, 20 mm external diameter, is inserted into a copper sleeve,
21 mm internal and 27 mm external diameter, one end of the tube being in contact
with the shoulder of the bolt-head. A rigid washer is placed on the other end of
the sleeve and a nut is screwed on the bolt until the compressive stress in the
sleeve is 80 MN/m
2
. It may be assumed that the washer slides freely on the end
of the sleeve and consequently that torsional stresses are negligible. Find the range
of external axial load that can be applied to the assembly if the stress in the sleeve
is never to be zero and that in the bolt never to be compressive. Osteel = 200 GN/m
2
;
^copper =
90
GN/m
2
.
(17. Lond.)
(Ans.: 80 kN (tensile) to 19-5 kN (compressive))
28. A steel rod of 320 mm
2
cross-sectional area and a coaxial copper tube of
8 00 mm
2
cross-sectional area, are rigidly bonded together at their ends. An axial
compressive load of 40 kN is applied to the composite bar, and the temperature
is then raised by 100 deg C.
Determine the stresses then existing in both steel and copper. The moduli of
elasticity for steel and copper at 200 GN/m
2
and 100 GN/m
2
, and the coefficients
of linear expansion 12 χ 10
_ e
/de g C and 16 χ 10~
e
/deg C respectively.
(I.O.E.)
(Ans.: 11-11 MN/m
2
; 45-55 MN/m
2
, both compressive)
29. A copper tube of mean diameter 120 mm, and 6-5 mm thick, has its open
ends sealed by two rigid plates connected by two steel bolts of 2 5 mm diameter,
initially tensioned to 20 kN at a temperature of 30°C, thus forming a pressure
vessel. Determine the stresses in the copper and steel at freezing point, and the
temperature at which the vessel would cease to be pressure tight.
Osteel = 200 GN/m
2
; as teel
= 11 X 10"
e
/deg C.
^copper = 100 GN/m
2
; aC Opper =
18 χ 10"
e
/deg C.
(Ü. Lond.)
(Ans.: 7-0 MN/m
2
; 17-4 MN/m
2
; -22-4°C )
30. A bar of brass 25 mm diameter is enclosed in a steel tube 50 mm external
diameter and 25 mm internal diameter. The bar and tube are both initially 1 m
long and are rigidly fastened together at both ends. Find the stresses in the two
materials when the temperature rises from 15°C to 95°C.
If the composite bar is then subjected to an axial tensile load of 50 kN, find
the resulting stresses and the increase in length from the initial state.
Osteel = 200 GN/m
2
; a stee i =
11-6 X 10"
e
/deg C.
^brass = 100 GN/m
2
; a bras s =
18-7 X 10~
e
/deg C. (U.
Lond.)
(Ans.: 48-7 MN/m
2
(comp); 16-23 MN/m
2
(tensile); 34-17 MN/m
2
(comp);
45-3 MN/m
2
(tensile); 1-115 mm) 20
STRENGTH OF MATERIALS
31. A steel tie-rod 25 mm diameter is placed concentrically in a brass tube
3 mm thick and 60 mm mean diameter. Nuts and washers are fitted on the tie-rod
so that the ends of the tube are enclosed by the washers. The nuts are initially
tightened to give a compressive stress of 30 MN/m
2
in the tube, and a tensile
load of 45 kN is then applied to the tie-rod. Assuming the rod and tube to have
the same effective length, find the resultant stresses in the tie-rod and tube;
(i) when there is no change of temperature ; (ii) when the temperature increases by
60 deg C.
Osteel = 200 GN/m
2
; as teel = M
X 10"
5
/deg C.
^brass = 80 GN/m
2
; a b rass =
1-89 x 10"
5
/deg C.
(U. Land.)
(Ans.: 97-2 MN/m
2
(tensile); 4-9 MN/m
2
(comp); 127-1 MN/m
2
(tensile);
30-9 MN/m
2
(comp))
32. A wagon, of mass 51 , is attached to a steel wire rope, the other end of which
is wound round a brake drum. The wagon is descending a slope at a uniform speed
of 5 km/h and the rope is taut. By the action of the brake the drum is suddenly
stopped and the wagon is brought to rest. If the length of the rope between the
wagon and the drum at that instant was 250 m and its cross-sectional area is
300 mm
2
, find the maximum stress in the rope.
Ε
= 200 GN/m
2
.
(U. Lond.)
(Ans.: 160 MN/m
2
)
33. A bar of certain material, 40 mm diameter and 1-2 m long, has a collar
securely fitted to one end. It is suspended vertically with the collar at the lower
end and a mass of 2 000 kg is gradually lowered on to the collar, producing an
extension in the bar of 0-25 mm. Find the height from which this load could be
dropped on to the collar if the maximum tensile stress in the bar is to be 100 MN/m
2
.
(U. Lond.) (Ans.: 3-58 mm)
34. A bar 10 mm diameter and 3 m long is fixed at the top and hangs vertically.
At the lower end is fixed a disc and on to this disc a mass of 10 kg is allowed to fall
freely through a distance of 50 mm. Calculate the maximum stress induced in the
rod, stating any assumptions made and proving any formula used. Ε = 200 GN/m
2
.
(U. Lond.) (Ans.: 92-5 MN/m
2
)
35. A steel wire 2-5 mm diameter is firmly held in a clamp from which it hangs
vertically. An anvil, the weight of which may be neglected, is secured to the wire
1 m below the clamp. The wire is to be tested by allowing a mass bored to slide
over the wire to drop freely from 0-5 m above the anvil. Calculate the mass
required to stress the wire to 1-2 GN/m
2
assuming the wire to be elastic to this
stress. Ε = 200 GN/m
2
.
(U. Lond.) (Ans.: 3-56 kg)
36. Define Resilience. Show how to calculate the resilience in a rod of uniform
section under a uniform direct stress.
A vertical tie, rigidly fixed at the top end, consists of a steel rod 2-5 m long and
20 mm diameter encased throughout in a brass tube 20 mm internal diameter and
25 mm external diameter. The rod and casing are fixed together at both ends.
The compound rod is suddenly loaded in tension by a mass of 1 Mg falling freely
through 3 mm before being arrested by the tie. Calculate the maximum stresses in
the steel and brass. E&tee\ = 200 GN/m
2
; #brass =
100 GN/m
2
.
(U. Land.)
(Ans.: 135 MN/m
2
; 67-5 MN/m
2
)
1.2. SIMPLE STRESS AND STRAIN
Longitudinal stress. The force tending to separate the right- and left
hand halves is the pressure multiplied by the area of one end, Fig. 1.8,
i.e. Pi = px-d
2
4
This is resisted by the stress acting on the circumferential section, YY,
. . . (1.9)
i.e,
σι
=
4*
FIG. 1.8
If the cylinder is made up from riveted plates and the efficiency of the
circumferential joints is r\
z
then the average stress in the joint is given by
(1.10)
It is evident from equations (1.8) and (1.10) that the efficiency of the
circumferential joints need only be half that of the longitudinal joints.
1.7 Stress in thin spherical shells. Let the internal diameter be d,
the thickness of metal be t and the internal pressure be p, Fig. 1.9. Then the
force tending to separate the two halves on a section XX is the pressure
multiplied by the projected area in the direction perpendicular to XX,
i.e. Ρ = ρχΐ ί
4
This is resisted by the stress acting on the section XX,
i.e.
π
<
ndt
pd
Tt
. (i-ii)
If the shell is made up from riveted plates and the efficiency of the joints
is η, then
pd
a —
. (1.12)
1.3. STRENGTH OF MATERIAL
1.6 Stresse s in thin cylindrical shells. When a thin cylinder is sub
jected to internal pressure, stresses are induced on the longitudinal section
X X , Fig. 1.5, due to the force tending to separate the top and bottom
halves, and on the circumferential section Y Y due to the force tending to
separate the right- and left-hand ends of the cylinder.
The stress on the longitudinal section is termed the circumferential stress
and that on the circumferential section is termed the longitudinal stress;
the type of stress is determined by the direction of the arrows.
In determining the stresses induced, it is assumed that the thickness is
small in comparison with the diameter so that the stress on a cross-section
may be taken as uniform* and also that the ends give no support to the
sides, an assumption which would be appropriate to a long cylinder such
as a pipe.
Let the internal diameter and length be d and I respectively, the thickness
of metal be t and the internal pressure be p.
Για
. 1.5 FIG. 1.6
Circumferential stress. The force tending to separate the top and
bottom halves is the pressure multiplied by the projected area in a direction
perpendicular to the diametral plane,f Fig. 1.6,
i.e. P
c
= pdl
This is resisted by the stress acting on the longitudinal section, XX ,
i.e.
_ pdl pd
°
=
2 α
=
2 ί
(1.7)
If the cylinder is made up from riveted plates and the efficiency of the
longitudinal joints is η
then the average stress in the joint is given by
pd
* See Chapter 14.
t The radial force on an element subtending an angle
άθ, Fig. 1.7, is ρ χ | άθ χ I The vertical component of
this force ie ^ dö.sin θ so that the total force normal to
(1.8)
de,
FIG. 1.7
X X is
J o
2
sin θ άθ = pdl