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1a. Torsion
CHAPTER OBJECTIVES
Shafts of circular section are generally used for power transmission from prime
mover (such as engine, electric motor) to operating machines such as lathe, shaper
or any manufacturing machine. In this chapter students are made to learn about the
development of shear stress and angular twist in a shaft or a combination of shafts
due to twisting moment applied on them. Then, the shafts are connected to gears
and pulleys with the help of keys and how stresses are developed in the keys.
Moreover, students will learn about the variation of stress along the radius and
angular twist along the length of the shafts.
Introduction
Shafts are generally made of ductile materials like mild steel. Due to the
application of a twisting moment, the development of stresses, strains and angular
twist will be discussed with in this chapter. Single shaft and a combination of
shafts as ‘shafts in series’ and compound shafts will also be discussed here. How
much horse-power is transmitted by a shaft transmitting torque at a particular
speed is the important criterion in the design of shafts.
Stress concentration due to key ways in shaft for providing key so that a shaft is
connected to a hub of machine component as gears/pulleys will also be discussed.
Development of Shear Stress and Angular Twist in a Shaft Due to Twisting
Moment
A relationship between the twisting moment, shear stress, angular twist, length of
the shaft and polar moment of inertia of shaft will be derived. For that the
following assumptions are taken into consideration:
1. Material is homogeneous and isotropic.
2. Shaft is not initially distorted.
3. Displacement at a point in the shaft is proportional to its distance from
centre of the shaft, that is, shear strain at any point is proportional to its
distance from the centre of the shaft.
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Figure 12.1 (a) Shaft subjected to twisting moment (b) variation of shear
strain and (c) variation of angular twist
4. Cross-sections perpendicular to the axis of the shaft which are plane
before the shaft is subjected to twisting moment remain plane after the
application of twisting moment.
5. Angular twist is uniformly distributed along the length of the shaft.
6. Shear stress developed in the shaft is within the proportional limit, that is,
shear stress α is shear strain.
Figure 12.1(a) shows a shaft of radius R, length L fixed at one end and a twisting
moment T applied at the other end. Say, a clockwise twisting moment T is applied
at A and a resisting moment T (ccw) develops at the other end. Axis of the shaft is
O′O. If a line CA, parallel to the axes O′O is drawn on the shaft before the
application of the twisting moment, it is twisted to the position CA′, making
∠ACA′ = ø (shear angle). Since the angle ø is extremely small, shear angle ø is
also known as shear strain ø.
At radius R (outer radius of shaft),
AA′ = Rθ, where θ is angle of twist.
At any radius r, shift BB′ = rθ
Shear strain ϕr at radius
or, shear strain ø α a radius r,
Similarly,
AA′ = Rθ
Shear strain,
or maximum strain, øαR = τ /G, where τ is maximum shear stress at radius R.
Taking Eq. (12.1) again,
Shear strain,
But,
or,
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From Eq. (12.2),
or,
From this relation, we learn that the shear stress at any point is directly
proportional to its distance from centre of the shaft.
Consider an elementary area at radius r and thickness dr (Fig. 12.2)
Area = 2πrdr
Shear stress = τr
Now, τ is the maximum shear stress at radius R,
Shear force in small area,
Figure 12.2
Twisting moment of δQ about the centre of the shaft,
Total twisting moment,
Resulting twisting moment,
where πR4/2 = J, polar moment of inertia of the section of shaft
Therefore,
or,
Combining the Eq. (12.5) and Eq. (12.8), we get,
The students must learn about the significance of each term as follows:
T = twisting moment applied on the shaft
= resisting twisting moment in shaft
J = polar moment of inertia of section of shaft
=
=
(for a solid shaft)
for a hollow shaft of external radius R2 and internal radius R1
τ = maximum shear stress at radius R in the case of solid shaft and at external
radius R2 in the case of hollow shaft
G = shear modulus of the maternal
θ = angular twist in shaft over length L of the shaft
τr = shear stress at any radius r of the shaft
Figure 12.3(a) shows the shear stress distribution in a solid circular shaft
and Figure 12.3(b) show the shear stress distribution in a hollow circular shaft, the
minimum stress occurs at inner radius R1 and maximum shear stress occurs at outer
radius R2.
Example 12.1 A hollow circular steel shaft of inner radius 30 mm and outer radius
50 mm is subjected to a torque of 10 kN m. Determine (a) the maximum and
minimum shear stresses developed in shaft and (b) the angular twist over 1 m
length of shaft. Given shear modulus, G = 82 kN/mm2 for steel.
Figure 12.3 Shear stress distribution in (a) solid circular shaft and (b) hollow
circular shaft
Solution
Outer radius,
R2 = 50 mm
Inner radius,
R1 = 30 mm
Polar moment of inertia,
Torque,
T = 10 kN m = 10 × 106 Nmm
Maximum shear stress,
Minimum shear stress,
Note that shear stress is proportional to radius.
Length of shaft,
L = 1 m = 1,000 mm
Angular twist,
Exercise 12.1 A circular steel shaft of 30 mm diameter is subjected to a torque of
0.56 kN m. Determine (a) the maximum shear stress developed in shaft, (b) the
angular twist over 1m long shaft and (c) the shear stress at a point 10 mm away
from the centre. Given G for steel = 82 kN/mm2.
Modulus of Rupture
Torsion formula, T/J = Gθ/L = τ/R, has been derived on the assumption that shear
stress is proportional to shear strain but when a specimen is tested in torsion, it
fails by fracture and stress developed in this material goes beyond the proportional
limit, that is, in the plastic stage. If shear stress and shear strain is drawn along
radius of the shaft, shear strain at any point remains proportional to radius but
shear stress at any point is no longer proportional to its radius,rather the curve
becomes non-linear as shown in Fig. 12.4. A torsion test piece is continuously
twisted, that is,θ is increased gradually till the shaft breaks. Say, Tmax is the
maximum torque at which shaft fails by fracture.
Figure 12.4 Shear stress-shear strain curves
Maximum shear stress,
This maximum shear stress calculated by using torsion formula is known
as modulus of rupture. It can be noted that it is not the actual shear stress on the
surface of the shaft but it is a hypothetical stress taking straight line relation
between shear stress and shaft radius. However, it is a useful information to know
the limiting torque at which the specimen would fail in torsion.
Example 12.2 A steel sample of a gauge length of 250 mm and a diameter of 25
mm is tested under torsion and it was observed that sample fractured at the torque
of 920 N m. Determine the modulus of rupture of the material.
Solution
Maximum torque,
Tmax = 920 N m = 9,20,000 Nmm
Shaft radius,
R = 12.5 mm
Modulus of rupture,
= 300 N/mm2
Exercise 12.2 A specimen of copper of a gauge length of 200 mm and a diameter
of 25 mm is tested under torsion. It was observed that the specimen was fractured
at the torque of 640 Nm. Determine the modulus of rupture of copper.
Horse-Power Transmitted by a Shaft
A shaft is transmitting torque T at N rpm, then
Power transmitted in 1 min = 2πNT
Horse-power transmitted by shaft
if T is in N m as 1 W = 1 N m/s
So,
If torque is in kgf m
Metric
, where Torque T is in kgf m
There is a minor difference between metric HP and HP.
If HP transmitted is known, then torque,
Maximum stress developed on the surface of the shaft is τ = 2T/πR3, in case of
solid shaft of radius,R,
radius, R2 and inner radius, R1.
, in case of hollow shaft of outer
Example 12.3 A hollow circular shaft is transmitting power at 300 rpm. Outer
diameter of shaft is 80 mm and inner diameter is 50 mm. The maximum shear
stress developed in shaft section is 65 N/mm2. Determine the horse-power
transmitted by the shaft.
Solution
Outer radius,
R2
= 40 mm
Inner radius,
R1
= 25 mm
rpm
= 300
Maximum shear stress, τ
= 65 N/mm2
Torque,
HP transmitted
Exercise 12.3 The maximum shearing stress developed in 60 mm steel shaft is 60
N/mm2. If the shaft rotates at 360 rpm, find the horse-power transmitted by the
shaft.
Shafts of Varying Diameters
(a) Uniformly tapered circular shaft:
Let us consider a circular tapered shaft of diameter d2, gradually decreasing to
diameter d1 over the axial length L of the shaft. Consider a small disc of
length dx at a distance x from the end A of the shaft as shown in Fig. 12.5.
Figure 12.5 Tapered circular shaft
Diameter of the shaft at the section
Polar moment of inertia of the disc,
Angular twist over this length dx, i.e.,
Total angular twist,
Figure 12.6 Stepped shaft
(b) Stepped shaft:
Consider a shaft with different diameters along different axial lengths as shown
in Fig. 12.6. Say, there are three steps of diameter, d3, d2 and d1 with axial
length L3, L2 and L1, respectively, as shown inFig. 12.6. Polar moment inertia of
these sections are:
Torque on each part of the stepped shaft is subjected to same twisting moment,
this is known asshaft in series. Maximum shear stresses on surface in three steps
are τ1 = 16T/πd13, τ2 = 16T/πd23 andτ3 = 16T/πd33 this shows that portion AB with
minimum diameter d1 is subjected to maximum shear stress. Angular twist between
ends A and D is:
Example 12.4 A tapered circular shaft of 1 m long having a diameter of 60 mm at
one end gradually reduces to a diameter of 36 mm at the other end is subjected to a
torque such that an angular twist of 1° develops between the ends. Determine the
torque applied on the shaft.
G = 82,000 N/mm2
Solution
Length, L
d1
= 1,000 mm
= 36 mm, d2 = 60 mm
or torque,
Exercise 12.4 A circular stepped shaft of a length of 750 mm, with diameters 80
and 40 mm over AB and AC, respectively, is shown in Fig. 12.7. This shaft is
subjected to a torque T = 1.5 kN m. Determine (a) maximum shear stress
developed in shaft and (b) angular twist between C and A, if G = 80 kN/mm2.
Compound Shaft
In a compound shaft, there are two distinct portions of shaft subjected to different
values of twisting moment but angular twist in the two shafts is the same. Fig.
12.8 shows two shafts A and B, of diameter dA and dB and length LA and LB,
respectively, subjected to a torque T at the junction of the two as shown in Fig.
12.8.
Torque T is shared by two shaft A and B,
Figure 12.7
T = TA + TB,
but angular twist
θA = θB
Say GA = shear modulus of shaft A
GB = shear modulus of shaft B
JA = polar moment of inertia of the shaft A
JB = polar moment of inertia of the shaft B
Figure 12.8
or,
or,
For the case shown in Fig. 12.8,
Another example of a compound shaft is shown in Fig. 12.9. A solid circular shaft
of material A is encased in a hollow circular shaft of material B. Both the shafts are
subjected to a twisting momentT, which is shared by A and B, that is, total twisting
moment T = TA + TB, in this case LA = LB = L.
θA = θB
so
or
Figure 12.9
where GA, GB = shear modulus of materials A and B, respectively.
Example 12.5 A composite shaft is made by joining an 800 mm long solid steel
shaft with 800 mm long hollow copper shaft as shown in Fig. 12.10. The diameter
of solid shaft is 40 mm, while internal and external diameters of hollow shaft are
25 and 50 mm, respectively. Determine the maximum shear stresses developed in
steel and copper shaft, if torque T applied at junction is 4 kN m. What is the
angular twist at the junction? Given G steel = 2.G copper = 82,000 N/mm2.
Solution
θCA = θCB
Torque,
So,
Figure 12.10
T
= TA + TB
LA
= LB = 800 mm
So
But
TA + TB = 4 kN m
0.8738 TB
=
4
TB
=
2.1.5 kN m
TA
=
1.865 kN m
Shear stresses
Maximum shear stress in steel,
Maximum shear stress in copper,
Angular twist at junction
Exercise 12.5 A solid circular steel shaft is encased in a copper hollow shaft so as
to make a compound shaft. The diameter of steel shaft is 40 mm and the outside
diameter of copper shaft is
60 mm. The length of compound shaft is 2 m and is subjected to an axial torque of
3.2 kN m. Determine (a) the maximum shear stress in steel and copper shafts and
(b) the angular twist per unit length, if Gsteel = 2Gcopper = 80 kN/mm2.
Stresses in a Shaft Subjected to Twisting Moment
A shaft in transmitting power, that is, the torque, is under pure torsion. Due to the
torque T, the maximum shear stress developed on the surface of the shaft is τ =
16T/πd3, where d is the diameter of a solid shaft. Figure 12.11(a) shows shear
stress τ at the surface of the shaft and a complementary shear stress τ at right angles
(i.e., in longitudinal direction). State of stress on a small element on the surface of
the shaft is shown in Figure 12.11(a). Figure 12.11(b) shows complementary shear
stress τ at right angles, variation of shear stress along the radius of the shaft. Figure
12.11(c) shows the enlarged view of the stresses on a small element. At angle θ =
±45°, principal stresses are developed, p1 = ±τand p2 = –τ as shown (derivation of
principal stresses) in a shaft has been dealt in Chapter 3 on principal
stresses. Figure 12.11(d) shows the development of principal stresses on a small
element with direction of principal stress ± 45° to the axes of the shaft.
So, principal stresses at a point are +τ, –τ, 0.
Principal strains,
Volumetric strain, ɛv = ɛ1 + ɛ2 + ɛ3 = 0
This shows that if a shaft is subjected to a pure twisting moment there will not be
any change is its volume. However, the shafts are subjected to bending moments
also due to the forces on elements mounted on shaft as gears, pulleys and due to
reactions from the supports. Therefore, a shaft is subjected simultaneously to a
bending moment M and a twisting moment T at any section shown inFig. 12.12.
Figure 12.11
Shafts Subjected to T and M
When a shaft transmits torque T, it is simultaneously subjected to bending
moments also due to belt tensions on a pulley, tooth load on gears and support
reactions in bearings. Figure 12.12 shows a shaft subjected to twisting
moment T and a bending moment M. If we take small elements on the surface of
the shaft, stresses acting on the element will be as shown in Figure 12.13.
Figure 12.12 Twisting and bending moments on a shaft
Figure 12.13 Stresses on a small element of shaft
σ = Normal stress due to bending moment M = 32 M/πd3
τ = Shear stress due to twisting moment M = 16 T/πd3
Maximum principal stress at element
or,
or,
The bending moment due to maximum principal stress in shafts is known as
equivalent bending moment, Me.
or,
Similarly, maximum shear stress developed on the surface of the shaft,
or,
equivalent twisting moment.
The twisting moment corresponding to maximum most shear stress on the surface
of the shaft is known as equivalent twisting moment, Te.
Example 12.6 A solid shaft of diameter d is subjected to a bending moment, M =
15 kN m and a twisting moment, T = 25 kN m. What is the maximum diameter of
the shaft if the maximum most shear stress in shaft is not to exceed 160 N/mm2 and
the maximum direct stress is not to exceed 200 N/mm2.
Solution
Bending moment,
M =15 kN m
Twisting moment,
T = 25 kN m
Equilateral bending moment,
Equilateral twisting moment,
Now
Maximum diameter of the shaft is 104 mm.
Exercise 12.6 A solid shaft of a diameter of 180 mm is transmitting 700 kW at 200
rpm. It is also subjected to a bending moment of 10 kN m. Determine (a) the
equivalent bending moment, (b) the equivalent twisting moment, (c) the maximum
principal stress on the surface of shaft and (d) the maximum most shear stress.
[Hint: 1 W = 1 Nm/s]
Torsional Resilience of a Shaft
A solid shaft of radius R and length L is subjected to a twisting moment, T.
Strain energy per unit volume τr2/2G
where
τr = shear stress at any radius
G = modulus of rigidity.
Consider an elementary area at radius r and of thickness dr,
Area, dA = 2πrdr
Volume, dv = 2πrdrL
Shear stress, τr = (τ/R)r, where τ is the maximum shear stress at outer radius R as
shown in Figure 12.14.
Figure 12.14
Strain energy per unit volume =
Strain energy for an elementary volume dv,
Strain energy for the solid shaft,
Note that τ is the maximum shear stress on the surface of the shaft.
Hollow shaft (inner radius R1 and outer radius R2)
τr, at any radius r = (τ/R2)r, R2 is maximum radius.
Strain energy for an elementary volume,
Example 12.7 A 1 m long hollow circular shaft of an outer diameter of 60 mm and
an inner diameter of 20 mm is subjected to a twisting moment of 4 kN m.
Determine the strain energy absorbed by the shaft, if G = 80 kN/mm2.
Solution
R2
=
30 mm, T = 4 × 106 Nmm
R1
=
10 mm
L
=
1,000 mm
Maximum shear stress,
Volume of the shaft,
V
=
π (R22 − R12) × L
=
π (302 − 102) × 1,000
=
8 π × 105 mm3
Ratio,
Strain energy,
Exercise 12.7 A solid circular steel shaft of diameter of 50 mm and length of 1 m
is subjected to a twisting moment of 2.5 kN m. Determine the strain energy
absorbed by the shaft. G = 78.4 kN/mm2.
Stresses Developed in a Key
Shafts transmit power through element like pulley, gear and flywheel. These
elements are keyed to the shaft with the different types of keys, but most common
key is rectangular section key as shown inFigure 12.15. The breadth and thickness
of the key are b and t, respectively, as shown. Half of the key is embedded in
keyway on shaft and another half fits into the keyway provided in hub of pulley,
flywheel, gear etc.
Say, torque transmitted = T
Shaft radius = R
The tangential force Q or shear force acting on the periphery of the
shaft, Q = T/R
Say, length of key = L
Section of the key in shear = bL
Shear stress developed in key = Q/bL = T/RbL
Same force Q acts as a bearing force for half of the key in shaft
Section of the key under compression = L(t/2)
Bearing pressure or stress developed in key,
σp = 2Q/Lt
Figure 12.15 Key fitting the shaft and hub
Example 12.8 A shaft of a diameter of 60 mm is transmitting 20 HP at 300 rpm.
The shaft is connected to a pulley of an axial width of 120 mm with the help of a
key 14 mm × 12 mm (deep). Determine the shear and compressive stresses
developed in the key. Take 1 HP = 746 W/s.
Solution
HP = 20
rpm = 300
Angular speed,
Torque,
Radius of the shaft, R = 30 mm = 0.03 m
Tangential force on periphery,
Length of the key = axial width of pulley = 120 mm
Breadth,
b = 14 mm
Thickness,
t = 12 mm
Shear stress developed in key,
Bearing pressure developed in key,
Exercise 12.8 A shaft of 50 mm in diameter and 1.5 m in length is transmitting 15
HP at 200 rpm. Shaft is connected to the pulley of an axial length of 100 mm by a
key of breadth 12 mm and depth 10 mm. Determine the shear and compressive
stresses developed in the section of the key.
Stress Concentration in Torsional Loading
Using the torsion formula, we determine the shear stress developed in a shaft of
uniform diameter. However, if the shaft has abrupt change in section such as fillet
radius, groove or notch, stress concentration will occur near the abrupt change in
diameter of shaft.
Maximum shear stress,
Where,
Ri = radius of smaller diameter at abrupt change.
T = torque.
J = polar moment of inertia for section for smaller diameter.
K = stress concentration factor due to abrupt change.
Magnitude of stress concentration factor K depends upon the ratio of D/d and the
ratio of r/d at the fillet as shown in Figure 12.16(a).
Similarly, for a grooved shaft the value of K depends on the ratio
of D/d and r/d where r is the groove radius as shown in Figure 12.16(b).
For shaft with keyways, which are generally used to connect gears, pulleys and
flywheels on shaft, ASME allows 25 per cent reduction in allowable stress using
the equation, τ = TR2/T, where R2 is outer radius of the shaft.
Figure 12.16 Stress concentration factors
Example 12.9 A stepped shaft transmits 100 HP as it rotates at 400 rpm. The
allowable shear stress in section of shaft is 50 MPa. The larger shaft has a diameter
of 100 mm and smaller shaft has a diameter of 75 mm. The fillet radius at the
junction of the shafts is 6 mm. Use the stress concentration factor value, to
determine the maximum shear stress in the smaller shaft.
Solution
Figure 12.17 shows the stepped shaft with a fillet, the dimensions are:
Figure 12.17
Fillet radius,
r = 6 mm
Ratio,
From Figure 12.16 (a),
Now,
HP = 100
rpm = 400
Torque,
Smaller diameter, d = 75 mm
Shear stress,
Maximum shear stress, τmax = K × t = 1.45 × 21.5 = 31.2
Exercise 12.9 A transmission shaft has a semi circular groove with a radius of 5
mm machined into a diameter of 100 mm. The maximum shear stress in the shaft
must not exceed 70 MPa, determine the maximum torque that can be transmitted
by the shaft (Refer to Figure 12.18)
Problem 12.1 A torsion bar, 1 m long, is to be designed. Shear modulus of the bar
= 84,000 N/mm2. Determine the required diameter of the shaft, so that resisting
torsional spring constant (or torsional stiffness) of the bar is 36 Nm/degree of
angular trust.
Solution
Figure 12.18
Twisting moment,
or,
T = 36 Nm
or,
Required shaft diameter, d = 22.36 mm.
Problem 12.2 A hollow circular steel shaft is required to transmit 300 HP at 200
rpm. The maximum torque developed is 1.5 times of the mean torque. Determine
the external diameter of the shaft if it is double the internal diameter. The
maximum shear stress is not to exceed 80 MPa. GivenG = 82 kN/mm2.
Solution
HP
N
=
300
=
300 × 746 Nm/s
=
200 rpm
Angular speed,
Mean torque,
Maximum torque,
Tmax
=
1.5 Tmean = 1.5 × 10,685.66
=
16,028.5 Nm
=
16.0285 − 106 Nmm
Say, external diameter = D
internal diameter = 0.5D
J = polar moment of inertia
Maximum shear stress, τ= 80 MPa, at radius D/2
or,
or,
Problem 12.3 A vessel having a single propeller shaft 200 mm in diameter running
at 300 rpm is re-engined to two propeller shafts of equal cross-section and
producing 40 per cent more HP at 500 rpm. If the working shear stress in reengined shafts is 15 per cent more than in the single shaft, determine diameter of
these shafts.
Solution
Say T1, torque transmitted by a single shaft
T2, torque transmitted by each of two shafts
H2 = 1.4 H1 (as given)
Now,
Diameter of each re-engined shaft = 143 mm.
Problem 12.4 A solid steel shaft of 60 mm diameter is made of low-carbon steel
which is assumed to be elasto-plastic with τy = 160 MPa and G = 84,000 N/mm2.
Determine the maximum shear stress due to an applied torque of (a) 2.5 kN m and
(b) 7.5 kN m.
Solution
1.
Shaft diameter,
d = 60 mm
Torque,
T = 2.5 kN m
= 2.5 × 106 Nmm
2.
Maximum shear stress,
3.
4. Torque,
T = 7.5 kN m
= 7.5 × 106 Nmm
Maximum shear stress,
But the behaviour of the material is elasto-plastic, that is, O–A, elastic and A–B,
fully plastic as shown in Figure 12.19.
So, maximum shear stress in shaft is only 160 N/mm2 but not 176.8 N/mm2 (as
calculated).
Figure 12.19
Problem 12.5 A solid marine propeller shaft is transmitting power at 1,200 rpm.
The vessel is being propelled at a speed of 18 km/h, for the expenditure of 4,000
HP. If the efficiency of propeller is 60 per cent, and the greatest shear stress is not
to exceed 50 MPa, calculate the shaft diameter and maximum shearing stress
developed in the shaft.
Solution
Say, thrust = P Newton
Useful work done per second
Efficiency
= 60 per cent
Input work
= 5P/0.6 = 8.33P
Work input per second
= 4,000 × 746 = 298.4 × 104 Nm
= 8.33P
Thrust,
Allowable direct stress,
σ = 50 N/mm2
Speed,
Torque,
Maximum shear stress,
Problem 12.6 A solid circular copper shaft is required to transmit 30 HP at 200
rpm. Determine the diameter of the shaft if the maximum shear stress is not to
exceed 60 N/mm2 in shaft.
The solid shaft is now replaced by a hollow copper shaft with the internal
diameter 0.7 times the external diameter. Determine the external diameter of the
shaft, if it is required to transmit same horse-power, at same rpm and maximum
stress produced in shaft also remains the same. Find the percentage saving of
material of shaft by using hollow shaft in place of solid shaft.
Solution
Torque,
Shear stress, τ = 16T/πd3 where T is shaft diameter
Hollow shaft
Inner diameter, d = 0.7D
Polar moment of inertia,
So,
D3
=
D3
=
119.365 × 103
D
=
49.23 mm
d
=
0.7 D = 34.46 mm
Area of cross-section,
Area of cross-section of solid shaft,
Percentage saving of material
Problem 12.7 A steel shaft of a diameter of 100 mm runs at 300 rpm. This steel
shaft has a 20-mm-thick bronze bushing shrunk over its entire length of 2 m. If the
maximum shearing stress in steel shaft is not to exceed 40 N/mm2, find (a) power
of the engine and (b) torsional rigidity of the shaft.
Given Gsteel = 84 kN/mm2, Gbronze = 42 kN/mm2
Solution
Maximum shear stress in steel shaft, τs = 40 N/mm2
Diameter of steel shaft, d = 100 mm
Torque transmitted by steel shaft,
Steel shaft has bronze sleeve of thickness 20 mm, say torque transmitted by bronze
shaft is TB, then
where polar moment of inertia of bronze shaft,
So,
TB
=
1.421 × TS = 1.421 × 7.854 × 106 = 11.16 × 106 Nmm
Total torque, T
=
TS + TB = (7.854 + 11.16) × 106 Nmm
=
19.014 × 106 Nmm = 19.014 × 103 Nm
Now, θs = θB, in a compound shaft
Length of shaft = 2 m = 2,000 mm
Torsional rigidity of shaft
=
=
1,000.74 × 106 Nmm/rad
Power of the engine
=
1,000.74 kN m/rad
=
2πNT Nm/min
=
2π × 300 × 19.014 × 103 Nm/min
=
35,840.5 × 103 Nm/min
=
597.74 × 103 Nm/s = 597.74 kW
Problem 12.8 A compound drive shaft consists of a 4-m-long solid circular steel
shaft of 50 mm in diameter surrounded by a 2-m-long aluminium tube as shown
in Fig. 12.20. Tube and shaft are both rigidly fixed to a machine. Pin D of diameter
15 mm fills a hole drilled completely through tube and the shaft. Shearing
deformation in pin and deformation in between pin and shaft can be neglected.
Calculate the maximum torque T which can be applied to the steel shaft without
exceeding an average shearing stress of
8 MPa in the pin.
Gsteel = 80 GPa, Galuminium = 28 GPa
Solution
Pin diameter,
dp = 15 mm
Shearing stress in pin = 8 MPa
Shearing force in pin,
(Shearing at two ends)
Figure 12.20 Problem 12.8
Radius of shaft B, R = 25 mm
Torque on shaft B in portion AB (Compound shaft)
=
Q × R = 2,827.4 × 25
=
70, 686 Nmm = 70, 686 Nm
Compound shaft
TA = torque shared by aluminium
TB = torque shared by steel
Now TB = 70.686 Nm
TA
=
1.4218 × 70.686 = 100.506 Nm
T = TA + TB
=
70.686 + 100.506
=
171.192 Nm
Problem 12.9 A circular stepped shaft shown in Fig. 12.21 is subjected to a
torque TA = 500 Nm andTB = 2,000 Nm. Determine the maximum shear stress in
both steel and brass section of the shaft. The shaft is fixed at left end.
Solution
For equilibrium, clockwise resisting couple is 2,000 + 500 = 2,500 Nm (cw).
Torque in two portions AB and BC will be as shown in Fig. 12.22.
Figure 12.21 Problem 12.9
Figure 12.22 Problem 12.9
Note that net torque at B is 2,500 – 500 = 2,000 Nm (ccw)
Stress
Steel shaft
Brass shaft
Problem 12.10 A 5-HP motor shown in Fig. 12.23 delivers 3 and 2 HP,
respectively, to accessory gears B and C. Determine (a) maximum shear stress in
steel shaft drive and (b) angular twist between C and A, if the angular speed of the
shaft is 1,000 rpm and G = 80 kN/mm2.
Figure 12.23
Solution
HP delivered by motor = 5 HP
HP consumed by gear B = 3 HP
HP received by gear C = 2 HP (as shown in Fig. 12.23)
Portion AB transmits 5 HP power.
Speed = 1,000 rpm
Angular speed,
Torque,
Diameter, d = 20 mm
Problem 12.11 A horizontal shaft AB, 1 m long, rigidly fixed at both the ends, A
and B, is subjected to axial twisting moments of T1 = 2.5 kN m and T2 = 4 kN m at
distances of 0.25 and 0.75 m from end A and are in anticlockwise direction looking
from end A as shown in Fig. 12.24. Determine end fixing couples.
Solution
Say, fixing couples at ends A and B are TA and TB, respectively.
Torque on portion CD = TA –2.5 kNm
Torque on portion DB = TA – 2.5 – 4 = TA – 6.5
Total angular twist between A and B
θAC + θCD + θDB = 0 as both the ends are fixed
G and J of shaft is the same throughout, so
or, TA × 0.25 + (TA − 2.5)0.5 + (TA − 6.5)0.25
=
0
or, 0.25TA + 0.5TA − 1.25 + 0.25TA − 1.625
=
0
TA − 2.875
=
0
TA = 2.875 kN m
=
TAC
TA − 2.5 = + 0.375 kN m
=
TCD
TA − 6.5 = 2.875 − 6.5
=
−3.625 kN m
Figure 12.24 shows the torque distribution diagram.
Figure 12.24 Torque distribution diagram
Problem 12.12 A pipe cantilever is subjected to a load at the end of a lever as
shown in the Fig. 12.25. The outer diameter of pipe is 100 mm and the inner
diameter of the pipe is 80 mm. Determine principal stresses at point D at top edge
of cantilever as shown in the figure.
Solution
Bending moment at point D,
M = 2 × 1.2 = 2.4 kN m
Twisting moment at point D,
T = 2 × 0.6 = 1.6 kN m
Outer diameter of pipe, D = 100 mm
Inner diameter, d = 80 mm
Moment of inertia,
Polar moment of inertia, Izz = 2Izz = 579.6 × 104 mm4
Shear stress at D,
Figure 12.25
Bending stress at D
State of stress at the point D is shown in Fig. 12.26.
Principal stress
Figure 12.26
Problem 12.13 A solid shaft of a diameter of 80 mm is transmitting 250 kW at 200
rpm. It is also subjected to a bending moment of 5 kN m and an end thrust. If the
maximum principal stress developed in shaft is 200 N/mm2, determine the
magnitude of end thrust.
Solution
Power transmission = 250 kW = 250 kN m/s
rpm = 200
Speed,
Torque,
Shaft diameter, d = 80 mm
Shear stress developed in shaft,
Bending moment, M = 5 kN m = 5 × 106 Nmm
σb = direct stress due to bending moment
Say, direct stress due to end thrust is σ
Resultant direct stress, σ = σb − σt = 99.47 − σt
= σb + σt = 99.47 + σt
Maximum principal stress,
Say σb and σt are both compressive
129.5 = 99.47 + σt (both are compressive)
σt = 129.5 – 99.47 = 30.03 N/mm2
Area of cross-section of shaft
End thrust,
P
=
σt × A = 5,026.55 σ 30.03
=
1,50,947 N
=
150.947 kN
Considering σ = σb – σt (tensile and compressive)
129.5 = 99.47 − σt which is not possible as σt will also be in tension
so, P = 150.947 kN
Problem 12.14 A flanged coupling has n bolts of a diameter of 19 mm arranged
symmetrically along a bolt circle of diameter of 300 mm. If the diameter of the
shaft is 80 mm and it is stressed up to 80 N/mm2, determine the value of n if the
shear stress in bolts is not to exceed 35 N/mm2.
Figure 12.27 Bolts joining flanges
Solution
A flange coupling is used to connect two shafts, so that the power can be
transmitted from one shaft to the other. Flanges connecting the shafts are joined
together with the help of bolts (Fig. 12.27)
Maximum stress in shaft, τ = 80 N/mm2
Shaft diameter,
d = 80 mm
Torque transmitted,
Bolt circle radius,
R = 150 mm
Shear force is both,
Bolt diameter,
db = 19 mm
Area of cross-section of each bolt
Allowable stress in each bolt
= 35 N/mm2
Shear force transmitted by each bolt
= 35 × 283.53
F = 9923.6 N
or number of bolts
n = 6 bolts to connect the two flanges of coupling.
Problem 12.15 A splined shaft connection is shown in Fig. 12.28 is 60 mm long
and is used to permit axial movement of the shaft relative to the hub during torque
transmission in order to facilitate axial movement in the connection, side pressure
on spline is not to exceed 10 N/ mm2. Calculate the power that can be transmitted
by shaft at 300 rpm.
Figure 12.28 Splined shaft
Solution
Length of splines, L = 60 mm
Outer diameter of splines = 60 mm
Inner diameter of splines = 48 mm
Width of each spline, w = 12 mm (as shown in Fig. 12.28)
Area under shear = w.L = 12 × 60 = 720 mm2
Side pressure = 10 N/mm2
Compressive force per spline
Angular speed,
Number of splines = 6
Mean radius of spline,
Torque transmitted = nPRm
= 6 × 3,600 × 27
= 583.2 Nm
P = ω.T = 31.416 × 589.2
Power,
= 18,322 Nm/s
= 18.322 kW
Shear stress in splines,
Key Points to Remember
o
Torsion formula,
where
T
= torque on shaft
J
= polar moment of inertia of shaft section
G
= shear modulus
θ
= angular twist over length L
τ
= maximum shear stress at radius R
τr
= shear stress at any radius r.
o
Modulus of rupture,
o
HP transmitted by a shaft
where
o
o
o
o
o
for a solid shaft
N
= speed in rpm
T
= Torque in Nm
If several shaft are in series, then same twisting moment acts on all the
shafts, but angular twist will be different and total angular twist will be the
sum of all the angular twists.
It two shafts are in parallel, then angular twist in both shafts will be the
same. Total torque T will be shared by the two, that is, T1 + T2 = T.
Strain energy per unit volume = τ2/4G, where τ is the maximum shear
stress (for a solid shaft). For hollow shaft shear strain energy per unit
volume τ2/4G (D22 + D12/D22), where D2 and D1 are outer and inner
diameters of the shaft.
If a key of breadth b, thickness t and length L connects a shaft and hub for
power transmission, shear stress in key = 2T/DbL
Bearing stress in key = 4T/DtL where D is shaft diameter and T is torque
transmitted by shaft.
Review Questions
1. A shaft is subjected to a twisting moment, show how shear strain varies
along the radius of the shaft.
2. A hollow shaft of outer radius R2 and inner radius R1 is subjected to a
twisting moment T. Derive expression for maximum shear stress in shaft.
3. A solid shaft is subjected to a twisting moment T, such that maximum
shear stress developed on the surface of the shaft is τ, if G is shear
modulus, prove that strain energy per unit is τ2/4G.
4. Take a hollow shaft subjected to twisting moment and with the help of a
sketch show the variation of shear stress along radial thickness of shaft.
5. What are equivalent twisting moment and equivalent bending moment in a
shaft? How these are obtained?
6. Make a simple sketch of a shaft subjected to twisting moment. Take a
small element on the surface of the shaft and mark directions of principal
stresses.
7. Show that volumetric strain for a shaft subjected to pure torsion is 0.
Multiple Choice Questions
1. A shaft of 20 mm diameter and length 1,000 mm is subjected to twisting
moment such that θ = 0.1 rad. What is the shear strain in the shaft at outer
surface?
1.
2.
3.
4.
0.001 rad
0.0001 rad
0.0005 rad
None of these
2. A hollow shaft of inner radius 30 mm and outer radius 50 mm is subjected
to a twisting moment. If the shear stress developed at inner radius of shaft
is 60 N/mm2. What is the maximum shear stress in shaft?
0. 60 N/mm2
1. 75 N/mm2
2. 100 N/mm2
3. None of these
3. Torsional rigidity of a shaft is given by
0. T/G
1. T/J
2. JG
3. None of these
where T is torque, G is shear modulus and J is polar moment of inertia.
4. A solid steel shaft is surrounded by a copper shaft, such
that,
Gsteel, what is the ratio of TS/TC if compound
shaft is subjected to a twisting moment?
0. 2
1. 1
2. 0.5
3. None of these
5. A shaft of diameter d and length L is subjected to twisting moment T,
shear angle developed in shaft is 0.001 rad. Now, the length of the shaft is
doubled, but the diameter and torque remain the same. What will be the
shear angle?
0. 0.0005 rad
1. 0.001 rad
2. 0.002 rad
3. None of these
6. A shaft is transmitting 2 HP at 100 rpm, maximum shear stress in shaft is
50 MPa. Now the shear stress remains the same, how much HP will be
transmitted at 200 rpm?
0.
1.
2.
3.
1 HP
2 HP
3 HP
None of these
7. What is the shear modulus of steel?
0.
1.
2.
3.
40–42 GPa
60–63 GPa
70–72 GPa
None of these
8. A solid shaft of diameter d is subjected to twisting moment T, such that
maximum shear stress developed is τ. Now a hole of 0.5d diameter is
drilled throughout the length of the shaft along the axis. How much
torque T must be reduced so that shear stress remain the same?
0. T/2
1. T/4
2. T/8
3. T/16
9. A solid circular steel shaft is subjected to twisting moment T, such that
maximum shear stress developed on shaft is 200 MPa. If E = 200 GPa and
n = 0.3, what is the maximum principal strain developed on the shaft?
0. Zero
1. 1,000 micro strain
2. 1,300 micro strain
3. None of these
10.A shaft of 100 mm diameter is keyed to a pulley transmitting power. The
dimension of key are b= 20 mm, t = 20 mm and L = 100 mm. Twisting
moment of shaft is 1,000 Nm. What is the shear stress developed in key?
0.
1.
2.
3.
10 N/mm2
20 N/mm2
40 N/mm2
None of these
Practice Problems
1. A torsion bar 1.2 m long is to be designed. Shear modulus of bar is 82 kN/mm2.
Determine the required diameter of the shaft so that the resulting torsional spring
constant (or torsional stiffness) of the bar is 40 Nm for 1° of angular twist.
2. A hollow circular steel shaft is required to transmit 200 HP at 360 rpm. The
maximum torque developed is 1.3 times the mean torque. Determine the external
diameter of the shaft if it is 1.6 times the internal diameter and the maximum shear
stress in shaft is not to exceed 75 MPa. GivenG = 82 kN/mm2.
3. A vessel having a single propeller shaft 250 mm in diameter running at 200 rpm is
re-engined to two propeller shafts of equal cross-section and producing 50 per cent
more horse-power at 500 rpm. If the working shear stress in these shafts is 20 per
cent more than that in the single shaft, determine diameter of these shafts.
4. A solid steel shaft of 50 mm diameter is made of a low carbon steel which is
assumed to be elasto-plastic with Ty = 150 MPa, and G = 84 × 103 N/mm2.
Determine the maximum shear stress due to an applied torque of
(a) 2 kN m and (b) and 4 kN m.
5. A solid marine propeller shaft is transmitting power at 1,000 rpm. The vessel is
being propelled at a speed of 20 km/h for the expenditure of 5,000 HP. If the
efficiency of propeller is 70 per cent, and the greatest thrust is not to exceed 60
MPa, calculate the shaft diameter and maximum shearing stress developed in the
shaft.
6. A solid circular steel shaft is required to transmit 60 HP at 200 rpm. Determine the
diameter of the shaft, if the maximum shear stress is not to exceed 60 N/mm2 in
shaft.
The solid shaft is now replaced by a hollow steel shaft with the internal
diameter equal to 75 per cent of the external diameter. Determine external
diameter of the shaft if it is required to transmit the same horse-power at same
rpm and the maximum shear stress produced is also the same. Find the
percentage saving of the material by using hollow shaft in place of solid shaft.
7. A solid alloy shaft of diameter 60 mm is coupled to a hollow steel shaft of same
external diameter. If the angular twist per unit length of hollow steel shaft is 80 per
cent of the angular twist of alloy shaft, determine the internal diameter of the steel
shaft. At what speed, the shafts will transmit power of 200 kW. The maximum
shearing stress in steel shaft is not to exceed 100 MPa and in alloy shaft it is not to
exceed 60 MPa, G for steel = 2G for alloy.
[Hint: It is a question of two shafts in series]
8. A compound drive shaft of solid circular steel of 40 mm diameter and 5 m long
surrounded by a copper tube 3 m long, outer diameter 80 mm as shown in Fig.
12.29. Both the tube and shaft are rigidly fixed to a machine. Pin C of diameter 12
mm fills a hole drilled completely through tube and shaft. Shearing deformation in
pin and shearing deformation in between pin and shaft can be neglected. Calculate
the maximum torque T which can be applied to the steel shaft without exceeding
an average shearing stress of 10 MPa in the pin. Gsteel = 80 GPa, Gcopper = 40 GPa
Figure 12.29
9. Figure 12.30 shows a hollow shaft subjected to torque TB = 3.5 kN m and Tc = 1.5
kN m. Determine the location and magnitudes of maximum shear stress if the outer
diameter of the hollow shaft is 60 mm and inner diameter is 40 mm.
[Hint: TCB = 1.5 kN m (ccw), TBC = 2 kN m (ccw)]
Figure 12.30
10. A 7.5-HP motor shown in the Fig. 12.31 delivers 5 and 2.5 HP responsively to
accessory gears B and C, respectively. Determine (a) maximum shear stress in
steel shaft drive, and (b) angular twist between C and A, if the angular speed of the
shaft is 500 rpm.
G = 84 kN/mm2
Figure 12.31
11. A horizontal shaft AD, 1.4 m long, rigidly fixed at both the ends A and D is
subjected to axial twisting moments of T1 = 3 kN m and T2 = 6 kN m at distances
of 0.4 and 1.0 m from end A and are in anticlockwise direction when looking from
end A. Determine end couples Fig. 12.32.
Figure 12.32
12. A cantilever made from a solid circular rod of diameter 50 mm is subjected to a
load P kN at point C as shown in the Fig. 12.33. Determine the magnitude
of P such that maximum principal stress at point is not to exceed 100 MPa.
13. A solid shaft of diameter 11 cm is transmitting 700 kW at 200 rpm. It is also
subjected to a bending moment of 15 kN m and an end thrust P. If the maximum
principal stress developed in the shaft is 200 N/mm2, determine the magnitude of
end thrust, P.
14. A flanged coupling has n bolts of 25 mm diameter arranged symmetrically along
a bolt circle of diameter 300 mm. If the diameter of the shaft is 100 mm and it is
stressed up to 100 N/mm2, determine the value of n, if the shear stress in the bolt is
not to exceed 50 N/mm2.
Figure 12.33
Special Problems
1. A hollow shaft of external diameter 120 mm is transmitting 400 HP at 200 rpm.
Determine the internal diameter if maximum shear stress in shaft is not to exceed
60 N/mm2.
2. How much torque is required to twist a 2-m-long steel shaft of an inside diameter
of 80 mm and an outside diameter of 120 mm by an angle of 1.5°? Calculate the
shear stress at outer and inner radii of hollow shaft.
3. A composite is fixed at one end and a torque of 11 kN m acts at the other end.
Determine the torque resisted by outer shaft. The outer radius is 40 mm and inner
radius is 25 mm. Shear modulus of outer shaft is 1.6 times the shear modulus of
inner shaft. What are the shear stresses developed in outer and inner shaft?
Figure 12.34 Special Problem 5
Figure 12.35
4. A circular steel shaft is transmitting 20 HP at 300 rpm. Determine the diameter of
the shaft if the maximum shear stress is not to exceed 80 N/mm2 and angular twist
per meter length of the shaft is not to exceed 1°. Given G = 80 kN/mm2.
5. A solid shaft is transmitting 2,000 kW at 200 rpm. The maximum torque
developed in shaft is 1.8 times of the mean torque. The distance between bearings
is 1.8 m as shown in Fig. 12.34. Weight of flywheel is 5,000 kg midway between
bearings. Determine shaft diameter if: (a) maximum permissible tensile stress is 60
N/mm2, and (b) maximum permissible shearing stress is 40 N/mm2.
6. A torque tube consists of two sections which are riveted together as shown in Fig.
12.35 by 36 rivets of 4 mm diameter each pitched uniformly and the radius of the
mating surface of the tubes is 100 mm. If the limiting shear stress for rivets is 50
MN/m2, determine the maximum torque that can be transmitted through the joint.
7. A motor shaft consists of a steel tube of 40 mm internal diameter and 5 mm thick.
The engine develops 15 HP at 2,000 rpm. What will be the maximum shear stress
in the tube when the power is transmitted through 4:1 gearing?
[Hint: N for tube is 500 rpm]
8. Figure 12.36 shows a vertical shaft with pulleys keyed on it. The shaft is rotating
with uniform velocity of 2,000 rpm. The belt tensions are indicated and three
pulleys are rigidly keyed to the shaft. If the maximum shear stress in the shaft is
not to exceed 50 N/mm2, determine the necessary diameter of the solid circular
shaft. Draw the torque distribution diagram along OABCD.
Figure 12.36
Answers to Exercises
Exercise 12.1: J = 7.9521 × 104 mm4, 105.63 N/mm2, 4.92°; 70.42 N/mm2
Exercise 12.2: 208.6 N/mm2
Exercise 12.3: 128.6
Exercise 12.4: 119.4 N/mm2; 0.021 rad; 1.2°
Exercise 12.5: 84.03, 63.0 N/mm2, 6.02°
Exercise 12.6: Me = 22.44 kN m, Te = 34.88 kN m,
p1 = 39.2 N/mm2,tmax = 30.46 N/mm2
Exercise 12.7: 13.233 kN m
Exercise 12.8: 17.8 N/mm2, 42.74 N/mm2
Exercise 12.9: 6.42 kN m
Answers to Multiple Choice Questions
1. (a)
2. (c)
3. (c)
4. (b)
5. (b)
6. (d)
7. (d)
8. (d)
9. (c)
10. (a)
Answers to Practice Problems
1. 23.1 mm
2. D = 74.45 mm
3. 157.5 mm
4. (a) 81.5 N/mm2, and (b) τ′ = 162.97 N/mm2 but shear stress is 150 MPa
5. 99.8 mm, 182.5 N/mm2
6. 56.6 mm, 64.25 mm, 43.6 per cent
7. Internal diameter = 46.95 mm, N=750 rpm
8. 384.53 Nm
9. τmax = 58.76 in portion AB
10.34.83 MPa, 0.0276 rad, 1.58°
11.TAB = 3.857 kN m, TBC = 0.857 kN m, TCD = –5.143 kN m
12.1.105 kN
13.33.8 kN
14.n = 6
Answers to Special Problems
1.
2.
3.
4.
5.
6.
7.
8.
Internal diameter = 88.8 mm
17.1 kN m, 41.87 MPa, 62.8 MPa
T0 = 9.887 kN m, 116 N/mm2, 45.35 N/mm2
43.14 mm
280.5 mm
2,262 Nm
14.75 N/mm2
d = 27.76 mm, TAB = 210 Nm, TBC = 210 – 100
= 110 Nm, TCD = 110 Nm
1b. Springs
CHAPTER OBJECTIVES
A spring is the most important element in a system or a mechanism. A mechanism
cannot work without the application of a spring which restores the mechanism to
its original configuration. In this chapter, students will learn about:
Stiffness of a spring, most useful characteristic.
o
Stresses developed in a spring due to applied axial load and or an
axial moment.
o
An open-coiled helical spring subjected to combined effect of
bending moment and twisting moment.
o
Angular rotation or angular twist in a spring.
o
Importance of graduated leaves in a leaf spring or a carriage spring.
o
Storage of energy in a plane spiral spring for the functioning of a
system as a time-watch.
o
Introduction
To start with, a close-coiled helical spring subjected to an axial load is considered
and relations for stress and stiffness are derived. The effect of curvature and direct
shear stress is taken into account while calculating shear stress in spring wire
section. The effect of axial moment on a close-coiled helical spring is also studied.
Open-coiled helical spring with a large angle of helix subjected to an axial load
and an axial moment is discussed. In this case, the axial moment or a moment due
to an axial load each has two components, that is, a component of twisting moment
and another component of bending moment. The effect of these moments on the
stresses in spring wire, on axial deflection and on axial rotation, is discussed.
Plane spiral springs used in watches or toys for storing energy for slow and
steady operation is also analysed.
Finally, carriage springs or leaf springs which are used in automobiles to absorb
shocks during motion of a vehicle over uneven roads and derivation of expressions
for deflection and stresses.
Helical Springs
When the axis of the wire of the spring forms a helix on the surface of a right
circular cylinder or a right circular cane, a helical spring is obtained.
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Figure 13.1 Coil of a helical spring
Figure 13.1 shows a coil of a helical spring subjected to an axial load W or an
axial moment M. Then ABCDE is the helix made by the spring wire. The axis of
the spring wire forms a right circular cylinder of radius R and axes of the cylinder
is OO′, ∠BPL = α, helix angle.
R =
mean radius of the coil of spring (radius of right circular cylinder)
α =
helix angle
n =
number of coils in the spring
Length of spring wire in one coil = 2πR/cos α
Length of spring wire in n coils = 2πnR/cos α
In the close-coiled helical spring, α is very small,
cos α ≅ 1,
Therefore, the length of spring wire = 2πnR
Close-Coiled Helical Spring
When the coils of the spring are very close to each other then the coils are regarded
as lying in planes at right angle to the axis of the helix, α very small (Fig. 13.2).
Say, wire diameter = d
Polar moment of inertia of section, J= πd4/32
Direct shear force at any section = W
Direct shear stress, τ0=4W/αd2
Torsional shear stress due to twisting moment WR,τ = 16WR /αd3
As shown in Fig. 13.3, direct shear stress τ0 is added to τ at inner radius and is
subtracted from τ at outer radius. So, the resultant shear stress at the inner radius
of the coil is more than the resultant shear stress at the outer radius of the coil.
Wahl’s Factor
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Figure 13.2
The coil of a spring is not contained in a plane as shown in Fig. 13.2, but the coil is
curved throughout its length. Shearing strain at the inner surface of the coil is more
than the shearing strain at the outer surface of the coil. Wahl studied the effect of
continuous curvature on spring and presented a research paper in a conference and
observed that the maximum shear stress in a spring is given by the equation,
where 16WR/αd3 = τ, torsional shear stress
Spring index,
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Figure 13.3 Resultant shear stress in spring wire section
Example 13.1 A close-coiled helical spring of mean coil radius R = 32 mm is
made of a steel wire of diameter d = 8 mm. If the axial load on spring is 80 N, then
determine the maximum shearing stress developed in spring wire.
Solution
Mean coil radius,
R = 32 mm
Mean coil diameter,
D = 64 mm
Wire diameter,
d = 8 mm
Spring index,
C = 64/8 = 8
Shear stress,
τ = 16WR/πd3
= 25.46 (1.107 + 0.077)
Maximum shear stress
= 25.46 (1.184) = 30.14 N/mm2
Exercise 13.1 A close-coiled helical spring of mean coil diameter 50 mm is made
of steel wire of 6 mm diameter. If the maximum shear stress developed is 90 MPa,
what is the axial load applied on the spring?
Close-Coiled Helical Spring Subjected to an Axial Load
Consider a small element mnn′m′ of the spring wire subtending an angle δø at the
centre of the spring axis. Say, under the action of twisting moment T = WR (acting
on every wire section along this coil), angular twist is δθ as shown in Fig. 13.4.
Figure 13.4
Vertical deflection, dδ = Rδθ
Angular twist, δθ = TRd/GJ
Because Rdφ = dL = length of wire considered cc′
T, twisting moment at any section is same, that is, WR
Total axial deflection,
n = number of turns of the spring
Total angle subtended by coils at the axis of the spring, φ = 2πn
Polar moment of inertia of wire, J = πd4/32
So, axial deflection,
Stiffness of the spring, k=W/δ = Gd4/64nR3, i.e. an axial force required to extend
or compress the spring by a unit axial deflection.
Example 13.2 A close-coiled helical spring made of round steel wire is required to
carry an axial load of 800 N for a maximum stress not to exceed 150 N/mm2 in
shear. Determine the wire diameter if the stiffness of the spring is 10 N/mm and
diameter of helix is 80 mm and also calculate the number of turns required in the
spring. Neglect the correction due to spring index. Given G for steel = 84 kN/mm2.
Solution
Mean coil radius,
Maximum shear stress,
R = 40 mm
τ = 150 N/mm2
Axial load,
W = 800 N
τ = 16WR/πd3, substituting the values
d = 10.28 mm.
Stiffness,
k = 10 N/mm
k = Gd4/64nR3, substituting the values
, number of coils.
Exercise 13.2 A close-coiled helical spring made of round steel wire with mean
coil radius 40 mm and the number of turns equal to 10 is subjected to an axial load
of 200 N. Determine (a) wire diameter, if the maximum shear stress is 160
N/mm2 and (b) axial deflection. Given G = 82 kN/mm2.
Closed-Coiled Helical Spring Subjected to an Axial Moment
Figure 13.5 shows a close-coiled helical spring subjected to an axial couple M. The
effect of the couple is to rotate end B of the spring with respect to end A.
Figure 13.5
Helix angle α is very small.
φ is the total angle through which one end of the spring is turned relative to the
other, when coupleM is applied.
Work done on spring = (1/2)Mφ
Figure 13.6 shows a bar AB of length or length of spring wire subjected to
bending moment M.
ρφ = L
φ = L/ρ, where ρ is radius of curvature
From Flexure formula,
So,
Therefore,
L = 2πnR,
I = πd4/64
φ = ML/EI
work done
where n = number of coils, R = radius of spring
for spring wire
Maximum stress in spring wire due to axial couple M is
where d = spring wire diameter.
Figure 13.6
Example 13.3 A close-coiled helical spring made of round steel wire 5 mm
diameter having 10 complete turns is subjected to an axial moment M. Determine
the magnitude of axial couple M if the maximum bending stress in spring wire is
not to exceed 240 N/mm2 and also calculate the angle through which one end of
the spring is turned relative to the other end, if the mean coil radius is 35 mm.
Given E for steel = 200 kN/mm2
Solution
Wire diameter,
d = 5 mm
Number of turn,
n = 10
Mean coil radius,
R = 35 mm
Maximum bending stress,
Angular rotation,
= 1.0555 rad = 60.47°
Exercise 13.3 A close-coiled helical spring made of round steel wire of diameter d,
having 12 complete turns is subjected to an axial moment of 3.6 Nm. Determine
wire diameter if the maximum bending stress in wire section is not to exceed 210
N/mm2 and also calculate the strain energy absorbed by spring when couple M is
applied. Take the mean coil radius as 40 mm.
Given, E = 210 kN/mm2.
Open-Coiled Helical Spring
In the case of open-coiled helical spring, helix angle α is sufficient and effect
of α on twisting moment or bending moment cannot be neglected. We have to
clearly identify the following four parameters in open-coiled helical spring.
1. Spring axis and coil axis, both inclined at helix angle, α.
2. Plane of spring and plane of coil, both are inclined to each other at helix
angle, α as shown inFig. 13.7.
Say, an open-coiled helical spring of mean coil radius R is subjected to an axial
load W (along axis YY of spring)
WR = total moment on spring wire.
WRcos α = T ′ = Twisting moment on a small element of length, δL at the centres
of coil (about y′y′ axis).
WRsin α = M ′, bending moment on small element of length, δL in the plane of
coil.
Due to twisting moment T ′, there will be an angular twist δø ′ and due to bending
moment M ′, there will be an angular rotation δø′.
δθ ′ is shown by ln and it has two components δθ ′cos α and δθ ′sin α.
Figure 13.7
δφ′= shown by pq has two components,
δφ′ cos α and δφ′ sin α.
δ θ = angular twist about XX axis
= δθ′ cos α + δφ′ sin α
δ ø = angular rotation about YY axis
Moreover,
δθ′
=
mm − qr = δθ′ sin α − δφ′ cos α
=
T′δL/GJ and δφ′ = M′δL/EI
Total angular twist θ, about XX axis,
When L = length of spring wire
= 2πnR sec α
So angular twist,
Axial deflection,
δ = Rθ
Total angular rotation about YY axis,
Substituting the values of T ′ and M ′
or
Length of spring wire,
Strain Energy Method
Total axial deflection of a spring can also be obtained by strain energy principle.
As
Example 13.4 An open-coded helical spring made from steel wire of circular
cross-section is to carry an axial load of 120 N. The wire diameter is 8 mm and
mean coil radius is 40 mm. Calculate the (a) axial deflection and (b) angular
rotation of the free-end with respect to the fixed-end if helix angle of spring is 25°,
and number of turns is 12. Given G = 84 kN/mm2, E = 210 kN/mm2
Solution
Helix angle, α = 25°
sin α = 0.4226, cos α = 0.9063
sin2α = 0.1786, cos2α = 0.8214
sin α cos α = 0.3830
d = 8 mm, number of coils, n = 12, W = 120 N
Angular rotation,
Exercise 13.4 An open-coiled helical spring made from steel wire of circular
cross-section is to carry a load of 100 N. The wire diameter is 8 mm and the mean
coil radius is 40 mm. Calculate
(a) the axial deflection and (b) the angular rotation of the free end with respect to
the fixed end if the helix angle of the spring is 30° and number of turns is 12.
Given E = 80 kN/mm2, E = 200 kN/mm2.
Open-Coiled Helical Spring Subjected to Axial Moment
An open-coiled helical spring of mean coil radius R, helix angle α and number of
turns n is subjected to an axial moment M as shown in Fig. 13.8. Components
of M are M′ = M cos α = bending moment and T′ = M sin α = twisting moment.
Take a small length, δL along coil of the spring.
Figure 13.8
δθ
=
angular twist due to T ′
δθ
=
T′δL/GJ, about x′x′
δø
=
angular rotation due to M′ about y′y′ axis = M′δL/EI
′
′
Taking components of δθ′ and δφ′ along xy direction
δ
=
angular twist about XX axis
=
δφ′ sin α − δθ′ cos α, substituting the value of δθ′, δφ′, T′, M
θ
δφ′ = angular rotation about YY axis
= δφ′ cos α+ δθ′ sin α
Total angular twist,
where L = spring wire length = 2πnR sec α
Total angular rotation,
as L = 2πnR sec α
Axial deflection,
Substituting the value of L = 2πnRsec α
Minus sign indicates winding couple, reduces the axial length of spring.
Example 13.5 An open-coiled helical spring made of 5 mm diameter steel wire, 25
mm mean coil radius, and 23° angle of helix is subjected to an axial moment of 2
Nm. Determine (a) angular rotation of one end with respect to the other end and (b)
axial deflection, if number of coils in the spring is 15. Given E = 210 GPa, G =
82.7 GPa.
Solution
Mean coil radius,
R = 25 mm
Wire diameter,
d = 5 mm, n = 15
Polar moment of inertia,
I = J/2=30.68mm4
Moment of inertia,
M = 2 Nm = 2 × 103 N mm
Moment,
EI = 2,10,000 × 30.68 = 6.44 × 106 N mm2
GJ
α
=
82,700 × 61.36 = 5.0744 × 106 N mm2
=
23°
sin α
=
0.390, sin2α = 0.1526
cos α
=
0.920, cos2α = 0.8473, sec α = 1.087
1.
Angular rotation,
2.
Axial deflection,
Exercise 13.5 An open-coiled helical spring made of steel wire 6 mm diameter and
36 mm mean coil radius with 65° inclination of coils with the spring axis is
subjected to an axial moment M. Determine the magnitude of M if the number of
turns in the spring increase by 1/8 and also calculate the change in the axial length
of spring if the original number of turns are 10. GivenGsteel = 84 kN/mm2, Esteel =
210 kN/mm2
Open-Coiled Helical Spring—Stresses Developed in Spring Wire
(a) Axial load W on spring
Total torque, T = WR (Fig. 13.9)
Twisting moment, T ′ = WR cos α
Bending moment, M ′ = WR sin α
Wire diameter = d
Maximum torsional shear stress, τs = 16T ′/πd3
Direct shear stress, τd = 4W/πd2
Maximum shear stress at inner coil radius, Ri
Minimum shear stress at outer coil radius, Ro
Maximum stress due to bending,
Principal stresses at inner coil radius
(b) Axial couple M acting on spring
Bending moment,
M′ = M cos α
Twisting moment,
T ′ = M sinα (Fig. 13.10)
Torsional shear stress on spring wire, τs = 16 M sin α/πd3
Maximum stress due to bending moment,
Principal stresses at extreme radius,
Figure 13.9
Figure 13.10
Example13.6 An open-coiled helical spring made of steel wire of 15 mm diameter,
mean coil radius 90 mm with helix angle 30° is subjected to an axial moment of 40
Nm. Determine the shear and direct stresses developed in the section of the wire of
the spring.
Solution
Axial moment,
M
=
40 Nm = 40 × 103 N mm
Helix angle,
α
=
30°
Wire diameter,
d
=
15 mm
Bending moment,
M′
=
Mcos α = 40 × 103 × 0.866
=
34.64 × 103 N mm
=
M sin α = 40 × 103 × 0.5
=
20 × 103 N mm
Twisting moment,
Shear stress,
T′
τs
=
=
Direct stress,
σ
30.18 N/mm2
=
=
104.54 N/mm2
Exercise 13.6 An open-coiled helical spring, wire diameter 10 mm, mean coil
radius 70 mm and helix angle 20°, carries an axial load of 400 N. Determine the
shear stress and direct stress at inner radius of coil.
Plane Spiral Spring
A plane spiral spring consists of a thin uniform strip wound in the form of spirals
as shown in Fig. 13.11. One end of the strip is connected to winding spindle A and
other end is hinged at point B. Winding couple is applied at spindle and the
number of turns of spirals are increased.
Figure 13.11 Plane spiral spring
Reaction at hinged end B has two components RH, along BA (joining hinged end
and centre of spindle) RV, perpendicular to line BA.
Consider a small element PQ of length δs and the co-ordinate of centroids of PQ
are x and y, taking origin at B and x along the line BA (Fig. 13.12).
Figure 13.12
Angle between the tangents P′P′ at P and Q′Q′ at Q is φ.
After applying the winding couple, the number of turns are increased
or φ increases to φ + dφ, radiusr1 decreases to r2.
Bending moment at element, Me = xRV − yRH
= EI × change of curvature
By integrating both the sides
Assuming that centre of the spring lies at the centre of the spindle.
when ∫ xds ≃ L × R, where L is the length of the strip making the spiral spring
and R is the outer radius of spring.
But RvR = M, winding couple
φ = ML/EI
Energy stored in spring,
U = (1/2) Mφ = M2l/2EI
Bending moment at any element,
Mc = xRv − yRH
Me will be maximum, where y = 0 along line BA and extended up to C. At the
point C, x = 2R
Mmax = maximum bending moment
= 2RRV
= 2M
Maximum bending stress,
σmax = 2 M/Z
where Z = bt2/6, section modulus of strip about plane of bending.
b = breadth of strip
t = thickness of strip
σmax = 12 M/Bt2
Energy stored,
Example 13.7 A plane spiral spring is made of 5 mm wide and 0.25 mm thick steel
strip. Torque applied at winding spindle is 5 N mm. Determine (a) the number of
turns if the length of the strip is
10 m and (b) the maximum stress developed at the point of greatest bending.
Given E = 210 GPa.
Solution
Winding moment,
M = 5 N mm
Width,
b = 5 mm
Thickness,
t = 0.25 mm
L = 10,000 mm = 10 m
E = 2,10,000 N/mm2
Angle of rotation,
= 36.57 rad
Number of turns,
Maximum stress at the point of greatest bending moment,
Exercise 13.7 A flat spiral spring is made of a strip 5 mm wide, 0.25 mm thick and
12 m long. The torque is applied at the winding spindle and seven complete turns
are given. Calculate the torque and the energy stored and the maximum stress
developed at the point of greatest bending moment. Given E = 210 kN/mm2.
Conical Spring
A conical spring is generally used for large deflections (i.e., very small stiffness),
when one coil sets into another coil and load is applied, as in the case of sofa sets
where deflection of top of a sofa set is of the order of several centimeters. Figure
13.13 shows a conical spring with minimum radius R1 is increasing
to R2 over n number of turns.
From A to B, total angle turned through by spring wire = 2πn
Consider a small element ab of length ds at radius R.
where
Figure 13.13
where φ is the angle turned through by spring wire from A to element ab of length
ds and dθ is the angular twist produced by an axial load W, over length ds.
dθ = WRds/GJ
Strain energy of the length ds of spring
when
So,
Total strain energy,
Axial deflection,
Length of this spring wire, L = πn(R2 + R1)
Axial deflection, δ = WL(R22 + R12)/2GJ
Polar moment of inertia, J = πd4/32, where d is the spring wire diameter
Maximum shear stress,
= maximum torsional shear stress +
(as per assumption)
× direct shear stress
Example 13.8 A conical spring of minimum diameter 50 mm and maximum
diameter 100 mm is made up of wire of 6 mm diameter. If the resultant shear stress
in wire is not to exceed 200 MPa, determine the load which the spring can carry if
it has six coils. What is axial deflection in spring? Given G = 84 kN/mm2.
Solution
R1 = 25 mm
R2 = 50 mm
Length of wire = πn(R1 + R2)
Axial deflection,
Exercise 13.8 A conical spring of minimum diameter 60 mm and maximum
diameter 120 mm of steel wire 10 mm diameter. If the maximum stress in the wire
is not to exceed 180 MPa, determine the load which the spring can carry if it has
six coils. What is axial deflection in spring. Given G = 82,000 N/mm2.
Leaf Spring
These springs are used in vehicles such as cars, trucks, railway wagons, trolley,
etc. and are connected between chassis and wheels. They are also termed
as Carriage springs. Generally they are of two types: (1) semi elliptic and (2)
quarter elliptic shapes. A number of plates of rectangular section of same width but
different lengths are clamped together with the help of bolts and nuts. At the ends
of largest leaf, there are eyes which connect the chassis to the spring with the help
of shackles in between.
Figure 13.14 Equilvalent carriage spring semi elliptical spring
A simple theory for deflection, stress and radius of curvature of a leaf spring is
developed taking the following assumptions:
1. Centre line of all the leaves is initially a circular arc of same radius R, so
that contact between the plates is only at the ends.
2. Each plates is of uniform thickness and each plate overlaps the plate below
it by an amountp=L/2n, where L is the length of longest leaf, n is the
number of plates.
3. The overlaps are tapered in width to a triangular shape. Figure
13.14 shows the Equivalent Carriage spring of width B = nb.
Since each plate is initially of same radius of curvature, each plate will touch the
one above it only at its ends when unloaded. After the application of a central load,
the change in radius of curvature is uniform and radius of curvature becomes the
same for all leaves (plates). They will continue to contact at the ends only.
Consider only two plates, top one of length L and bottom one of length (L – 2p),
loaded at ends byW/2 load. Bending moment diagram varies linearly from A to B
and C to D, while bending moment is uniform from B to C and equal to (W/2)p.
This shows that each triangular overhanging end (refer toFig. 13.15) is loaded as a
cantilever while portion of uniform width carries uniform bending momentWp/2.
So on this tapered portion, M and I (moment of inertia) vary linearly and
proportional to this distance from end, so M/I remains constant. Moreover in the
central portion M and I are constant. This proves that M/I remains constant
throughout, the length of the plate (with triangular ends as shown).
Figure 13.15
Now, M/EI = 1/R′–1/R, change in curvature. If M/EI is constant, then 1/R is also
constant. This 1/R′ is also constant with changed radius of curvature R′, contact
between the leaves continues to be at ends only. Taking the friction between the
plates as negligible, each plate is of same radius of curvature. So, these plates can
be considered to be arranged side by side forming a beam of same thickness t with
width B = nb at centre as shown in Fig. 13.14.
In a carriage spring as shown in Fig. 13.14
Maximum bending moment,
Mmax = WL/4
Maximum width,
B = nb
Moment of inertia,
Bending stress,
Initial central deflection,
y0 = L2/8R (using properties of a circle),
Final radius of curvature,
(a constant)
R′ is constant, if all the plates are bent into the arcs of a circle of radius R′.
Say R′ = ∝ (infinity) plates will become straight under Proof load, W0
Example 13.9 A carriage spring centrally loaded has eight steel plates, 5 mm thick
and 50 mm wide. If the largest plate is 800 mm long, find the initial radius of
curvature if the maximum stress is 150 N/mm2, when the plates become straight
under the central load. Given, E = 210 GPa.
Solution
Initial radius of curvature,
Exercise 13.9 A carriage spring centrally loaded has six steel plates, 6 mm thick
and 60 mm wide. If the largest plate is 960 mm long and the load required to
straighten the spring is 3 kN. Determine (a) the initial radius of curvature, (b) the
initial central deflection and (c) the bending stress under proof load.
Cantilever Leaf Spring (Quarter Elliptic Spring)
Proceeding in the same manner as for semi elliptic spring (Fig. 13.16), we get:
Bending moment = WL
Moment of inertia, I = nbt3/12
Bending stress, σ = 6WL/nbt2
Figure 13.16 Equivalent cantilever spring
Say, R′ = final radius of curative
R = initial radius of curative
y0 = L2/2 (using property of a circle)
y = final deflection under load
or,
Say, 1/R′=α (infinity)
From Eq. (1),
Example 13.10 A cantilever leaf spring of length 0.6 m has six leaves of thickness
8 mm each. The width of each plate is 48 mm. If an end load of 1 kN is applied at
its end, determine (a) the end deflection under this load, (b) the initial radius of
curvature if deflection provided is 100 mm and (c) the bending stress developed
under the load. Given E = 210 kN/mm2.
Solution
1.
W
=
1,000 N
L
=
600 mm
b
=
48 mm
t
=
8 mm
n
=
6
Deflection under load,
2.
Initial radius of curvature,
3.
Bending stress,
Exercise 13.10 A cantilever leaf spring of length 500 mm has five leaves of
thickness 10 mm. If an end load of 2 kN produces a deflection of 30 mm, find the
width of the leaves. Given E = 200 GPa.
Problem 13.1 Design a close-coiled helical spring to have a mean coil diameter
120 mm and an axial deflection of 110 mm under axial load of 2,750 N, so that the
maximum shear stress developed in the spring is not to exceed 330 N/mm2. The
steel wires are available in the following diameter: 10, 12, 14 and 16 mm.
Determine the most suitable diameter of the wire and the number of coils required.
Also calculate the maximum shear stress in the designed spring (neglect direct
stress). Given G = 84,000 N/mm2.
Solution
Axial load, W = 2,750 N
Mean coil radius = 60 mm
τmax, allowable = 330 N/mm2
Wire diameter,
Number of coils,
d
=
13.65
=
14 mm available size of the wire
Actual shear stress,
Problem 13.2 A close-coiled helical spring is made of round steel wire. It carries
an axial load of 200 N and is to just get over a rod of 35 mm. Deflection in the
spring is not to exceed 20 mm. The maximum allowable shearing stress developed
in spring wire is 165 N/mm2. G for steel = 80 kN/mm2. Find the mean coil
diameter, wire diameter and number of turns in spring (neglect direct stress in
spring).
Solution
Say wire diameter = d
Spring is just get over a rod of 35 mm, i.e.,
Mean coil diameter,
Now,
D
=
35 + d
W
=
200 N
δ
=
20 mm
δ
=
8WD3n/Gd4, substituting the value
or
From Eqs. (13.2) and (13.4),
From Eq. (13.3),
=
d3(0.3239)
35 + d
=
d3(0.3239)
0.3239d3 − d − 35
=
0
d3 − 3.0873d
=
108.058
d3
=
3.0873d + 108.058
d
=
5,
=
125, RHS = 123.4945
=
4.95,
=
121.87375, RHS = 123.5
=
4.98 mm,
D
If,
LHS
d
LHS
d
LHS
=
123.5, RHS = 123.47
LHS
=
RHS
d
=
4.98 mm
D
=
0.3239, d3 = 0.3239 × 4.983 = 40 mm
n
=
Wire diameter,
So, wire diameter = 4.98 mm, D = 40 mm, n = 9.8
Problem 13.3 A close-coiled helical spring is made of round wire with n turns and
spring index is eight. Show that the stiffness of such a spring is (D/n)× constant.
Determine the constant, if G = 84,000 N/mm2; such a spring is required to support
a load of 500 N with an extension of 40 mm, and maximum shear stress is not to
exceed 230 N/mm2. Determine the mean coil diameter and the number of turns.
Solution
Stiffens,
k = GD4/8nD3
where D is mean coil diameter.
Now, W = 500 N,
or
Problem 13.4 A safety valve 60 mm diameter is to blow off at a pressure of 3
N/mm2 gauge.
The safety valve is held by a close-coiled helical spring of steel with a mean coil
radius equal to 60 mm. Determine the diameter of the steel wire and number of
coils necessary if the maximum shear stress in wire is not to exceed 220 N/mm2,
and spring is initially compressed by 30 mm. Given G = 84 kN/mm2. Take into
account the effect of direct shear stress also.
Solution
Diameter of safety valve = 60 mm
Pressure (at which valve open) = 3 N/mm2
Total force on spring wire,
Say wire diameter = d mm
Mean coil radius,
Compression in spring,
R = 60 mm
δ = 30 mm
Value is to open, when extension in spring is 30 mm, due to shear pressure.
Maximum shear,
For
By trial and error,
(Wire diameter)
Number of turns
d3
=
49.09d + 11,782
d
=
25,
RHS
=
13,009, LHS = 15,625
d
=
24,
LHS
=
13,824, RHS = 129
d
=
23.5,
LHS
=
12,978, RHS = 12,935
d
=
23.5 mm
Problem 13.5 A weight of 200 N is dropped onto a close-coiled helical spring
from a height of 600 mm which produces a maximum instantaneous stress of 120
N/mm2 in the spring. If the mean radius of coil is six times the wire diameter,
determine (a) instantaneous compression in spring and (b) number of coils in the
spring. Given wire diameter is 15 mm and G for steel = 84 kN/mm2.
Solution
W = 200 N
Height,
h = 600 mm
Maximum instantaneous stress, τ = 120 N/mm2
Wire diameter,
d = 15 mm
R = 6d = 6 × 15 = 90 mm
We = Equivalent static load on the spring
Say, instantaneous compression in spring is δ mm.
Potential energy lost by weight = W (h + δ)
Energy stored in spring = (1/2)Weδ
Instantaneous compression in spring = 496.3 mm.
Number of turns
Problem 13.6 A semi elliptical spring, 960 mm long, carries a central load of
4,000 N. Determine the number, breadth and thickness of the leaves, if central
deflection is 45 mm. Assume breadth is eight times the thickness. The leaves are
available in breadths in multiples of 5 mm and thickness in multiples of 1 mm. The
maximum bending stress is also limited to 320 N/mm2. Given E = 210 kN/mm2.
Solution
Substituting the value,
Bending stress,
From Eqs. (13.9) and (13.10),
Available thickness,
t
=
8 mm
Breadth,
b
=
8 × 8 = 64 mm
=
65 mm available width
=
8 mm
so, b = 65 mm,
t
(13.11)
Now, nbt2 = 18,000
Substituting the value of b and t,
Let us check for maximum stress,
Problem 13.7 A rigid bar AB, weighing 150 N carries a load W = 40 N as shown
in Fig. 13.17. The bar rests on three springs of stiffnesses k1 = 15 N/mm, k2 = 7
N/mm and k3 = 10 N/mm as shown in the figure. If the unloaded springs are of the
same length, determine the value of distance x such that the bar remains horizontal.
Solution
Say, the bar moves downwards by deflection δ, reactions from spring will
be k1δ, k2δ and k3δ, respectively.
Figure 13.17
Taking moments of forces about A,
400 × (250 − x) + 150 × 250 = k2δ × 250 + k2δ × 500
Substituting the values of k2 and k3
1,00,000 − 400x + 37,500 = 7δ × 250 + 10δ × 500
= 1,750δ + 5,000δ = 6,750δ
1,37,500 − 400x = 6,750δ
or
1,375 − 4x = 67.5δ
(13.12)
Taking the moment of forces about point B
k2 × δ × 250 + k1 × δ × 250 = 150 × 250 + (250 + x) × 400
Substituting the value of k2 and k1, we get
7 × 250δ + 15δ × 500 = 37,500 + 1,00,000 + 400x
9,250δ = 1,37,500 + 400x
1,375 + 4x = 92.5δ
or,
(13.13)
By adding the Eqs (13.12) and (13.13),
2,750 = 160δ
Deflection,
δ = 2,750/160 = 17.1875
Substituting the value of d in Eq. (13.13),
Problem 13.8 A rigid chute plate, 2 m in length, is supported horizontally at a
height of 250 mm above a hopper by a spring at one end of stiffness 23 N/mm and
a second spring at mid length of stiffness 15 N/mm as shown in Fig. 13.18.
Determine the position on the chute to which a component of 2 kN reaches when
the unsupported end of the chute just touches the edge of the hopper.
Solution
When the chute touches the hopper, extension in spring 1 is δ1 and compression in
spring 2 is δ2 as shown in Fig. 13.18(b).
As it is obvious from the figure
(250 + δ1) = 2 (250 − δ2)
2δ2 + δ1 = 500 − 250
(13.14)
Figure 13.18 Problem 13.8
Say that the load shared by spring 1 is W1 and load shared by spring 2 is W2
−W1 + W2 = 1.5 kN= 1,500 N
(W1 is in tension and W2 is in compression).
k1 = 23 N/mm
k2 = 15 N/mm
Numerically,
From Eq. (13.14),
23 × 2W2 + 15W1 = 15 × 23 × 250
46W2 + 15W1 = 86,250
But
(13.16)
−W1 + W2 = 1,500
W2 = (1,500 + W1)
(13.17)
Substituting the values in Eq. (13.16),
46(1,500 + W1) + 15W1 = 86,250
69,000 + 46W1 + 15W1 = 86,250
61W1 = 17,250
W1 = 2,828 N
W2 = 1,500 + 282.8 = 1782.8 N
Taking moments about A,
At this position, chute just touches the edge of the hopper.
Problem 13.9 In designing a valve spring, it is estimated that the mass of the valve
is 0.8 kg and it requires an acceleration of 200 m/s2, when lifting through a height
of 6 mm. The free length of the spring is 210 mm and the axial length of the spring
is 170 mm, when the valve is shut. If the total lift is 10 mm, determine the
maximum force on the spring.
The diameter of the spring wire is 2.8 mm and the maximum shear stress is not
to exceed 320 MPa. Determine the mean coil diameter and the number of coils.
Given G = 84 kN/mm2.
Solution
Mass of the valve = 0.8 kg
Acceleration = 200 m/s2
Force on valve, F = 0.8 × 200 = 180 N
Total valve left = 10 mm
Valve lift through opening and closing of valve = 6 mm
Free length of the spring = 210 mm
Axial length when valve is shut = 170 mm
Initial compression in length of the spring = 210 − 170 = 40 mm
Further compression when valve is shut = 6 mm
Total compression in length = 40 + 6 = 46 mm
For a force of 100 N,
Stillness of the spring = 160/46 = 3.48 N/mm
Maximum valve lift = 10 mm
Maximum change in the length of the spring = 40 + 10 = 50 mm
Maximum force on the spring Wmax = 3.48 × 50 = 174 N
Maximum shear stress,
Mean coil radius,
Mean coil diameter,
D = 2R = 15.86 mm
Number of coils,
Problem 13.10 A close-coiled helical spring of 17 mm mean coil diameter and 10
turns is arranged with in and concentric with an outer spring. The free length of the
inner spring is 5 mm more than that of the outer spring. The outer spring has 12
coils of mean diameter 25 mm and a wire diameter of 3 mm. The spring load
against which a valve is opened is provided by the inner spring. The initial
compression in outer spring is 5 mm when the valve is closed, find the stiffness of
inner spring if the greatest force required to open the valve by 8 mm is 138 N.
Also, find the wire diameter of the inner spring. Given G = 80,000 N/mm2.
Solution
Initial compression in outer spring = 6 mm
Initial compression in inner spring = 6 + 5 = 11mm
(As the free length of the inner spring is 5 mm more than the free length of the
outer spring)
Say, k1 = stiffness of inner spring in N/mm
k2 = stiffness of outer spring in N/mm
Fi, initial load on valve = 11k1 + 6k2
Outer spring
The valve is opened by 8 mm, additional force required to open the valve, Fo =
8k1 + 8k2 Total load to lift the valve,
Ft = Fo + Fi
So,
=
11k1 + 6k2 + 8k1 + 8k2
=
19k1 + 14k2
19k1 + 14k2
=
138 N
19k1 + 14 × 4.32
=
138 N
d1 = 2.116 mm (wire diameter of the inner spring).
Key Points to Remember
o
For a close-coiled helical spring, W = axial load, R = mean radius, C =
spring index and D/d = ratio of mean coil diameter and wire diameter.
o
Stiffness of a close-coiled helical spring,
G = shear modulus, n = number of coils, δ = axial
change in length.
o
o
o
o
In a close-coiled helical spring, angular rotation due to axial
moment M,
where E = Young’s modulus.
For a open-coiled helical spring, W = axial load, R = mean coil
radius, Twisting moment, T ′ =WRcos α and Bending
moments, M′ = WRsin α, where α = helix angle.
Axial deflection,
where J = 2I = Polar moment of inertia = πd4/32
Angular rotation,
Open-coiled helical string, subjected to an axial moment M,
Twisting moment = M sin α, Bending moment = M cos α
Angular rotation,
o
Axial deflection,
Plane spiral spring, b = breadth, t = thickness, M = moment
Maximum stress, σmax = 12M/bt2
o
Energy stored
× volume of trip
Carriage spring, n = number of leaves, b = breadth, t = thickness, L =
Length of longest leaf; R = radius of curvature.
Initial central deflection,
y0
=
L2/8R
Proof load,
W0
=
Enbt3/3LR
σmax
=
3WL/2nbt2.
Review Questions
1. What is the effect of spring index on the stresses developed in spring
wire?
2. What is Wahl’s factor and how it accounts for the curvature effect on
stress in spring wire?
3. Show by a sketch, how resultant shear stress at inner coil radius is
maximum.
4. Derive relation ϕ = ML/MI, for a close-coiled helical spring subjected to
axial moment M.
5. What are the twisting moment and bending moment components on an
open-coiled helical spring subjected to axial load W?
6. What are the applications of a plane spiral spring? Make a simple sketch
and mark the point where bending stress is maximum.
7. Where the conical springs are used? Derive the expression for maximum
shear stress in a wire of conical spring.
8. In a leaf spring, explain the assumptions that a leaf touches the adjoining
leaf only at ends.
9. What is proof load in a cantilever spring?
10.What is meant by resilience of a spring?
Multiple Choice Questions
1. A close-coiled spring absorbs 50 N mm energy is extending by 5 mm,
what is the stiffness of spring?
1. 10 N/mm
2. 5 N/mm
3. 2 N/mm
4. None of these
2. A carriage spring is with longest leaf 800 mm long and radius of curvature
is 2,000 mm. What is the central deflection?
0. 80 mm
1. 40 mm
2. 20 mm
3. None of these
3. A carriage spring subjected to a central load such that leaves become
straight. What is this load called?
0. Safe load
1. Proof load
2. Ultimate load
3. None of these
4. A close-coiled helical spring of stiffness 4 N/mm is in series with another
spring of stiffness 6 N/mm. What is the stiffness of composite spring?
0. 5 N/mm
1. 4 N/mm
2. 2.4 N/mm
3. None of these
5. An open-coiled helical spring with α = 45° is subjected to an axial
load W such that shear stress due to twisting moment is 100 MPa. What is
the bending stress due to bending moment?
0.
1.
2.
3.
50 MPa
100 MPa
200 MPa
None of these
6. A flat spiral spring is subjected to winding couple producing σon at the
point of greatest bending. What is the strain energy per unit volume?
0.
1.
2.
3. None of these
7. A closed-coil helical spring of wire, diameter 6 mm, is made by taking
mean coil radius as 16 mm, what is its spring index?
0. 2.66
1. 5.33
2. 8
3. None of these
8. Stiffness of a close-coiled helical string of wire diameter d, modulus of
rigidity G, number of coils n, mean coil radius R, is
0.
1.
2.
3. None of these
9. A close-coiled helical spring is subjected to an axial moment M, producing
an angle of rotation 90° at free end with respect to fixed end, the strain
energy absorbed is 100p N mm, what is M ?
0. 100 Nm
1. 200 N mm
2. 300 N mm
3. None of these
10.Wahl’s factor takes into account
0. Curvature of the helical wire
1. Direct shear stress
2. Both curvature and direct shear effect
3. Neither (a) nor (b)
Practice Problems
1. Design a close-coiled helical spring to have a mean coil diameter of 120 mm and
an axial deflection of 150 mm under an axial load of 4,050 N, so that the maximum
shear stress developed in the spring is not to exceed 320 N/mm2. Steel wires are
available in the following diameters: 10, 12 and 16 mm. Determine the most
suitable diameter of the wire and number of coils required. Also calculate the
maximum shear stress developed in designed spring. Given, G = 84,000 N/mm2.
2. A close-coiled helical spring is made of round steel wire. It carries an axial load of
150 N and is just to get over a rod of 36 mm. The deflection in the spring is not to
exceed 25 mm. The maximum allowable shear stress developed in spring wire is
200 N/mm2 (neglecting the effect of direct shear stress). G for steel = 80,000
N/mm2. Find the mean coil diameter, wire diameter and number of turns.
3. A close-coiled helical spring is made of round wire having n turns and mean coil
radius is five times the wire diameter. Show that stiffness of wire spring is (R/n) ×
constant. Determine constant, if G = 82 kN/mm2. Such a spring is required to
support a load of 1 kN with 100 mm compression and maximum shear stress is 245
N/mm2, determine (a) mean coil radius and (b) number of turns.
4. A safety valve 80 mm diameter is to blow off at a pressure of 2 N/mm2 gauge. The
safety valve is held by a close-coiled helical spring of steel with a mean coil radius
equal to 75 mm. Determine the diameter of the steel wire and the number of turns
necessary, if the maximum shear stress in the wire is not to exceed 200 MPa and
the spring is initially compressed by 25 mm. Given, G = 84 kN/mm2. Take into
account the effect of direct shear stress also.
5. A weight of 250 N is dropped onto a close-coiled helical spring through a height
of 800 mm, which produces a maximum instantaneous stress of 200 N/mm2 in the
spring. If the mean radius of the coil is five times the wire diameter, determine (a)
instantaneous compression in the spring and (b) number of coils in the spring.
Given, wire diameter, d = 20 mm and G = 8,410 N/mm2.
6. A laminated carriage spring made of 12 steel plates is 1 m long. The maximum
central load is 6 kN. If the maximum allowable stress in steel is 200 MN/m2 and
the maximum deflection is 40 mm, determine the thickness and width of plates.
Given, E = 200 GPa.
7. A rigid bar AB, weighing 100 N carries a load W = 300 N as shown in Fig.
13.19. The bar rests on three springs of stiffnesses: k1 = 20 N/mm, k2 = 8 N/mm
and k3 = 10 N/mm as shown in Fig. 13.19.If the unloaded springs are of the same
length, determine the value of distance x such that the bar remains horizontal.
[Hint: Reaction k1δ, k2d, k3δ]
Figure 13.19
8. A rigid chute plate, 2.4 m in length, is supported horizontally at a height of 300
mm above a hopper by a spring at one end of stiffness 30 N/mm and a second
spring of stiffness 20 N/mm. Determine the position on the chute to which a
component of 2 kN reaches the unsupported end of chute just touches the edge of
the hopper (Fig. 13.20).
9. In designing a valve spring, it is estimated that the mass of valve is 1 kg and it
requires an acceleration of 150 m/s2; when lifting through a height of 5 mm. The
free length of the spring is 200 mm and the axial length of the spring is 160 mm
when the valve is shut. If the total valve lift is 10 mm, determine the maximum
force on the spring. The diameter of the spring wire is 3 mm and the maximum
shear stress is not to exceed 300 MPa, determine the mean coil diameter and
number of coils. Given, G = 84 kN/mm2.
Figure 13.20
Special Problems
1. A close-coiled cylindrical spring is of 80 mm mean coil diameter. The spring
extends by 40 mm when axially loaded by a weight of 530 N. When the same
spring is subjected to an axial couple, M= 28 Nm, there is an angular rotation of
free end by 60°. Determine Poisson’s ratio of the spring.
2. Determine the stiffness of a close-coiled helical spring consisting of 10 turns of 4
mm diameter steel wire coiled on a mandrel of 60 mm in diameter. Given, G = 84
GPa.
[Hint: D = 60 + d = 64 mm]
3. An open-coiled helical spring made of 10 mm diameter steel rod, 45 mm mean
coil radius and 20° angle of helix is subjected to an axial load W. Determine the
magnitude of W if the maximum shear stress in the wire is limited to 135 MPa.
Calculate the number of turns in the spring if axial extension in the spring under
load is 40 mm. Given, G = 80 kN/mm2 and E = 200 kN/mm2 for steel.
4. Two close-coiled helical springs wound from the same wire but with different core
radii are compressed between two rigid plates at their ends. Calculate the
maximum shear stress induced in each spring, if the wire diameter is 10 mm and
the load applied between the rigid plates is 500 N. The core radii of the springs are
75 and 100 mm, respectively. The number of turns in both the springs is the same.
[Hint: k1 + k2 = k, W = W1 + W2]
5. A semi-elliptical carriage spring made of steel leaves, 1 m long is to support a
central load of 8 kN with a maximum deflection of 60 mm and maximum bending
stress of 320 N/mm2. E = 200 GPa. Calculate the thickness of the leaves and decide
their number and breadth.
6. An open-coiled helical spring made of round steel bar 10 mm diameter has 10
coils of 80 mm mean diameter and the pitch is 60 mm. If the axial moment is 40
Nm, find the deflection in spring. Given, E = 210 kN/mm2, v = 0.28.
7. A wagon weighing 3,000 kg is moving at a speed of 1 m/s. How many springs
each of 20 coils will be required in a buffer stop to absorb energy of impact during
a compression of 150 mm? The mean coil radius is 100 mm and diameter of steel
rod comprising the coil is 25 mm. Given, G = 84 kN/mm2.
Answers to Exercises
Exercise 13.1: 129.8 N
Exercise 13.2: d = 6.34 mm, 61.83 mm, axial deflection
Exercise 13.3: d = 5.6 mm, 1.928 Nm (strain energy)
Exercise 13.4: 16.45 mm, 2.1 °
Exercise 13.5: 4.029 Nm, d = 2.585 mm
Exercise 13.6: 5.093 + 134 = 139.093 N/mm2,97.54 N/mm2
Exercise 13.7: 5.01 N mm, 110.2 N mm, 192.4 N/mm2
Exercise 13.8: 501.62 N, 23.78 mm
Exercise 13.9: (a) 1.89 m, (b) 60.95 m, (c) 333.33 N/mm2
Exercise 13.10: 50 mm
Answers to Multiple Choice Questions
1. (d)
2. (b)
3. (b)
4. (c)
5. (c)
6. (c)
7. (b)
8. (a)
9. (d)
10. (c)
Answers to Practice Problems
1.
2.
3.
4.
5.
6.
7.
8.
9.
16 mm, 14.75 turns, 302.15 N/mm2
40.14 mm, 4.248 mm, 8.4 turns
constant 2.05, R = 51 mm, n = 10.46
d = 27.6 mm, n = 4.5
151.4 mm; n = 10.12
t = 6.25 mm, b = 96 mm
x = 122.8 mm
x = 1.65 m, W1 = −750 N, W2 = 2,750 N
19.08 mm, 36.7 turns
Answers to Special Problems
1.
2.
3.
4.
5.
6.
7.
0.2612
1.025 N/mm
592 N, 8.92 turns
W1 = 351.63 N, W2 = 148.37 N 134.3 N/mm2, 75.56 N/mm2
t = 6.66 mm, n = 10, b = 85.54 mm
6.56 mm
6 springs
2a. Columns and Struts
CHAPTER OBJECTIVES
Any strut or a column member in a machine, or a structure is designed on the basis
of failure against buckling. Professor Euler initially provided a theory of buckling
of such members.
In this chapter, students will learn about:
o
o
o
o
o
o
o
o
o
Theory of buckling developed by Euler for long columns.
Euler considered the columns in the buckled state and then derived
expressions for buckling loads. Professor. Rankine modified the Euler’s
buckling load formula by combining the crushing load for short columns
and Euler’s buckling load for long columns.
Johnson’s parabolic formula for buckling load—an empirical formula.
Secant formula providing information on maximum stress developed in a
column section due to eccentric load on column.
Professor Perry’s approximate formula for eccentrically loaded columns.
Long columns with eccentricity in its geomentry.
Professor Perry Robertson’s formula for combining eccentricities in
geometry of columns and in loading of columns.
Columns subjected to lateral loads in addition to axial buckling loads.
Energy approach for determination of buckling loads in columns with
different end conditions.
Introduction
Collapse of long structural members under compressive axial loads is a common
occurrence. Through an efficient design of strut or column members in a structure,
the collapse must be prevented; otherwise, in addition to damage to structure, there
will be loss of human lives.
Euler has developed a buckling theory for long column members; however, for
short columns, the theory gives absurd results. However, he provided the basic
formula for buckling of columns. Rankine modified the Euler’s buckling load
formula by combining the crushing load for a short column and the buckling load
(provided by Euler) for a long column. In Rankine’s formula, there are two
experimentally determined constants. This formula is applicable for long as well as
medium-long columns.
Many a times, load is eccentrically applied on a column and buckling takes place
at a smaller load than the load given by Euler or Rankine. For eccentrically loaded
columns, expressions for maximum stress have been derived in the form of a
secant formula. Many researchers such as Professor Perry and Perry–Robertson
have modified the secant formula for eccentrically loaded columns.
Higher-order differential equations have been used for determining the buckling
loads of columns with different end conditions. Moreover, energy approach is
briefly discussed to determine the buckling loads of columns, taking various shape
functions for the deflection curve.
Euler’s Theory of Buckling
Euler has developed a theory of buckling for columns or struts, but considered the
column/strut already buckled under a particular load while developing the theory
of buckling. He has made the following assumptions:
1.
2.
3.
4.
5.
6.
7.
material of the column is homogeneous and isotropic,
compressive load acting on the column is fully axial,
column has failed only due to buckling,
weight of the column is negligible,
column is initially straight and buckles suddenly at a particular load,
fixed ends are rigid and
hinged ends are frictionless.
Example 14.1 Consider a column AB with one end fixed and the other end free,
buckled under the load P as shown in the Fig. 14.1. Length of the column is L.
Deflection at free end B is a as shown, and the buckling load is P. Consider a
section of the column at a distance x from end A.
Solution
Bending moment at section XX = P(a − y),
Solution of this differential equation is
y = A sin mx + B cos mx + a
where
and A and B are constants.
(One can verify the correctness of the solution by differentiating y two times
with respect to x.)
End conditions
y = 0, x = 0, dy/dx = 0 at x = 0, at end A.
y = a, x = L, at end B
so,
0 = A sin 0 + B cos 0 + a
0=0+B+a
Figure 14.1 Column fixed at one end and free at the other end
or constant,
B = −a
dy/dx = Am cos mx − Bm sin mx
0 at x = 0
0 = A mx cos 0 − B × sin 0
0 = Am
m # 0, because
then the load P will become zero, if m = 0
so, constant A = 0
Finally,
y = −a cos mx + a
At end B,
x = L, y = a
a = −a cos mL + a
or,
a cos mL = 0
a # 0, because if a = 0, then no buckling takes place
so,
mL=π/2 (minimum significant value)
Euler’s buckling load,
Pe = π2EI/4L2.
Example 14.2 Let us take another case of a column hinged at one end and fixed at
the other end as shown in Fig. 14.2. The column is under buckled condition. P is
the buckling load, H is the horizontal reaction at B. Considering a section at a
distance x from fixed end A as shown.
Figure 14.2 Column with one end fixed, other end hinged
Solution
Bending moment at sections XX
Solution of this differential equation is
End conditions x = 0, y = 0
Constant,
B = −HL/P
(14.4)
Then,
So,
Am = H/P
Constant,
A = H/mP
(14.5)
Putting the values of A and B in Eq. (14.3)
At the end B,
x = L, y = 0,
so
or
tan mL = mL
(14.6)
Exercise 14.1 A column of length L and flexural rigidity EI is hinged at both the
ends. Derive the expression of Euler’s buckling load.
[Hint: M = −Py, y = A sin mx + B cos mx, End conditions x = 0, y = 0, x = L, y = 0]
Exercise 14.2 Derive the expressions for Euler’s buckling load of a column fixed
at both the ends. Length of column L, buckling load P (Fig. 14.3). (Fixing
moments at fixed ends are MA and MB, respectively)
M at section shown = –Py + MA
Figure 14.3
y = A sin mx+ B cos mx + MA/ P
End conditions
x = 0, y = 0, dy/dx = 0
x = L, y = 0, dy/dx = 0
Equivalent Length
Buckling load of a column is given by Euler’s formula and it depends on the end
conditions of a column. However, to express the buckling load formula in a
simplified manner, equivalent length of column is determined that is equivalent to
the length of column with both ends hinged. This length can be obtained by
completing the bending curve of the column with different end conditions similar
to the bending curve of a column with both the ends hinged as shown in Fig.
14.4(a–d).
Pe = Euler’s buckling load = π2EImin/Le2
Figure 14.4 (a) A column with both ends hinged, Le = L; (b) a column with both
ends fixed, Le = L/2; (c) a column with one end fixed and the other end
hinged,
and (d) a column with one end fixed and the other end free, Le =
2L.
Example 14.3 An allowable axial load for a 3-m-long column with hinged ends is
30 kN. Another column of the same material, same cross-section and same length,
but with one end fixed and the other end hinged, suffers buckling; what is buckling
load for the column?
Solution
Euler’s buckling load,
where Le = L, both ends hinged.
other column with one end fixed and the other end hinged
Exercise 14.3 A column of a circular section, diameter d and length L, buckles at a
load of 25 kN, when the column is fixed at one end and free at the other end. If
both the ends of the column are now fixed, what will be its buckling load?
Limitations of Euler’s Theory of Buckling
Euler’s theory of buckling is applicable for long columns. If there is a limit on the
slenderness ratio of the column and it is less than this limiting value, then Euler’s
formula gives the value of buckling load that is greater than the crushing load,
which is not possible.
Euler’s buckling load,
where
E = Young’s modulus
Imin = Ak2min, where A is the area of cross-section
Le = equivalent length, kmin = the minimum radius of gyration
so
Pe/A < σc (the crushing stress of a short column)
Le/Kmin = the slenderness ratio of a column
Mild stress
E = 208 GPa = 2,08,000 N/mm2
σc = crushing strength
= 320 N/mm2
For mild steel column, the slenderness ratio should be greater than 80, so that
Euler’s formula can be used to predict the buckling load of a column.
Example 14.4 For what length of a mild steel bar of 60 mm in diameter used as
strut, the Euler’s theory is applicable, if the ultimate compressive strength is 0.33
kN/mm2 and E = 210 kN/mm2, when one end of the strut is hinged and the other
end is fixed?
Solution
End conditions, one end hinged and the other end fixed
Equivalent length,
kmin = the minimum radius of gyration
Length of the column,
Length should be greater than 1.68 m.
Exercise 14.4 For what length of a cast iron column of 80 mm in diameter, the
Euler’s theory is applicable, if σc = 550 N/mm2 for CI and E = 102 kN/mm2, the
column is hinged at both the ends?
Higher-Order Differential Equation
For any end conditions, the buckling load of a column/strut can be determined by a
higher order differential equation as follows:
Solution of this differential equation is
y = C1sin kx + C2cos kx + C3x + C4
(14.7)
Constants C1, C2, C3 and C4 are determined using end conditions.
Example 14.5 Let us consider a column with both the ends fixed as shown in Fig.
14.5.
Figure 14.5
Solution
Column AB of length L is fixed at ends A and B. At the ends, both slope and
deflection are zero, that is, the boundary conditions of the column are:
At end A, x = 0, y = 0, dy/dx = 0
At end B, x = L, y = 0, dy/dx = 0
Differentiating Eq. (14.7) with respect to x, we get
dy/dx = C1k cos kx − C2k sin kx + C3
(14.8)
Taking x = 0, y = 0,x = L, y = 0, we get,
0 = C1sin 0 + C2cos 0 + C3x 0 + C4
0 = 0 + C2 + 0 + C4
(14.9)
0 = C1sin kL + C2cos kL + C2L + C4
(14.10)
Taking dy/dx = 0 at x = 0, and x = L we get,
0 = C1k cos 0 − C2k sin 0 + C3
0 = C1 k − 0 + C 3
(14.11)
0 = C1k cos kL − C2sin kL + C3
(14.12)
Using Eqs (14.9), (14.10), (14.11) and (14.12), the following matrix is made:
Let us take 1 − cos kL = 0,
cos kL = 0
kL = 0, 2π, 4π - - So, kL = 2π (minimum significant value)
or,
Squaring both the sides,
Buckling load,
Moreover, sin kL = sin 2π = 0(also), which will also give P = 4π2EI/L2.
Example 14.6 Let us take another case of a column fixed at one end and hinged at
the other end as shown in Fig. 14.6. End A is fixed as shown. End B is hinged.
Figure 14.6
Solution
Boundary conditions
at
x = 0, end A, y = 0, dy/dx = 0
at
x = L, end B, y = 0, d2y/dx2 = 0 (moment at hinged end is zero).
Equation of deflection,
y = C1sin kx + C2cos kx + C3x + C4
(14.3)
At x = 0, y = 0
0 = C1 × sin 0 + C2cos 0 + C3 × 0 + C4
= 0 + C2 + 0 + C4
At
x = 0, dy/dx = 0
0 = C1kcos 0 − C2ksin 0 + C3
(14.16)
0 = C1k + C3
At
x = L, y = 0
0 = C1sin kL + C2cos kL + C3L + C4
At
(14.17)
(14.18)
x = L,d2y/dx2 = 0, that is, bending moment is zero at the hinged end
0 = −C2k2sin kL − C2k2cos kL
(14.19)
Using Eqs. (14.16)–(14.19), the following matrix is made:
0 = kL coskL − sin kL
or
or
tan kL = kL
or
tan 0 = θ
Buckling load,
Exercise 14.5 Using higher-order differential equation, determine the buckling
load for a column hinged at both the ends.
Exercise 14.6 Using higher-order differential equation, determine the buckling
load for a column fixed at one end and free at the other end.
[Hint: say, y = C1sin kx + C2cos kx + C3x + a, where a = deflection at free end]
Rankine Gordon Formula
Euler has developed a theory for the buckling load of long columns, but short
columns get crushed under the compressive load and crushing stress (σc) is much
greater than the stress (σe) given by Euler’s buckling load. Some columns are of
medium length and cannot be classified as short columns. Rankinehas combined
the two loads, that is, the crushing load for short column and the Euler’s buckling
load for long columns to determine the buckling load for any long or short column.
Rankine’s load, PR is
or
Rankine’s load,
where σc = crushing stress,
Rankine has taken σc/π2E a constant that is determined experimentally
Rankine’s load,
both σc and a are determined experimentally.
Le = equivalent length of column
kmin = minimum radius of gyration of the section of the column
Le/Kmin is known as slenderness ratio of the column, an important parameter, and
the buckling load depends on the slenderness ratio; more the slenderness ratio, less
is the buckling load. Table 14.1 gives the values of σc and a for various materials.
Table 14.1 Rankine’s constants for different materials
Example 14.7 A cast iron column of a hollow circular section with an external
diameter of 250 mm and a wall thickness of 45 mm is subjected to an axial
compressive load. The column is 7 m long with both ends hinged. Taking factor of
safety (FOS) as 8, determine safe value of P.
Rankine’s constants are, σc = 560 N/mm2, a = 1/1,600
Solution
External diameter,
D = 250mm
Wall thickness,
t = 45mm
Internal diameter,
d = 250 − 2 × 45 = 160 mm
Area of cross-section,
Length,
L = 7 m = 7,000 mm
End condition: both the ends are hinged
Rankine’s load,
Safe load,
P′R = 286.6/8 = 35.83 kN
Exercise 14.7 A hollow cast iron column has 200 mm outside diameter, 150 mm
inside diameter and is 6 m long with both the ends fixed. It is subjected to an axial
compressive load P. Taking a FOS as 6, determine safe Rankine’s buckling load.
Constants are σc = 550 N/mm2, a = 1/1,600 for both ends hinged.
Johnson’s Parabolic Formula
We have learnt so far that buckling loads depend upon the slenderness ratio
(Le /kmin) of the column. As the slenderness ratio increases, buckling required for
the column decreases. Based on this principle, Johnson has suggested a formula:
Working stress, σw = σc′[1 − φ(Le/k)] as function of Le/k, slenderness ratio
Taking φ(Le/k) = b(Le2/k2), where b is a constant
σw = σc′[1 − b(L2e/k2)], an equation of a parabola.
σ’c = allowable stress in compression taking into account the FOS (factor of
safety)
= 110 N/mm2 for mild steel.
Constant,
b = 0.00003 for hinged ends
= 0.00002 for fixed ends.
Straight line formula
σw′ = σc′[1 − C(Le/k)], applicable for column for which slenderness ratio is
greater than 90.
For structural steel,
σc′ = 140 N/mm2 in compression
C = 0.0054 for hinged ends
= 0.0038 for riveted ends.
Example 14.8 A strut is built-up of two 100 mm × 45 mm channels placed backto-back at a distance of 100 mm apart and riveted to two flange plates each 200
mm × 10 mm symmetrically, properties of one channel are:
Area, A = 7.41 cm2,Ixx = 123.8 cm4,Iyy = 14.9 cm4, and = 1.4 cm (distance of CG
from outer edge of web). If the effective length is 5 m, calculate the working load
for the strut using Johnson’s parabolic formula:
where σw = σc′[1 − b(Le/k2)], where constant b = 0.00003 for hinged ends and
σc′ = 110 N/mm2
Solution
Area of cross-section of built-up section, A
= 2 × 200 ×10 + 2 × 741
= 4,000 + 1,482 = 5,482 mm2
Ixx of built-up section
= 2 ×123.8 ×104 + 2 × 200 ×10 × (55)2
= 247.6 ×104 +1210 ×104
= 1457.6 ×104 mm4
Iyy of built-up section (Fig. 14.7)
Iyy = 2 ×14.9 ×104 + 2 × 741× (50 +14)2 + 2 ×10 × (2002/12)
= 29.6 ×104 + 607 ×104 +1,333.33×104
= 1,970 ×104 mm4
Figure 14.7 Example 14.8
Now,
Exercise 14.8 A stanchion is built up of 3 − 200 × 100 mm Rolled Steel Joist
(RSJ) as shown inFig. 14.8. If the length of the stanchion is 6 m, calculate the
working load. The working stress is given by:
Properties of one RSJ, area = 2,527 mm2, Ixx = 1,696.6 ×104 mm4, Iyy = 115.4
×104 mm4 and web thickness = 5.4 mm.
Figure 14.8 Exercise 14.8
Eccentric Loading of Columns
A column AB of length L loaded eccentrically e (eccentricity with respect to axis
of column) buckled under load P as shown in the Fig. 14.9. Say, deflection at end
B is a, when the column has buckled. Consider a section at a distance x from end
A, where deflection is y.
Figure 14.9
Bending moment at the section,
M = P(a + e − y)
or,
EI(d2y/dx2) = P(a + e − y)
or,
Solution of this differential equation is,
y = A sin mx + Bcos mx + (a + e)
where
m=
At end A,
x = 0, y = 0 dy/dx = 0
A and B are constants
Pulling
x = 0, y = 0
0 = Asin 0 + Bcos 0 + (a + e)
0 = 0 + B + (a + e)
B = −(a + e)
or constant,
Moreover,
dy/dx = Am cos mx − Bm sin mx
dy/dx = 0, at x = 0
0 = Am cos 0 − Bm sin 0
0 = Am
m ≠ 0, because m =
therefore, constant A = 0.
Finally,
At the end B,
y = −(a + e) cos mx + (a + e)
x + L, y = a, putting this boundary condition
a = −(a + e)cos mL + (a + e)
or,
−e = −(a + e) cos mL
(a + e) = e sec mL
Maximum bending moment occurs at fixed end A.
Mmax = P(a + e) = Pe sec mL
σb = maximum stress due to bending
= P e sec mL/ , secant formula,
where
= section modulus = I /yc,
I = moment of inertia
yc = distance of extreme layer in compression from neutral layer
σc = direct compressive stress.
Resultant stress at fixed end,
= P/A
In this case, we have considered that one end of the column is fixed and the other
end is free. In this case, Le = 2L, that is, equivalent length, Le = 2L. Secant formula
can be generalized for any column with any type of end conditions as follows:
σr = resultant maximum stress
where Le is the equivalent length.
Example 14.9 A steel tube of 80 mm outer diameter, 50 mm inner diameter and 3
m long is used as a strut with both ends hinged. The load is parallel to the axis of
the strut but is eccentric. Find the maximum value of eccentricity so that the
crippling load on strut is equal to 50 per cent of the Euler’s crippling load.
Yield strength = 320 N/mm2, E = 210 kN/mm2.
Solution
End conditions: both ends hinged
Length,
Le = L = 3 m
E = 210 kN/mm2
Steel tube
Euler’s buckling load,
Crippling load due to eccentricity, P = Pe /2 = 1,96,184.7 N
Eccentricity, e = (320 - 64.05)/10.37 = 24.68 mm
Exercise 14.9 A steel tube of 80 mm outer diameter, 60 mm inner diameter and 2.8
m long is used as a strut with the ends hinged. The load is parallel to the axis of the
strut but is eccentric. Find the maximum value of the eccentricity so that the
crippling load on the strut is 60 per cent of the Euler’s buckling load. Given yield
strength = 320 N/mm2.
E = 210 kN/mm2
Professor Perry’s Approximate Formula
Professor Perry has given an approximate formula for the buckling load of
eccentrically loaded columns.
In the secant formula for maximum stress due to buckling load,
If Euler’s buckling load,
Pe = π2EI/L2e
putting the value in secant formula,
Professor Perry found through experiments that the expression,
where
Pe = Euler’s buckling load (without eccentricity)
P = buckling load of column with eccentricity.
Moreover,
σe = Pe /A, σo = P/A
where A is the cross-sectional area of the column.
Now,
Section modulus, = Ak2/yc, where yc is distance of extreme layer in
compression from neutral layer.
However,
This is Professor Perry’s approximate formula, knowing the value e, yc and k,
one can determine the stress σo due to load P on the eccentrically loaded column.
Example 14.10 A column is of I-section, 125 mm × 250 mm. Find the safe load
for this column of length of 5 m, hinged at both the ends using Professor Perry’s
formula if maximum compressive stress is limited to 80 N/mm2 (Fig. 14.10).
For RSJ,
Ixx = 37.18 ×106mm4, A = 3,553 mm2, Iyy = 1.93×106mm4
E = 200 GPa, eccentricity from y-axis is 30 mm.
Solution
Eccentricity is about x-x axis, e = 30 mm, we will take Iyy as moment of inertia
Length,
L = 5,000 mm
Iyy = 1.93 × 106 mm4
Figure 14.10
Safe load
= 38.10 × 3,553
= 1,35,369 N
= 135.37 kN
Exercise 14.10 A 400 mm × 140 mm RSJ is used as a sturt with hinged ends,
having a length of 6 m. Using Prof. Perry’s formula, determine the safe load for:
(a) eccentricity along x-x axis, e = 24 mm, (b) minimum allowable permissible
compressive stress is 75 MPa and (c) for the joist A = 7,846 mm2, Ixx = 20,458.4 ×
104 mm4,
Iyy = 622.1×104 mm4, E = 210 kN/mm2
Long Columns with Eccentricity in Geometry
A column AB of length L with both the ends hinged has initial eccentricity e′ in the
centre. Assuming deflection curve to be sinusoidal (Fig. 14.11).
Eccentricity at any section,
e′ = maximum initial eccentricity at centre.
Taking a section at a distance of x from end A, considering column buckled
under load P.
Final deflection = y
Change in deflection = y-y′ (as shown)
BM at the section at a distance x from A as shown
Figure 14.11
as given in Eq. (14.21)
Putting this value in Eq. (14.22)
Let us assume the solution as
Putting this value in Eq. (14.24),
Putting the assumed value of y
However,
π2EI/L2 = Pe, Euler’s buckling load
Finally, the deflection curve equation will be,
Maximum deflection occurs at the centre, where
Maximum bending moment,
σmax = maximum compressive stress at central section of column
where
k = radius of gyration
yc = distance of extreme layer
However, Pe = σeA, Euler’s load.
P = σeA, putting these values; load applied.
Example 14.11 A 5-m-long hollow circular steel strut having an outside diameter
of 120 mm and an inside diameter of 80 mm with both the ends hinged is initially
bent. Assume that the centre line of the strut as sinusoidal with maximum deviation
of 6 mm. Determine the maximum stress developed due to an axial load of 100 kN.
E = 210 kN/mm2
Solution
Length,
L=5m
Eccentricity,
e′ = 6 mm
Area of cross-section,
A = (π /4)(1202 - 802) = 6283.2mm2
Moment of inertia,
I = (π /64)(1204 — 804)
= 816.816 × 104 mm4
Distance,
yc = 120/2 = 60 mm
e′ = 6 mm
Exercise 14.11 A 4-m-long hollow circular aluminium strut having an outside
diameter of 150 mm and an inside diameter of 100 mm with both ends hinged is
initially bent. Assuming that the centre line of the strut as a sinusoidal curve with
maximum deviation of 8 mm. Determine the maximum stress developed due to an
axial load of 60 kN.
E = 70 kN/mm2
Professor Perry Robertson Formula
Professor Perry gave the relationship between σe, σo and σmax (discussed in the last
section ‘Long Columns with Eccentricity in Geometry’) for a long column with
eccentric loading as follows:
where = eccentricity in loading (load is not axial)
Moreover,
where e′ = eccentricity in geometry of strut
e1 = 1.2e + e′
say,
For an initially bent and eccentrically loaded column, formula has been modified
as
where σmax = allowable stress, σ
say
e1 yc/k2 = λ, then
Rearranging the terms, we get,
σo2 − σo [σ + σe(1 + λ)] + σ σe = 0,
(14.29)
a quadratic in σo.
where λ = e1 yc / k2 and σe = πE/(L2 k2)for both ends hinged.
Professor Robertson after having experimental observations came to the
conclusion that λ = 0.003(L/k) is valid for a large number of observations.
However, λ depends on allowable stress in tons/in2 and σe = π2E/(L/k)2 in
tons/in2.
However, in the Eq. (14.28), there are ratios of σ/σ0, and σo/σe,
dimensionless λ = e1 yc/k2 also dimensionless, the expression for λ remains the
same for stresses in N/mm2.
Example 14.12 Two 200 × 70 mm 6-m-long MS channels are welded together at
their toes to form a box section of 200 × 140 mm.
The box section is used as a strut with both the ends hinged. Estimate the safe load
for this strut using Prof. Perry Robertson’s formula taking allowable stress, σ = 250
MPa, λ = 0.003(L/K)(Fig. 14.12).
Figure 14.12 Built-up section
For each channel
Area of cross-section,
A = 1,777 mm2
Ixx = 1,161.9 × 104 mm4, Iyy = 84.2 × 104 mm4
= 19.7 mm, E = 210 kN/mm2
Solution
Total,
A = 2 × 1,777 = 3,554 mm2
Ix′x′ = 2 × 1,161.9 × 104 mm4
= 2,323.8 × 104 mm4
Iy′y′ = 2 × 84.2 × 104 + 2 × 1,777(70 − 19.7)2
= 168.4 × 104 + 899.2 × 104
= 1,067.6 × 104 mm4
Note that
Length of the strut,
Allowable stress,
Exercise 14.12 A stanchion is built up of a 500 mm × 180 mm RSJ with 200 mm ×
20 mm plates riveted to each flange as shown in Fig. 14.7 of example 14.8.
Estimate the safe load for this stanchion of a length of 5 m, with both the ends
hinged using Prof. Perry Robertson’s formula, taking allowable stress, σ = 280
MPa, λ = 0.003(L/k) for the joist.
A = 9,550 mm2,
Ixx = 3.8529 × 106 mm4
Iyy = 10.639 × 106 mm4
E = 208 GPa
Hint: [Taking Iyy, calculate, k and σe]
Lateral Loading of Strut with Point Load
A strut or column has buckled under the combined action of axial thrust P and
transverse load (at centre) as shown in Fig. 14.13. Lateral loading produces
deflection in the strut and axial thrust produces a bending moment on account of
deflection. Figure 14.13 shows a column AB of length L, hinged at both the ends
and carrying a central transverse load W as shown. Due to W at C, there are
horizontal support reactions, W / 2each.
Consider a section at a distance x from end As
Bending moment at the section,
or
Figure 14.13 Lateral point load on a strut
or
M = −Py − (Wx/2)
Solution of this differential equation is
where
Boundary conditions are x = 0, y = 0,
0 = A cos 0 + B sin 0 − 0
0=A+0−0
or constant,
A=0
Moreover,
Equation of deflection curve becomes
Maximum deflection occurs at centre at x = L/2
Maximum bending moment occurs at the centre
putting the value of m
yc = distance of extreme layer in compression from neutral layer
or,
Example 14.13 A 4-m-long horizontal pin-ended strut is formed from a standard
T-section of 150 mm × 100 mm × 12.5 mm. The axial compressive load is 60 kN.
A lateral concentrated load of 6 kN acts at the centre of the strut. Find the
maximum stress developed if the xx axis is horizontal and the table of the T-section
forms the compressive face. The centre of gravity is 24 mm away from the edge of
the table (Fig. 14.14).
Given
Ixx = 250 × 104mm4, A = 3,100 mm2, E = 200 GPa
Figure 14.14
Solution
Axial load,
P = 60 kN
Lateral load,
W = 6 kN
Ixx = 250 × 104 mm2
A = 3,100 mm2
Length,
L=4m
T-section,
yc = 24 mm
Exercise 14.13 A circular steel strut of a diameter of 25 mm and length of 1 m is
subjected to an axial thrust of 12 kN. In addition a lateral load W acts at the centre
of the strut. If the strut is to fail at σmax = 320 N/mm2, determine the magnitude
of W. Given E = 210 kN/mm2.
Strut with an Uniformly Distributed Lateral Load
A strut subjected to axial thrust P and lateral load w per unit length is shown
in Fig. 14.15. The length of the strut is L and it is hinged at both the ends A and B.
Consider a section at a distance x from end A. The bending moment at the section,
Figure 14.15
Solution of this differential equation, y = complementary function + particular
integral.
Complementary function = Acos mx + Bsin mx, where
.
Particular integral,
or constant,
A = wEI / P2
(14.31)
Then,
dy/dx = 0 at x = L/2, at the centre of the beam due to symmetrical loading about
centre C,
However,
A = wEI / P2
So constant,
Finally, the equation of y
At the centre, deflection is maximum, therefore, taking
Bending moment at the centre,
Maximum stress,
Example 14.14 A circular rod of diameter 50 mm is supported horizontally
through pin joints at its ends and carries an uniformly distributed load of 1 kN/m
run throughout its length and an axial thrust of 25 kN. If its length is 2.4 m,
estimate the maximum stress induced in the rod, E = 200 GPa.
Solution
Axial thrust, P = 25,000 N
Diameter, d = 50 mm
Area,
A = (π/4)d2 = 1963.5 mm2
w = 1 kN/m = 1 N/mm
E = 2,00,000 N/mm2
yc = 25 mm
Length,
L = 2.4 m
Exercise 14.14 A rod of rectangular section 80 mm × 40 mm is supported
horizontally through pin joints at its end and carries a vertical load of 3,000 N/m
and an axial thrust of 100 kN. If its length is 2.0 m, estimate the maximum stress
induced.
E = 208 × 103 N/mm2
Energy Approach
To determine the critical load at which a column ceases to be in a stable
equilibrium, energy criterion can be used. Moreover, in a situation where the exact
solution of a differential equation is not possible or difficult to obtain, energy
approach can be used to get a good approximate solution. However, this approach
converges to the exact solution if a large number of terms are taken to represent the
deflection curve.
Now,
Energy stored in a system = work done by external loads
δU + δVe = 0
δVe = external work done due to virtual displacement.
δ(U+Ve) = 0
δ(Kp) = 0
U + Ve = Kp = a constant referred to as total potential of the system.
Energy stored,
External work done,
For small values of P, Kp is positive for any non-trivial admissible function y(x).
The critical load can be obtained by assuming suitable admissible function.
Critical condition is reached when constant Kp = 0
Example 14.15 Let us consider a case of column that is fixed at one end and free
at the other end, with boundary conditions (Fig. 14.16).
Figure 14.16
Solution
Deflection, y = 0 and x = 0
slope, dy/dx = 0, x = 0
Moment at end B is zero, that is,
Let us take a shape function,
Putting these values in Eq. (14.32)
Pcr = 3EI / L2, but the exact value given by Euler’s theory is π2EI / 4L2 =
2.467(EI /L2)
Let us take more terms in the deflection shape function
Putting these values in the equation of total potential Kp,
Exercise 14.15 Consider a column with both ends hinged as shown in Fig. 14.17.
Take the shape function y = A(x4 – 2Lx3 + L3x), using energy approved, that is,
total potential, derive expression for its buckling load.
Problem 14.1 A column made of an aluminium alloy of rectangular
section b × d and a length of 600 mm is fixed at end B as shown in Fig. 14.18. Two
smooth and rounded fixed plates at the end A restraint the end A to move in one of
the vertical planes of symmetry of the column. Determine the ratio of b/d of two
sides of cross-section for the efficient design against buckling.
E = 67 GPa, P = 15 kN, FOS = 2
Figure 14.17 Exercise 14.15
Solution
For the efficient design, the buckling load for two possible modes of failure should
be the same. There are two types of end conditions in the present case:
1. one end is fixed and the other end is free,
I1 = bd 3/12.
2. one end is fixed and the other end is hinged, I2 = db 3/12.
Figure 14.18
Taking Euler’s formula for buckling
Design load = 2 × 15 = 30 kN
Length = 600 mm
Dimensions of the section are 13.63 mm × 8.6 mm.
Problem 14.2 A long strut AB of length L is of uniform section throughout. A
thrust P is applied at the ends eccentrically on the same side of the centre line with
eccentricity at the end B twice than that at end A (Fig. 14.19). Show that the
maximum bending moment occurs at a distance x from end A, where
Figure 14.19
Solution
The strut is buckled under the thrust load P.
P at two ends produce a couple P(2e − e) (cw) that is to be balanced by
anticlockwise couple of horizontal reactions F each at A and B, as
shown F × L (ccw).
Force F is unknown.
so 2Pe − Pe − FL = 0
or,
Consider a section at a distance x from end A,
Bending moment at the section, M = –P(2e + y) + F(L – x)
or
Solution of this differential equation is,
where A and B are constants,
End conditions
At end A, x = 0, y = 0, putting this in Eq. (14.36)
0 = A cos 0 + B cos 0 – e – 0
Constant,
A=e
Moreover, at end B,
k = L, y = 0
0 = Acos kL+ Bsin kL – 2e
However,
A=e
so,
0 = e cos kL + Bsin kL – 2e
Constant,
Finally, the equation of deflection becomes,
Bending moment at the section,
For maximum bending moment, dMx /dx = 0
Problem 14.3 A strut of length L is fixed at its lower end, its upper end is
eccentrically supported against a lateral deflection (through a spring), so that the
resisting force is k times the end deflection, kis spring constant. Show that the
crippling load P is given by 1 – (P/kL) = tan mL/mL.
where,
Figure 14.20
Solution
Figure 14.20 shows a strut AB of length L, fixed at end A and free at end B. At the
end B, a horizontal reaction H = ka, given by a spring and P is the crippling load.
Consider a section, at a distance x from end A.
Bending moment at the section, M = P(a − y) − ka(L − x)
Solution of the differential equation
At end A, fixed end, x = 0, y = 0
Then,
or constant, B = –(ka/mP)
Finally, the equation of deflection curve is
At the end B,
or
Problem 14.4 A thin steel bar of a rectangular section 6 mm × 4 mm is axially
compressed by 200-N load between two plates that are fixed at a constant distance
of 160 mm apart as shown in Fig. 14.21. The assembly is made at 24°C, how high
can the temperature of the bar rise so as to have an FOS of 2 with respect to
buckling. E = 200 GPa, α = 15 × 10−6/°C?
Figure 14.21
Solution
Section = 4 × 6 mm
End conditions: both ends hinged.
L = 160 mn
Euler’s buckling load,
Initial compressive load = 200 N
Balance load to be applied through expansion of column (prevented by fixed ends)
= 1,233.7 − 200 = 1,033.7 N
Say temperature rise is ∆T
Thus
αΔT EA = 1033.7 N
where A = area of cross-section of column
15 × 10−6 × ΔT × 2,00,000 × 24 = 1,033.7 N
Final temperature,
= 24 + 14.36 = 38.36°C.
Problem 14.5 A 6-m-long RSJ of a rectangular section of 325 mm × 165 mm is
used as a strut, with one end is fixed and the other end is hinged. Calculate the
crippling load by Rankine’s formula. Compare this with the load obtained by
Euler’s formula. For what length of this strut will the two formulae give the same
crippling load for the joist?
Area of cross-section,
A = 5,490 mm2
Ixx = 9,874.6 × 104 mm4
Iyy = 510.8 × 104 mm2
E = 210 kN/mm2
σc = 320 N/mm2
a=
for both hinged ends.
Solution
Iyy < Ixx
L=6m
(one end is fixed and the other end is hinged)
Euler’s buckling load.
Rankine’s load,
Equating two loads,
Putting the values,
L = 8.48 × 103m, for this length both Pe and PR are equal
L = 8.48 m
Problem 14.6
A round bar in a vertical position is clamped at the lower end and is free at the
other end. The effective length is 2 m. If a horizontal force of 400 N at the top
produces a horizontal deflection of 15 mm, what is the axial buckling load for the
bar under the given conditions?
Figure 14.22 Problem 14.6
Solution
The bar is fixed as a cantilever beam (as shown Fig. 14.22).
Deflection at the top
Buckling load, Pe = π2 × EI /4L2, as one end is fixed and the other end is free
Problem 14.7
A 3-m-long straight steel bar of a rectangular section of 30 mm × 20 mm is used as
a strut with both the ends hinged. Assuming that Euler’s formula is applicable and
the material attains its yield strength at the time of buckling, determine the central
deflection. E = 210 kN/mm2. Yield strength of steel is 320 N/mm2.
Solution
Section 30 × 20 mm2,
A = 600 mm2
E = 2,10,000 N/mm2
L = 3,000 mm (both the ends are hinged)
Euler’s buckling load,
Say, central deflection, ymax.
Problem 14.8
A steel column of a hollow circular section of 200 mm external diameter, 160 mm
internal diameter and 7-m-long has to take a 20-kN load at an eccentricity of 30
mm from the geometrical axis. If the ends are fixed, calculate the maximum and
minimum stress intensities induced in the section, taking E= 210 kN/mm2.
Moreover, calculate the maximum permissible eccentricity so that no tension is
induced any where in the section.
Solution
Section
D = 200 mm
d = 160 mm
A=
(2002 − 1602) = 1.131 × 104mm2
I=
(2004 − 1604) = 0.4637 × 108mm4
P = 200 kN
E = 2,10,000 N/mm2
L=7m
Ends are fixed,
Le = L/2 = 3.5 m = 3,500 mm
yc = 100 mm
Eccentricity,
e = 30 mm
For no tension
Eccentricity,
= 34.3 mm
Key Points to Remember
o
Euler’s buckling load, Pe = π2EImin /Le2
where Imin = minimum moment of inertia
o
Le = equivalent length of a column depending upon end conditions.
Euler’s formula is applicable for the slenderness ratio
,
where σc is ultimate compressive strength of the material.
Higher-order differential equation
o
Solution, y = C1sin kx + C2 cos kx + C3x + C4
o
using different end conditions, the buckling load is determined.
o
Rankine’s load =
where
σc = crushing strength of column
Le = equivalent length,
kmin = minimum radius of gyration
o
Johnson’s parabolic formula for working stress
where σc′ = allowable stress in compressive taking into account FOS
b = constant
o
Eccentrically loaded column
where
P = axial applied load
Z = section modulus
e = eccentricity
I = moment of inertia
o
Professor Perry’s formula
where
σ = maximum stress allowed
σ0 = P/A, applied load/area
σe = Pe /A, Euler’s load/area
e = eccentricity
yc = distance of extreme layer in compression from neutral layer
o
If the column has initial eccentricity e′ along central section
o
Energy approach, total potential, Kp = 0
to determine Pcr, critical load.
Review Questions
1. What are the drawbacks of Euler’s theory of buckling?
2. What is limiting value of the slenderness ratio beyond which Euler’s
formula is applicable?
3. What are the merits of Rankine’s load over Euler’s load in buckling?
4. What do you mean by equivalent length of a column?
5. Discuss the effect of the slenderness ratio of a column over buckling load?
6. Why the value of ‘a’ Rankine’s constant varies for different materials?
7. What is straight line formula for working stress under buckling, where it is
used?
8. What approximation is taken by Professor Perry to modify the secant
formula for eccentrically loaded column?
9. What do you understand by total potential constant in energy approach?
Multiple Choice Questions
1. A column fixed at one end and free at the other end buckles at a load P.
Now, both the ends of the column are fixed. What is the buckling load for
these end conditions?
1. 16 P
2. 8 P
3. 4 P
4. 2 P
2. In Rankine’s formula ‘a’ is used, what is its value for cast iron?
0.
1.
2.
3. None of these
3. What is the approximate value of σc for cast iron/σc for mild steel?
0. 0.6
1. 1.0
2. 1.7
3. None of these
4. A hollow circular column, with D = 100 mm, d = 80 mm, what is radius of
gyration?
0. 32
1. 24
2. 19.4
3. None of these
5. In Johnson’s parabolic formula, allowable stress in compression for mild
steel is
0. 270 N/mm2
1. 110 N/mm2
2. 80 N/mm2
3. None of these
6. A 3-m-long column is hinged at one end and fixed at the other end, what is
its equivalent length?
0.
1. 3 m
2.
3. None of these
7. A column of a length of 2.4 m, an area of cross-section of 2,000 mm2 and
moment of inertia of Ixx= 720 ×104mm4 and Iyy = 80×104mm4 is subjected
to buckling load. Both the ends of the column are fixed. What is the
slenderness ratio of column?
0. 120
1. 80
2. 60
3. 40
8. The ratio of equivalent length of a column with one end fixed and the
other end free to its own length is
0. 2
1. 1.0
2. 0.5
3. None of these
9. Euler’s buckling theory is applicable for
0. Short columns
1. Long columns
2. Medium long columns
3. All of them
10.In Johnson’s parabolic formula, what is constant b for hinged ends, for
mild steel
0. 2 × 10−5
1. 3 × 10−5
2. 0.005
3. None of these
Practice Problems
1. What size of a steel pipe should be used for the horizontal member of a jib crane
shown in Fig. 14.23 for supporting the maximum force of 20 kN. Use an FOS of
2.5. The internal diameter of the pipe is 0.8 times the external diameter. E = 200
GPa.
Figure 14.23 Practice problems 14.1
2. Find the shortest length for a steel column with hinged ends having a crosssectional area of 30 mm × 60 mm for which the elastic Euler’s formula is
applicable. E = 200 GPa and assume proportional limit to be 250 MPa.
3. A short-length of a tube of 30 mm internal diameter and 40 mm external diameter
failed under a compressive load of 180 kN. When a 2-m length of the same tube is
tested as a strut with both the ends hinged, the buckling load was found to be 40.8
kN. Assuming that σc, for the Rankine’s constant is given by first test, determine
Rankines’ constant ‘a’. Moreover, estimate the crippling load for a piece of 3-mlong tube when used as strut with fixed ends.
4. A strut of a 2-m-long aluminium alloy has a rectangular section of 20 mm × 50
mm. A bolt through each end secures the strut so that it acts as a hinged column
about an axis perpendicular to the 50-mm dimension and as fixed-ended column
about an axis perpendicular to the 20-mm dimension. Determine the safe central
load using a FOS of 2. E = 70 GPa.
5. A simple beam of flexural rigidity EIB is propped up at the middle by a slender rod
of flexural rigidity EIC. Estimate the deflection of the beam at the centre if a
force P double the Euler’s buckling load for the column is applied to the system as
shown in Fig. 14.24.
Figure 14.24
6. A thin vertical strut of uniform section and length L is rigidly fixed at its bottom
end and its top end is free. At the top end, there is a horizontal force H and a
vertical load P acting through the centroid of the section as an axial load. Prove
that horizontal deflection at the top is
7. A horizontal bar CD is supported by a pin-ended column AB, 2.5 m long and 40
mm diameter as shown in Fig. 14.25. Calculate the allowable load P, if FOS with
respect to buckling of column is 3.0.
E = 200 GPa
Figure 14.25
8. A hollow cast iron column of external diameter of 200 mm and length of 4 m with
both ends fixed supports an axial load of 800 kN. Find the thickness of the metal
required, use Rankine’s constants,a = 1/1,600 for both ends hinged, working
stress, σw = 80 N/mm2.
9. A 1.2-m-long hollow circular steel pipe of 60 mm outside diameter and 50 mm
inside diameter is fixed at both the ends so as to prevent any expansion in its
length. The pipe is unstressed at the normal temperature. Calculate the temperature
stress in the pipe and the FOS against failure as a strut if the temperature rises by
40°C. Use Rankine’s formula.
10. A 5-m-long steel strut, with I = 50 × 104 mm4, carries thrust load P = 20 kN, with
eccentricities e= 10 mm on one side and 2e = 20 mm on the other side. E = 200
GPa. Determine the distance from one end, where the bending moment on strut is
maximum.
Answers to Exercises
Exercise 14.1:
Exercise 14.2:
Exercise 14.3: 400 kN
Exercise 14.4: L > 0.856 m
Exercise 14.5:
Exercise 14.6:
Exercise 14.7: 516.3kN
Exercise 14.8: 466 kN
Exercise 14.9: 12.07 mm
Exercise 14.10: 128 kN
Exercise 14.11: 8.05 N/mm2
Exercise 14.12: 2,180 kN
Exercise 14.13: W = 1.340 kN
Exercise 14.14: 161.70 N/mm2
Exercise 14.15: 9.88 EI/L2
Answers to Multiple Choice Questions
1. (a)
2. (c)
3. (c)
4. (a)
5. (b)
6. (c)
7. (c)
8. (a)
9. (b)
10. (b)
Answers to Practice Problems
1. D = 55.47 mm; d = 44.37 mm
2. L = 769 mm
3.
4. safe load = 11.50 kN
5.
6.
7.
8.
9.
5.29 kN
32.3 mm
92.35 MPa, 3.17
x = 2.859 m
2b. Bending of Curved Bars
CHAPTER OBJECTIVES
There are many engineering components which are not straight but curved and
possessing large initial curvatures. Flexure formula of theory of simple bending
cannot be used to determine stresses in such components. A theory has been
developed for the determination of stresses and deflections in such components. In
this chapter, students will learn about stresses and deflections in components as:
Curved bars of different sections as rectangular, circular,
trapezoidal and triangular.
o
Chain links.
o
Rings.
o
Crane hooks.
o
Frames of power presses, C-clamps.
o
Components with hollow section.
o
Since the derivations are lengthy and cumber some, students are advised to derive
the relationships by themselves after studying the chapter.
Introduction
Components with a large curvature are of engineering use in lifting machine and
conveyor equipments. Components such as crane hook, chains and links used for
lifting machines are designed with safety considerations. Therefore, determination
of stresses in such components and location of critical sections are of utmost
importance.
The analysis of stresses in such components is quite complex. The theory of
simple bending cannot be used for such components because in simple bending,
the component is considered as initially straight.
In this chapter, we will analyse stresses and deflection in components such as
rings, links, crane hooks, frame of punch presses and eccentrically loaded members
with large initial curvature, etc.
In the chapter on ‘Theory of simple bending’, we assumed the beam to be
initially straight before application of a bending moment and derived the
relationship M/I = E/R = σ/y, and studied about the stresses and deflections
developed in beams. But, in this chapter, we will study about the effect of bending
moment on bars of large initial curvature.
Stresses in a Curved Bar
Figure 17.1 shows a portion of a curved bar of initial radius of curvature R,
subtending an angle q at the centre of curvature O. This curved bar is subjected to a
bending moment M tending to increase the curvature of the bar. To find out the
stresses developed in the bar, a relationship between bending moment M, radius of
curvature R and dimensions of the section of the bar are derived by taking the
following assumptions:
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Figure 17.1
1. The transverse sections of the bar, which are plane before application of a
bending moment, remain plane after the application of bending moment.
2. The material obeys Hooke’s law and stress is directly proportional to
strain.
Consider a small portion IJHG of the curved bar in its initial unstrained position,
where AB is a layer at a radial distance of y from the centroidal layer CD, that is, a
layer passing through the centroidal axis of the sections. At layer AB, stresses due
to the bending moment M are to be determined.
After application of the bending moment, say, I′J′H′G′ is the final shape of the
bar. The centroidal layer is now C′D′ and the layer AB takes the new position A′B′.
Say, the final centre of curvature is O1and final radius of curvature is R1 and θ1 is
the angle subtended by the length C′D′ at the centre.
Say, σ is the stress in the strained layer A′B′ under the bending
moment M tending to increase the curvature (or tending to reduce the radius of
curvature), and e is the strain in the same layer.
Strain,
where y1 is the distance between centroidal layer C′D′ and layer A′B′, in the final
position.
or,
Moreover, ε0 is the strain at the centroidal layer, that is, when y = 0
or,
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and
Dividing Eq. (17.1) by Eq. (17.2),
or,
or,
Now, y1 ≅ y considering that change in thickness is negligible.
Therefore, strain,
Here, we have added and subtracted a term
or,
The stress in the layer AB, which is tensile as is obvious from the diagram, that
is, layers below the centroidal layer are in tension and layers above the centroidal
layer are in compression for the bending moment shown.
Stress,
where E is the Young’s modulus of the material.
Total force on the section, F = ∫σ dA
Considering a small strip of elementary area dA, at a distance of y from the
centroidal layer CD.
where A is the area of cross-section of the bar.
Now, the total resisting moment will be given by
Because ∫ ydA = 0, that is, first moment of any area about its centroidal layer is
zero.
Therefore,
Let us assume that
a quantity which depends upon the
disposition of section and the radius of curvature.
Therefore,
From Eq. (17.4),
because the bar is in equilibrium and the net force on the section is zero as no
force is applied, only moment is applied.
Now,
Considering Eq. (17.4) again,
Substituting this value of
in Eq. (17.5)
Substituting the value of ε0 in the equation for stress,
Substituting the value of ε0 again from Eq. (17.7),
On the other side of the centroidal layer y will be negative as for the
layer EF shown in the figure.
σ′ = stress where y is negative
The expressions given in Eqs (17.8) and (17.9) are for the stresses due to the
bending moment which tends to increase the curvature. If the bending moment
tends to straighten the bar or tends to decrease the curvature,
then θ1 < θ and R1 >R and the stresses will be reversed.
Bending moment tending to decrease the curvature
For y to be positive (away from centre of curvature)
On the other side of the centroidal layer, where y is negative (towards centre of
curvature)
Ah2 for a Rectangular Section
Figure 17.2 shows the rectangular cross-section of breadth B and depth D of a
curved bar with radius of curvature R, i.e., the radius from the centre of
curvature C to the centroid G of the section. Consider a strip of thickness dy at a
distance y from the centroidal layer. Area of the strip, dA = Bdy.
Now, area,
A = BD
Figure 17.2
where R is the radius of curvature up to the centroidal layer, R1 is the radius up to
the inner surface of the curved bar and R2 is the radius up to the outer surface of the
curved bar.
Example 17.1 A circular ring of rectangular section with a slit is loaded as shown
in Fig. 17.3. Determine the magnitude of the force P if the maximum resultant
stress along the section ab does not exceed 100 N/mm2. Draw the stress
distribution diagram along ab.
Figure 17.3
Solution
Mean radius of curvature,
R = 11 cm
Radius of curvature of inner surface,
R1 = 8 cm
Radius of curvature of outer surface,
R2 = 14 cm
Breadth,
B = 4 cm
Depth,
D = 6 cm
Maximum resultant stress will occur at the inner radius along section ab, that is, at
the point b.
Bending moment,
M = P × R = 11P N-cm
Direct stress,
Resultant stress at the point
Therefore,
Stress distribution (along Gb)
where y varies from 0 to 3, compressive stressees
Stresses distribution Along Ga where y varies from 0 to 3
Resultant stress,
Since, the direct stress is compressive
tensile stresses
Figure 17.4 shows the stress distribution along the radial thickness ab of the
section which has maximum bending moment PR. In this case the resultant stress
at centroid G is zero.
Figure 17.4
Exercise 17.1 A curved bar of rectangular cross-section 40 mm × 60 mm is
subjected to a bending moment, tending to increase the curvature of the bar. Radius
of curvature of the bar is 200 mm. If the maximum stress developed in the bar is
100 N/mm2, what is the magnitude of bending moment? Due to this bending
moment what is the stress at the CG of the section?
Value of h2 for Sections Made Up of Rectangular Strips
Sections such as T, I and channel section are made up of rectangular strips, the
value of h2 for each section can be determined by considering each strip separately.
For a single strip,
where B is the breadth, R2 is the radius at outer fibres and R1 is the radius of the
inner fibres of the section from the centre of curvature. Using this expression, let
us determine h2/R2 for various sections.
1. T-section: Figure 17.5 shows a T section with following dimensions:
Breadth of the flange = B
Breadth of the web = b
Radius of curvature up to centroid G of the section = R
Radius up to extreme outer edge of web = R1
Radius up to inner edge of flange = R2
Radius up to outer edge of flange = R3
Area of cross-section of T section = A
Figure 17.5
2. I-section: Figure 17.6 shows an I section with flange and web of
breadths B and b, respectively.
R = Radius of curvature up to centroid G of the section
R1 = Radius up to outer edge of inner flange
R2 = Radius up to inner edge of inner flange
R3 = Radius up to inner edge of outer flange
R4 = Radius up to outer edge of outer flange
A = area of cross-section = B(R4 − R3) + b(R3 − R2) + B(R2 − R1)
Figure 17.6
3. Channel Section: Figure 17.7 shows a channel section with
B = breadth of web
b = breadth of flanges
R = Radius of curvature up to centroid G of the section
R1 = Radius up to inner surface
R2 = Radius up to outer edge of web
R3 = Radius up to the outer edge of flange
A = Area of cross-section = B(R2 − R1) + 2b(R3 − R2)
Figure 17.7
Example 17.2 A curved beam whose centroidal line is a circular arc of 12 cm
radius. The cross-section of the beam is of T shape with dimensions as shown
in Fig. 17.8. Determine the maximum tensile and compressive stresses set up by a
bending moment of 70,000 N cm; tending to decrease the curvature.
Solution
Figure 17.8 shows the curved bar with T section subjected to a bending
moment M tending to decrease the curvature. Therefore, there will be tensile
stresses between A to G, and compressive stresses between G to B.
Let us first calculate the distance of centroid from the outer edge of web.
Figure 17.8
Radius of curvature,
R = 12 cm (given).
Radius up to inner surface,
R1 = 12 −1.864 = 10.136 cm
Radius up to outer edge of flange,
R2 = 11.136 cm.
Radius up to outer edge of web,
R3 = R1 + 6 = 10.136 + 6 = 16.136 cm
Maximum compressive stress at point B
Maximum tensile stress at point A
Example 17.3 Figure 17.9 shows a press applying a 150-kN force on a job.
Determine the stresses at the points a and b. The section is hollow as shown.
Figure 17.9
Solution
Let us first determine the position of the centroid
Radius of curvature,
R = 24 + 13 = 37 cm
Area of cross-section,
A = 24 × 6 + 2 × 4 × 20 + 4 × 16 = 398 cm2
Bending moment,
M = Force × (60 + R) = 150 × 97 kN cm, where R = 37 cm
Direct tensile stress,
Bending stress due to M at
Bending stress due to M at
Resultant stress at the point a
= 6.869 + 0.408 = 7.277 kN/cm2
= 72.77 N/mm2 (tensile)
Resultant stress at the point b
= 5.682 − 0.408 = 5.274 kN/cm2
= 52.74 N/mm2 (compressive)
Exercise 17.2 An open ring of channel section is subjected to a compressive force
of 25 kN as shown in Fig. 17.10. Determine the maximum tensile and maximum
compressive stress along the section ab.
Exercise 17.3 A load P = 5 kN is applied on a C-clamp as shown is Fig. 17.11.
Determine the stresses at the points a and b.
Figure 17.10 Example 17.2
Figure 17.11 Example 17.3
Ah2 for a Trapezoidal Section
Figure 17.12 shows a trapezoidal section of a curved bar with breadths B1 and B2,
depth D and radius of curvature R. Say, C is the centre of curvature and G is the
centroid of the section. Then,
Figure 17.12
Consider a strip of depth dy at a distance of y from the centroidal layer.
If b is the breadth of the strip
Area of the strip,
Now
Example 17.4 Determine the maximum compressive and tensile stresses in the
critical section of a crane hook lifting a load of 40 kN. The dimensions of the hook
are shown in Fig. 17.13. The line of application of the load is at a distance of 8 cm
from the inner fibre (rounding off the corners of the cross-section are not taken into
account).
Figure 17.13 Example 17.4
Solution
Figure 17.13 shows a crane hook and the trapezoidal section. The load line KK′ is
away from the centre of the curvature C.
Position of CG of the section
So,
Radius of curvature,
Area of cross-section,
Now,
Substituting the values,
Distance,
Bending moment,
The bending moment tends to reduce the curvature, so the portion GA will be in
compression and portion GB will be in tension.
Direct stress,
Maximum compressive stress at A,
Maximum tensile stress at B
Exercise 17.4 The section of a crane hook is a trapezium. At the critical section,
the inner and outer sides are 40 mm and 25 mm, respectively and depth is 75 mm.
Centre of curvature of the section is at a distance of 60 mm from the inner fibres
and the load line is 50 mm from the inner fibres. Determine the maximum load the
hook can carry if the maximum stress does not exceed 120 N/mm2.
Ah2 for a Circular Section
Figure 17.14 shows the circular section of diameter d of a curved bar of radius of
curvature R, from the centre of curvature C up to the centroid G of the section.
Area of cross section,
Consider a strip of depth dy at a distance of y from the centroidal layer as shown.
Breadth of the layer,
Figure 17.14
Area of the strip,
Now,
or,
Example 17.5 A curved bar is formed of a tube of an outside diameter of 8 cm and
a thickness of 0.5 cm. The centre line of this beam is a circular arc of radius 15 cm.
Determine the greatest tensile and compressive stresses set up by a bending
moment of 1.2 kN m tending to increase its curvature.
Solution
Figure 17.15 shows the cross-section of a curved bar of radius of curvature R = 15
cm.
Area of cross-section,
Area of inner circle,
Area of outer circle,
Bending moment,
M = 1.2kNm = 1.2 × 105 N cm
Figure 17.15 Example 17.5
For a circular section
For inner circle,
For outer circle,
Maximum tensile stress at b,
Maximum compressive stress at a,
Figure 17.16 Example 17.5
Exercise 17.5 A bar of circular cross-section is bent in the shape of a horse shoe.
The diameter of the section is 8 cm and mean radius R is 8 cm as shown in Fig.
17.16. Two equal and opposite forces P = 15 kN each are applied so as to
straighten the bar. Determine the maximum tensile and compressive stresses along
the central section.
Ring Subjected to a Diametral Load
Figure 17.17 shows a circular ring of mean radius R subjected to a diametral
pull P. Consider a section CD at an angle θ from the line of application of the load,
that is, Y1Y2, and determine the bending moment and stresses in this section. Due to
symmetry, the ring can be divided into four equal quadrants. Say, M1 is the
bending moment on the section AB along the line of symmetry X1X2.
Taking moments about CD,
Bending moment at the section CD,
From Eq. (17.5) of the section ‘Stresses In A Curved Bar’,
Figure 17.17
where E is the Young’s modulus, ε0 is the strain in centroidal layer, R is the initial
radius of curvature and R1 is the radius of curvature after bending
Therefore,
Multiplying this equation throughout by Rdθ and integrating for one quadrant
That is,
However,
[Eq. (17.2) of the section ‘Stresses in a Curved Bar’]
Now for one quadrant, initial angle θ = 90° = π/2 and final angle θ1 = 90° = π/2
due to symmetry, that is, ∠Y1OX1 remains 90° even after the application of
diametral load.
Therefore
Substituting this in Eq. (17.12), we get,
Again by the Eq. (17.4) of the section ‘Stresses in a Curved Bar’
Normal force on the section,
where
Therefore, normal force,
Normal force on the section CD,
(refer to Eq. (17.5) of the section ‘Stresses in a Curved Bar’ again)
Therefore,
Substituting the value of ε0 in Eq. (17.13)
Now,
M will be maximum when θ = 0°.
M will be zero when
So, there will be four sections, one in each quadrant where the bending
moment M will be zero and consequently the stress due to bending moment will be
zero.
Now, substituting the value of M1 in the following equation.
and
Stress,
‘Stresses in a Curved Bar’]
[Refer to Eq. (17.3) of the section
Direct stress at any section,
Resultant stress at any point on the section
Stress along Y1Y2axis, where θ = 0°
at the point K,
where d = diameter of the rod of the ring
stress,
at the point J,
stress,
It can be observed that the maximum stress occurs at the point J, where the
diametral load is applied. Stresses along X1X2axis, where θ = 90°
At the point B,
At the point A,
Example 17.6 A ring is made of round steel bar, 2 cm diameter and the mean
diameter of the ring is 12 cm. Determine the greatest intensities of tensile and
compressive stresses along a diameter XXif the ring is subjected to a pull of 5 kN
along diameter YY.
Solution
Figure 17.18 shows a ring of a mean diameter of 12 cm, a bar diameter of 2 cm,
subjected to a diametral pull P.
Radius of curvature,
R = 6 cm
Bar diameter,
d = 2 cm
Pull,
Area of cross-section,
P = 10 kN
Figure 17.18
Stresses
Exercise 17.6 A ring is made of round steel rod of a diameter of 24 mm. The mean
diameter of the ring is 240 mm. The ring is pulled by a force of 1 kN. Determine
the greatest intensities of tensile and compressive stresses along the diameter of the
loading.
Chain Link Subjected to a Tensile Load
Figure 17.19 shows a chain link of mean radius R, length of the straight portion l,
subjected to a pullP. Consider a section CD at an angle θ from the line of
application Y1Y2 of the pull P. Determine the bending moment and stresses in this
section. Due to symmetry, the ring can be divided into four equal parts as shown.
Say, M1 is the bending moment on the section AB along the line OX1.
Taking moments at the section CD,
From Eq. (17.5) of the section ‘Stresses In A Curved Bar’.
where E is the Young’s modulus, ε0 is the strain in the centroidal layer, R is the
initial radius of curvature and R1 is the final radius of curvature
Figure 17.19
Therefore,
Multiplying throughout by Rdθ and integrating from 0 to π/2
Now,
[Eq. (17.2) of the section ‘Stresses In A Curved Bar’]
In this case, initial angle θ = ∠ X1OY1 = 90°
The final angle θ1 will be a slight change from 90°.
Slope at X1
where I is the moment of inertia of the section.
Therefore,
Substituting in Eq. (17.17)
Again by Eq. (17.4) of the section ‘Stresses In A Curved Bar’.
Normal force on the section CD,
Now,
Therefore, normal force,
Therefore,
(by Eq. (17.5) of the section Stresses In A Curvedd Bar’)
Therfore,
Substituting the values of ε0 in Eq. (17.18)
where I = Ak2; k = radius of gyration of the section
Dividing throughout by π/2
However,
Substituting the value of M1 in Eq. (17.20)
Stress at any layer at a distance of y from the neutral layer is
Moreover,
Therefore,
Substituting the values of M and ε0, stress due to bending moment
Direct stress due to F,
Resultant stress, σR = σb + σd
This is an equation for resultant stress in any section along the curved
portions X2Y1X1 and X2′Y2X1′ of the chain link.
The bending moment M1 on the straight portion X1X1′ and X2X2′ will remain
constant and for the straight portion bending stress will be found with the help of
general flexural formula. To obtain the resultant stress in this straight portion,
direct tensile stress P/2A will be added to the bending stress.
On the inner surface of the ring which is also called intrados stress can be
obtained by substituting y= −d/2 in Eq. (17.21) where d is the diameter of the bar
of the chain link. Similarly, for the outer surface which is also known
as extrados the resultant stress is obtained by replacing y by +d/2 in Eq. (17.22).
Maximum stress along Y1OY2axis,
At the intrados,
At the extrados,
Maximum stresses along X1OX2axis,
At the intrados,
At the extrados,
Maximum stress in straight portion,
X1X1′ or X2X2′
Bending moment,
Stress due to bending,
Direct stress,
Resultant stress at intrados and extrados, respectively,
Example 17.7 A chain link is made of round steel rod of 1 cm diameter. If R = 3
cm and l = 5 cm, determine the maximum stress along the section where tensile
load is applied. If P = 0.5 kN.
Solution
R = 3 cm, d = 1 cm, l = 5 cm, and P = 1 kN
Area,
Radius of gyration,
Now,
Then,
Maximum stress at intrados,
Maximum stress at extrados,
Maximum stress occurs at the intrados, i.e., where the load is applied.
Exercise 17.7 A chain link is made of round steel rod, 10 mm diameter. If R = 30
mm and L = 50 mm, determine the maximum stresses along the section at the end
of the straight portion. A load of 1 kN (tensile) is applied on the chain link.
Deflection of Curved Bar
In order to estimate the stiffness of a curved beam, subjected to a bending moment.
It is necessary to determine the deflection of the curved beam and in such cases the
influence of the initial curvature of the beam on its deflection is
considerable. Figure 17.20 shows the centre line ABCD of a curved bar subjected
to variable bending moment. Consider a small portion BC of length ds along the
centre line. Say, the bending moment at B is M and at C is M + δM. Due to the
bending moment say the centre line of the curved beam takes new
position ABF and the element BC rotates by an angle dɸ = DBF at the point B. The
angular rotation is small and the displacement of the point D is also small.
Displacement DF ≅ BD dϕ
∠ BDF = 90° for very small displacement DF
Components of the displacement are DE perpendicular to the
chord AD and EF parallel to the chordAD, that is, the line joining the ends of the
centre line of the curved beam considered. FE shows negative displacement
towards the point A.
Deflection of the point D with respect to A is δDA and considering the small
length ds only, say the deflection is
ΔδDA
∠ADF
=
−EF = −DF cas α where ∠DEF = α
=
−(BD dϕ) cos α
=
α
Figure 17.20 Deflection of curved bar
Therefore,
∠BDG = 90° − α or ∠DBG = α
and
BD cos α = BG
Therefore,
ΔδDA = −(BD cas α)dϕ = − BG dϕ = −h dϕ
where h is the perpendicular distance of the point B from the chord AD.
Moreover,
Therefore,
Total defelection of D with respect to A
(1) Deflection of a closed ring
Figure 17.21 shows the quadrant of a ring of mean radius of
curvature R subjected to diametral pull Palong OY1. We have to determine the
deflection along the load line or along the chord Y2Y1. (Point Y2not shown in Fig.
17.19).
Figure 17.21 Closed ring-quadrant
OY1 is half of the chord Y2Y1. Consider a small length ds at C at an angular
displacement θ.
CC1 = perpendicular distance on chord from the point C
= h = Rsin θ
Bending moment at the section C,
[see Eq. (17.15) of the section ‘Ring Subjected To A
Diametral Load’]
Note that we have considered only one quadrant and when we consider the
complete ring, the deflection along the load will be δY1Y2
(2) Deflection perpendicular to load line
Refer to Fig. 17.19 again, now the chord is OX1 and perpendicular
distance CC2 = h on the chord OX1from C; h = Rcos θ
δx10 = deflection perpendicular to load line
Total deflection,
Example 17.8 A ring with a mean diameter of 120 mm and a circular cross-section
of 40 mm diameter is subjected to a diametral compressive load of 20 kN.
Calculate the deflection of the ring along the load line. E = 200 GN/m2.
Solution
Since the diametral load is compressive, there will be reduction in diameter along
the load line and increase in diameter perpendicular to the load line.
R = 60 mm = 6 cm; d = 40 mm = 4 cm; P = 20 × 103 N
E = 2,000 GN/m2 = 200 × 105 N/cm2
Deflection along the load line
Exercise 17.8 A Ring with a mean diameter of 150 mm and a circular crosssection of 30 mm diameter is subjected to a diametral tensile load of 8 kN.
Calculate the deflection of the ring along the direction perpendicular to load
line E = 200 kN/mm2.
Deflection of a Chain Link
Figure 17.22 shows the quarter of a chain link subjected to axial tensile load P. The
radius of curvature of the link is R and length of the straight portion is l. Consider a
section at an angle θ from the axis OY1. We have to determine the deflection along
the load P or along the chord Y2Y1.
CC1 = h = perpendicular from C to the chord
= Rsinθ
Bending moment at the section
Figure 17.22
[Refer to Eq. (17.6) of article 19.7.]
Over the length l/2, the bending moment is constant and is equal to M1 = PK′
where the value of K′ is as above.
Deflection along the line of loading
(with the help of theory of simple bending)
Deflection due to direct load P/2 is
Total deἀection,
where
(2) Deflection perpendicular to load line
In this case chord is XX and CC2 = h = Rcos θ + l/2
From theory of simple bending deflection in straight portion
where M1 is bending moment on straight portion and l is the length of the straight
portion.
where
Example 17.9 A chain link is made of a steel rod of 12 mm diameter. The straight
portion is 60 mm in length and the ends are 60 mm in radius. Determine the
deflection of the link along the load line when subjected to a load of 1 kN.
Given E = 200 × 103 N/mm2.
Solution
Rod diameter,
d = 1.2 cm
Area of cross-section,
A = π2/4 = 1.31 cm2
Length of straight portion,
l = 6 cm
Radius of curvature,
R = 6 cm
Load,
P = 1.0 kN, E = 200 × 105 N/cm2
Radius of gyration,
k = d/4 = 0.3 cm
Deflection along the load line
Exercise 17.9 A chain link is made of steel rod, 12 mm diameter. The straight
portion is 60 mm in length and the ends are 60 mm in radius. Determine the
deflection in the link along the direction perpendicular to the load line if the chain
link is subjected to a load of 1 kN. Given E = 200 kN/mm2.
Problem 17.1 For the frame of a punching machine shown in Fig. 17.23.
Determine the circumferential stresses at A and B on a section inclined at an
angle θ = 45° to the vertical.
Force,
P = 200 kN.
Solution
Force,
P = 200 kN.
Perpendicular force on the section AB = Psin 45° =
Tangential force on the section AB = Pcos 45° =
Figure 17.23
Area of cross-section, A = 30 × 10 + 5 × 20 + 15 × 10 = 550 cm2
Location of G
Radius of curvature, R = R1 + y2 = 20 + 15.9 = 35.9 cm
Bending moment on the section,
Direct force on the section,
Tensile stress at point A
Compressive stress at point B
Problem 17.2 The radius of the inner fibres of a curved bar of trapezoidal section
is equal to the depth of the cross-section. The base of the trapezium on the concave
side is four times the base on the convex side. Determine the ratio of the stresses in
the extreme fibres of the curved bar to the stresses in the same fibres of a straight
bar subjected to the same bending moment.
Solution
See Fig. 17.24
Now,
R2 = D
Therefore,
R1 = 2R2 = 2D
Therefore, y2 = 0.4D
Area,
Figure 17.24
Let us consider that this curved bar is subjected to a bending moment, M tending to
reduce the curvature.
The maximum tensile stress,
(when y = −y2)
The maximum compressive stress,
(when y = +y1)
Stresses in the straight bar
Let us divide the section in two triangles as shown, so as to calculate the moment
of inertia IYY.
Area of triangle I,
Area of triangle II,
IYY of triangle I, about its CG =
Distance,
IYY of triangle II, about its CG
Distance,
IYY of the whole section
The maximum tensile stress,
The maximum compressive stress,
(y = + y1)
Therefore, the ratios
and, the ratios
Problem 17.3 A chain link is made of round steel rod, diameter 12 mm. If R = 40
mm and l = 60 mm, determine the extreme stresses along the intrados if the link is
subjected to a tensile load of 1 kN.
Solution
Bar diameter,
d = 12 mm
Radius of curvature,
R = 40 mm
Length of straight portion,
Load,
l = 60 mm
P = 1,500 N
Area of cross-section,
Radius of gyration,
Moreover,
Along the intrados y = −d/2, the equation for the resultant stress is
Substituting the values
Problem 17.4 A chain coupling is made of a 20-mm-diameter steel rod bent into
an ‘S’ shape as shown in Fig. 17.25. If the coupling is attached to a tensile load of
1.0 kN as shown; calculate the maximum stress developed in chain coupling.
Solution
Maximum tensile stress will occur either at point A of smaller loop or at point B of
larger loop.
Rod diameter,
d = 20 mm
Radius of curvature of smaller loop,
R1 = 40 + 10 = 50 mm
Radius of curvature of longer loop,
R2 = 60 + 10 = 70 mm
Load,
W = 1.0 kN = 1,000 N
Area of cross-section of rod,
Figure 17.25
and
Stress at A,
Stress at B,
Multiple Choice Questions
1. A bar of square section 6 cm × 6 cm is curved to a mean radius of 12 cm.
A bending moment Mis applied on the bar. The moment M tries to
straighten the bar. If the stress at the innermost fibres is 60 N/mm2 tensile,
then the stress at the outermost fibres is
1.
2.
3.
4.
60 N/mm2 (compressive)
60 N/mm2 (tensile)
More than 60 N/mm2 (compressive)
Less than 60 N/mm2 (compressive).
2. The most suitable section of a crane hook is
0. Square
1. Round
2. Hollow round
3. Trapezoidal
3. A bar of square section 4 cm × 4 cm is curved to a mean radius of 80 m. A
bending moment M, tending to increase the curvature is applied on the bar.
If the stress at the outermost fibres is 80 MPa tensile, then the stress at the
innermost fibres will be approximately equal to
0. 120 MPa (compressive)
1. 100 MPa (compressive)
2. 90 MPa (compressive)
3. 80 MPa (compressive)
4. A ring is subjected to a diametral tensile load. The variation of the stress at
the intrados surface from the point of loading up to the section of
symmetry is
0. Maximum tensile stress to maximum compressive stress.
1. Throughout tensile stress.
2. Maximum compressive stress to maximum tensile stress.
3. Throughout compressive stress.
5. The distribution of stress along a section of a curved bar subjected to a
bending moment tend to increase its curvature is
0. Linear
1. Uniform
2. Parabolic
3. Hyperbolic
Practice Problems
1. A sharply curved beam of rectangular section is 10 mm thick and 50 mm deep. If
the radius of curvature, R = 60 mm, compute the stress in terms of the bending
moment M at a point 20 mm from the outer surface.
2. Determine the diameter d of a round steel rod that is used as a hook to lift a 9-kN
load acting through the centre of curvature of the centroidal axis of the hook.
Assume that
, and the maximum stress permitted is 125 N/mm2.
3. The cross-section of a triangular hook has a base of 5 cm and altitude of 7.5 cm
and a radius of curvature of 5 cm at the inner face of the shank. If the allowable
stress in tension is 100 N/mm2and in compression is 80 N/mm2, what load can be
applied along a line 7.5 cm from the inner face of the shank?
4. Three plates are welded to form the curved beam of or, I-section shown in Fig.
17.26. If the moment M = 1 kN m, determine the stresses at points A and B and at
the centre of the section (Fig. 17.26).
5. A steel link of rectangular section 24 mm × 8 mm is shown in Fig. 17.27. If the
angle β = 90° and the allowable stress in link is 100 MPa, determine the largest
value of P which can be applied on the link.
6. A curved bar of rectangular section with breadth B and depth D = 2B, is bent to a
radius of curvature equal to 1.2D. It is subjected to a bending moment of 1 kN m,
tending to increase its curvature. Determine the size of the section if the maximum
stress does not exceed 80 N/mm2.
Figure 17.26
Figure 17.27
7. Determine the maximum stress along, the section A-A in the crane hook. Section
of the hook is circular of diameter 25 mm. Load 2 kN (Fig. 17.28).
8. Section of a punch press is in a T section of dimensions shown in Fig. 17.29.
Centre of curvature of section is at a distance of 400 mm from inner side and load
line passes through O, at a distance of 600 mm from inner side. Punch press frame
is made of CI. Determine the safe load for a factor of safety of 3 based on ultimate
strength.
σut of CI = 150 MPa
σuc of CI = 600 MPa
9. A proving ring is 250 mm diameter, 40 mm wide and 6 mm thick. The maximum
allowable stress in ring is 200 N/mm2. Find the load to cause this stress and the
load should give 1 mm deflection of the ring in the direction of loading. (E = 200
GPa).
10. The curved beam with a circular centre line has a trapezoidal cross-section as
shown in Fig. 17.30 and is subjected to pure bending in its plane of symmetry. The
face b1 is on concave side of the beam. If h = 100 mm and a = 100 mm, find the
ratio of b1/b2 of base widths so that the extreme fibre stresses in tension and
compression are numerically equal.
11. A chain link is made of round steel rod of diameter 12 mm. If R = 40 mm and l =
60 mm draw the stress distribution diagram along the extrados for an angle of 90°
starting from the outermost edge (along the direction of loading) of the link, if the
link is subjected to a tensile load of 10 kN.
Figure 17.28
Figure 17.29
Figure 17.30
Answers to Exercises
Exercise 17.1: 2.16 kN m, +4.50 N/mm2
Exercise 17.2: −93.8; +84.05 MPa
Exercise 17.3: 86.70, −37.873 MPa
Exercise 17.4: 30.60 kN
Exercise 17.5: −39.22, +99.83 MPa
Exercise 17.6: +26.1 N/mm2, −30.33 N/mm2
Exercise 17.7: 42.99, −30.26
Exercise 17.8: 0.055 mm
Exercise 17.9: 0.090 mm
Answers to Multiple Choice Questions
1. (d)
2. (d)
3. (d)
4. (c)
5. (d)
Answers to Practice Problems
1. 72.92 M
2. 43.52 mm
3. 15.7 kN
4. +22.755, −16.71,−3.55 MPa
5. 4.37 kN
6. 29.7, 59.4 mm
7. 93.6 MPa
8. 148.433 kN
9. 375 N, 496 N
10.1.87
11.79.56 to −17.37 MPa
4a. Unsymmetrical Bending and Shear Centre
CHAPTER OBJECTIVES
When a section of a beam is not symmetrical about the plane of bending, an
unsymmetrical bending takes place, i.e., in addition to bending, due to applied
loads twisting is observed in the beam. Then there are principal axes of the section
where the product of inertia is zero. In this chapter students will learn about:
o
o
o
o
o
o
o
Principal axes and their directions with respect to centroidal axes of the
section.
Moment of inertia of the section about the principal axes.
Stresses developed at various points of the section due to unsymmetrical
bending.
Product of inertia of the section about any co-ordinate axes.
Position of neutral axis with respect to centroidal axes of the
unsymmetrical sections.
Deflection in a beam due to unsymmetrical bending.
Location of shear center of sections which are symmetrical about one
centroidal axes and sections which are not symmetrical about any
centroidal axes.
Introduction
Every section is not symmetrical about both the centroidal axes. Some sections are
symmetrical only about one axis, whereas many sections as angle sections are not
symmetrical about both the centroidal axes. In theory of simple bending, the
section of the beam is symmetrical about the plane of bending. The simple flexural
formula derived in theory of simple bending is not applicable when the section is
not symmetrical about the plane of bending. In such sections, the principal axes
and principal moments of inertia and the product of inertia are determined. Stresses
developed in such sections of a beam are dependent on these parameters.
If the load line on a beam does not coincide with one of the principal axes of the
section, the bending takes place in a plane different from the plane of principal
axes. This type of bending is known as unsymmetrical bending. The two reasons of
unsymmetrical bending are as follows:
1. The section is symmetrical about two axes like I-section, rectangular
section, circular section but the load-line is inclined to both the principal
axes.
2. The section itself is unsymmetrical like angle section or a channel section
(with vertical-web) and load-line along vertical any centroidal axes.
Figure 18.1(a) shows a beam with I-section with load-line coinciding
with YY principal axis. I-section has two axes of symmetry and both these axes are
the principal axes. Section is symmetrical aboutYY plane, i.e., the plane of bending.
This type of bending is known as symmetrical bending.
Figure 18.1(b) shows a cantilever with rectangular section, which has two axes
of symmetry which are principal axes but the load-line is inclined at an angle α
with the YY axis. This is the first type of unsymmetrical bending. Then, Fig.
18.1(c) shows a cantilever with angle-section which does not have any axis of
symmetry but the load-line is coinciding with the YY axis. This is the second type
of unsymmetrical bending.
Figure 18.1(d) shows a channel section subjected to a vertical load passing
through its centroid G. This member has been subjected to bending and twisting
under the applied vertical load W. Now, the question arises; is it possible to apply
the vertical load W in such a way that the channel member will bend without
twisting and, if so, where the load W should be applied?
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Figure 18.1 (a) Symmetrical bending, (b) unsymmetrical bending symmetrical
section but oblique load, (c) unsymmetrical bending (unsymmetrical section) and
(d) unsymmetrical bending(section not symmetrical about bending plane) (e)
Channel section(not symmetrical about yy axis, (f) Bending without twisting
Shear force in the flanges and web of the channel section is F1, F2 and F1,
respectively, as shown in Fig. 18.1(e). Forces F1 constitute a couple F1 × h about
centroid G. This couple is responsible for twisting of the member. Now, if the
vertical load W or the shear force in the section is shifted from G, such
thatW × e = F1 × h, then the twisting couple is eliminated. So, it can be concluded
that if the vertical loadW, or vertical shear F is moved to the left in the channel
section through a distance e, such that, F1 ×h = We = Fe, the member will bend
without twisting as shown in Fig.18.1(f).
Before proceeding further, let us study about the principal axes of a section.
Principal Axes
Figure 18.2 shows a beam section which is symmetrical about the plane of
bending Y–Y, a requirement of the theory of simple bending or symmetrical
bending. G is the centroid of the section.XX and YY are the two perpendicular axes
passing through the centroid. Say, the bending moment on the section (in the
plane YY of the beam) is M, about the axis XX. Consider a small element of area
dA with (x, y) co-ordinates.
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Figure 18.2 Plane of bending yy
Stress on the element,
Force on the element,
Bending moment about YY axis,
Total moment,
If no bending take place about YY axis, then
M1 = 0
or
or
or
The expression
is called the product of inertia of the area
about XX and YY axes, represented byIxy. If the product of inertia is zero about the
two co-ordinate axes passing through the centroid, then the bending is symmetrical
or pure bending. Such axes (about which product of inertia is zero) are called
principal axes of the section and moment of inertia about the principal axes are
called principal moments of inertia.
The product of inertia may be positive, negative or zero depending upon the
section and co-ordinate axes. The product of inertia of a section with respect to two
perpendicular axes is zero if either one of the axis is an axis of symmetry.
Example 18.1 Show that product of inertia of a T-section about a centroidal axis is
zero.
Solution
Figure 18.3 shows a T-section with flange B × h1 and web b × h2. The section is
symmetrical about YYaxis. Say G is the centroid of the section on the axis YY,
and XX and YY are the centroidal axes.
Ixy = I′xy for flange + I″xy for web
For flange, x varies from –(B/2)to +(B/2)
For web, x varies from –(b/2) to +(b/2)
Now,
Figure 18.3 Example 18.1
Example 18.2 Determine the product of inertia about axes X and Y for a triangular
section shown in Fig.18.4.
Figure 18.4 Example 18.2
Solution
Consider a small element of area dA at co-ordinates x, y.
Product of inertia about XY axis,
Note that limiting value of x = 40 mm = 2 × limiting value of y
Figure 18.5 Example 18.2
Exercise 18.1 Consider an I-section with flanges B × t1, and web H × t2 and show
that the product of inertia about the centroidal axes is zero.
Exercise 18.2 Figure 18.5 shows a rectangular section with breadth, b and
altitude, h. Determine the product of inertia of the section about X–Y axes.
Parallel Axes Theorem for Product of Inertia
Figure 18.6 shows a section with its centroid at G, and GX′ and GY′ are the two
rectangular co-ordinates passing through G. Say, the product of inertia about X′Y′
is . Let us determine the product of inertia about the axis OX and OY, i.e., Ixy.
Say, distance of G from OX axis = , and distance of G from OY axis = .
Consider a small element of area
dA = dxdy.
Figure 18.6
Say, co-ordinates of the element about the centroidal axis GX′, GY′ are x′, y′.
Then, co-ordinates of the element about X–Y axis are,
x=
+ x′ and y =
+ y′
Therefore, the product of inertia,
i.e., the product of inertia of any section with respect to any set of co-ordinate axes
in its plane is equal to the product of inertia of the section with respect to the
centroidal axes parallel to the co-ordinate axes plus the product of the area and the
co-ordinates of the centroid of the section with respect to the given set of coordinate axes.
Example 18.3 Figure 18.7 shows an unequal channel section, determine its
product of inertia Ixy and .
Solution
Let us break up the section into three rectangular strips I, II and III as shown in the
figure and write the coordinates of their centroids with respect to the given set of
axes YOX.
Remember that the product of inertia of these rectangular strips about their
principal axes passing through the respective centroids is zero, because these
rectangular strips have two axes of symmetry.
(Ixy) I = 0 + 81 cm4 (using the parallel axis theorem for product of inertia)
Figure 18.7 Example 18.3
(Ixy) II = 0 + 24 cm4
(Ixy) III = 0 + 386 cm4
Ixy = 501 cm4
To determine
, let us first determine the position of the centroid of the section.
Area of cross-section,
A = 18 + 8 + 12 = 38 cm2
= Ixy − A
= 501 − 38 × 3.184 × 5.21
= 501 − 630.37 = −129.37 cm4
Figure 18.8 Exercise 18.3
Exercise 18.3 Figure 18.8 shows one unequal angle section, determine its product
of inertia Ixyand
(through the centroidal axes).
Determination of Principal Axes
In the section ‘Introduction’, we have learnt that principal axes pass through the
centroid of a section and product of inertia of the section about principal axes is
zero. Figure 18.9 shows a section with centroid G and XX and XX are two coordinate axes passing through G. Say, UU and VV is another set of axes passing
through the centroid G and inclined at an angle θ to the X–Y co-ordinate. Consider
an element of area dA at point P having co-ordinates (x, y). Say, u, v are the coordinates of the point P in U–V co-ordinate axes.
Figure 18.9 Co-ordinates along principal axes
So,
where
u = GA′ = GD + DA′ = GD + AE
GD = GA cosθ = x cosθ
AE = DA′ = y sinθ
or,
u = x cosθ + y sinθ
v = GB′ = PA′ = PE − A′E
= PE − AD since A′E = AD
= PA cosθ − x sinθ = y cosθ − x sinθ
Similarly, x, y co-ordinates can be written in terms of u, v co-ordinates.
x = GC − AC = GC − A′F = u cosθ − v sinθ
(as PA′ = v and GA′ = u)
y = GB = PA = AF + FP = A′C + FP = u sinθ + v cosθ
Second moment of area about U–U axis,
Second moment of area about V–V,
From Eqs.(18.4) and (18.5),
Iuu + Ivv = Ixx (sin2θ + cos2θ) + Iyy (sin2θ + cos2θ) = Ixx + Iyy
(18.6)
Product of inertia about UV axes,
However, as per the condition of pure bending or symmetrical bending Iuv = 0,
then U and V will be the principal axes
or,
2Ixy cos 2θ + (Ixx − Iyy)sin 2θ = 0
or,
Say, θ1 and θ2 are two values of θ given by Eq. (18.7)
Substituting these values of sin 2θ1 and cos 2θ1 in Eq. (18.4)
Similarly,
Now, for
θ2 = θ1 + (π/2)
sin 2θ2 = sin(2θ1 + π) = −sin 2θ1
cos 2θ2 = cos(2θ1 + π) = −cos 2θ1.
Substituting these values in Eqs.(18.4) and (18.5)
From Eqs.(18.8) to (18.11), we learn that
Maximum and minimum values of Iuu and Ivv
For maximum value of Iuu,
This shows that the values of (Iuu)θ1 and (Iuu)θ2 are the maximum and minimum
values of Iuu and Ivv.These values are called the principal values of moment of
inertia as Iuv = 0. The directions θ1 and θ2are called the principal directions.
Moment of Inertia About Any Axis
If the principal moments of inertia Iuu and Ivv are known, then moment of inertia
about any axis inclined at an angle θ to the principal axes can be determined.
Say u, v are the co-ordinates of an element of area dA in the U–V principal axes
system. X and Y are the co-ordinate axes inclined at an angle θ to the U–V axes.
x co-ordinate of element = u cos θ − v sin θ
y co-ordinate of element = u sin θ + v cos θ
Moment of inertia,
Similarly,
Ixx = Iuu cos2θ + Ivv sin2θ
From Eqs.(18.12) and (18.13),
Ixx + Iyy = Iuu + Ivv = J,
(18.13)
where J is polar moment of inertia about an axis passing through G and normal to
the section.
Example 18.4 Determine the principal moments of inertia for the equal angle
shown in Fig.18.10.
Figure 18.10 Example 18.4
Solution
Let us consider the angle section in two portions, I and II, as shown and determine
the position of the centroid
(due to symmetry
Moment of inertia,
)
Ixx = Iyy
Co-ordinates of centroid of portion I
= [(5 − 2.87) − (2.87 − 0.5)]
= (2.13, − 2.37)
Co-ordinates of centroid of portion II
= [−(2.87 − 0.5), (5.5 − 2.87)]
= (−2.37, 2.63)
Product of inertia,
Ixy = 10(2.13)(−2.37) + 9(2.63)(−2.37)
(as the product of inertia about their own centroidal axis is zero, since portions I
and II are rectangles).
Therefore,
Ixy = – 50.481– 56.098 = –106.579 cm4
If θ = angle of principal axis UU with respect to XX-axis
Principal angles are
θ1 = 45°, θ2 = 90° + 45° = 135°
Principal moments of inertia
Example 18.5 Figure 18.11 shows an I-section 15 cm × 20 cm. Axis X′X′ and Y′Y′
are inclined at an angle of 30° to the axis of symmetry. Determine the moment of
inertia about these axes. Calculate also the product of inertia Ix′y′.
Figure 18.11 Example 18.5
Solution
The I-section shown has two axes of symmetry, i.e., UU and VV passing through
the centroid G. Therefore, these are the principal axes and Iuu and Ivv are the
principal moments of inertia. The angle of inclinations of UU and VV axes with
respect to X ′X ′ and Y ′Y ′ axes is θ = 30°.
sin2θ = 0.25, cos2θ = 0.75
Iy′y′ = Ivv cos2 θ + Iuu sin2θ
since
Iuv = 0
Ix′x′ = Iuu cos2θ + Ivv sin2θ
Now,
Therefore, Iy′y′ = 1,126.333 × 0.75 + 5,221.333 × 0.25
= 844.749 + 1,305.333 = 2,150.082 cm4
Ix′x′ = 5,221.333 × 0.75 + 1,126.33 × 0.25
= 3,915.999 + 281.583 = 4,197.582 cm4
Now,
Figure 18.12 Exercise 18.5
Exercise 18.4 Consider a rectangular section of 6 cm width and 12 cm depth.
Determine Ixx, Iyyand Ixy about XX and YY axes inclined at an angle of 45° to the
principal axes.
Exercise 18.5 Determine the principal angles and principal moments of inertia of
a z-section shown in Fig.18.12.
Stresses due to Unsymmetrical Bending
When the load-line on a beam does not coincide with one of the principal axes of
the section, unsymmetrical bending takes place. Figure 18.13(a) shows a
rectangular section, symmetrical aboutXX and YY axes or
with UU and VV principal axes. Load-line is inclined at an angle ø to the principal
axis VV, and passing through G (centroid) or C (shear centre) of the section.
Figure 18.13
Figure 18.13(b) shows an angle section which does not have any axis of
symmetry. Principal axes UUand VV are inclined to axes XX and YY at an angle θ.
Load-line is inclined at an angle ø to the vertical or at an angle (90 − ø − θ) to the
axis UU. Load-line is passing through G (centroid of the section).
Figure 18.13(c) shows a channel section which has one axis of symmetry,
i.e., XX. Therefore, UU andVV are the principal axes. G is the centroid of the
section while C is the shear centre. Load-line is inclined at an angle ø to the
vertical (or the axis VV) and passing through the shear centre of the section.
Shear centre for any transverse section of a beam is the point of intersection of
the bending axis and the plane of transverse section. If a load-line passes through
the shear centre there will be only bending of the beam and no twisting will occur.
If a section has two axes of symmetry, then shear centre coincides with the centre
of gravity or centroid of the section as in the case of a rectangular, circular or Isection. For sections having one axis of symmetry only, shear centre does not
coincide with centroid but lies on the axis of symmetry, as shown in the case of a
channel section.
For a beam subjected to symmetrical bending only, following assumptions are
made:
1. The beam is initially straight and of uniform section throughout.
2. Load or loads are assumed to act through the axis of bending.
3. Load or loads act in a direction perpendicular to the bending axes and
load-line passes through the shear centre of transverse section.
Figure 18.14, shows the cross-section of a beam subjected to bending moment M,
in the plane YY. Gis the centroid of the section and XX and YY are the two coordinate axes passing through G. Moreover, UU and VV are the principal axes
inclined at an angle θ to the XX and YY axes, respectively. Let us determine the
stresses due to bending at the point P having the co-ordinates u, vcorresponding to
principal axes. Moment applied in the plane YY can be resolved into two
components M1 and M2.
Figure 18.14
M1, moment in the plane UU = M sin θ
M2, moment in the plane VV = M cos θ
The components M1 and M2 have their axis along VV and UU, respectively.
Resultant bending stress at the point P,
The exact nature of stress (whether tensile or compressive) depends upon the
quadrant in which the point P lies. In other words sign of co-ordinates u and v is to
be taken into account while determining the resultant bending stress.
The equation of the neutral axis can be determined by considering the resultant
bending stress. At the neutral axis bending stress is zero, i.e.,
or,
where
This is the equation of a straight line passing through the centroid G of the
section. All the points of the section on one side of the neutral axis have stresses of
the same nature and all the points of the section on the other side of the neutral axis
have stresses of opposite nature.
Example 18.6 A 40 mm × 40 mm × 5 mm angle section shown in Fig.18.15 is
used as a simply supported beam over a span of 2.4 m. It carries a 0.100-kN load
along the line YG, where G is the centroid of the section. Determine the resultant
bending stresses on points A, B, C, i.e., outer corners of the section, along the
middle section of the beam.
Figure 18.15 Example 18.6
Solution
Let us first determine the position of the centroid
Moment of inertia,
Co-ordinates of G1 (centroid of portion I)
= +(20 − 11.83), −(11.83 − 2.5) = (8.17, −9.33) mm
Co-ordinates of centroid G2
= −(11.83 − 2.5), +(22.5 − 11.83) = −9.33, +10.67 mm
Product of inertia, Ixy = 40 × 5(8.17) × (−9.33) + 35 ×5(−9.33) × (10.67)
(Product of inertia about their own centroidal axes is zero because portions I and
II are rectangular strips.)
Ixy = −15,245.22 − 17,421.44 = −32,666.6 mm4
= −3.266 × 104 mm4.
Principal angle, θ
Principal moments of inertia
Bending moment,
Components of bending moment,
M1 = M sin 45° = 0.060 × 0.707 × 106 = 42.42 × 103 N mm
M2 = M cos 45° = 0.060 × 0.707 × 106 = 42.42 × 103 N mm
u–v co-ordinates of the points
Point A,
x = −11.83, y = 40 − 11.83 = 28.17 mm
u = x cos θ + y sin θ = −11.83 × 0.707 + 28.17 × 0.707 = 11.55 mm
v = y cos θ − x sin θ = 28.17 × 0.707 + 11.83 × 0.707 = 28.28 mm
Point B,
x = −11.83, y = −11.83
u = −11.0083 × 0.707 − 11.83 × 0.707 = −16.727 mm, v = 0
Point C,
x = 40 − 11.83 = 28.17, y = −11.83
u = 28.17 × cos 45° − 11.83 sin 45°
= 28.17 × 0.707 − 11.83 × 0.707 = 11.55 mm
v = −11.83 × 0.707 − 28.17 × 0.707 = −28.28 mm.
Resultant bending stresses at points A, B and C
Figure 18.16 Exercise 18.6
Exercise 18.6 Figure 18.16 shows I-section of a cantilever1.2 m long subjected to
a load W = 40 N at free end along the direction Y′G inclined at 15° to the vertical.
Determine the resultant bending stress at corners A and B, on the fixed section of
the cantilever.
Deflection of Beams Due to Unsymmetrical Bending
Figure 18.17 shows the transverse section of a beam with
centroid G. XX and YY are two rectangular co-ordinate axes and UU and VV are the
principal axes inclined at an angle θ to the XY set of co-ordinate axes. Say the
beam is subjected to a load W along the line YG. Load can be resolved into two
components, i.e.,
Wu = W sin θ (anlong UG direction)
Wv = W cos θ (anlong VG direction)
Say, deflection due to Wu is GA in the direction GU
i.e.,
where K is a constant depending upon the end conditions of the beam and position
of the load along the beam.
Deflection due to Wv is GB in direction GV
Figure 18.17
i.e.,
Total deflection,
Total deflection δ is along the direction GC, at angle γ to VV axis.
Comparing this with Eq. (18.15) of the section ‘Stresses Due to Unsymmetrical
Bending’
where α is the angle of inclination of the neutral axis with respect to UU axis
and
where γ is the angle of inclination of direction of δ with respect to VV axis.
γ = α, showing thereby that resultant deflection δ takes place in a direction
perpendicular to the neutral axis.
Example 18.7 A simply supported beam of a length of 2 m carries a central load of
4 kN inclined at 30° to the vertical and passing through the centroid of the section.
Determine (a) maximum tensile stress, (b) maximum compressive stress, (c)
deflection due to the load and (d) direction of neutral axis. Give, E = 200
×105 N/cm2.
Solution
Let us first determine the position of the centroid of the T-section shown
in Fig.18.18
The section is symmetrical about vertical axis; therefore, the principal axes pass
through the centroidG and are along U–U and V–V axes shown.
Therefore,
Figure 18.18 Example 18.7
Load
W = 4,000 N
Components of W,
Wv = 4,000 × cos 30° = 4,000 × 0.866 = 3,464 N
Wu = 4,000 × sin 30° = 4,000 × 0.500 = 2,000 N
Bending moment,
Bending moment,
Due to Mv, there will be maximum compressive stresses at A and B and
maximum tensile stress at Cand D.
Due to Mu, there will be maximum compressive stresses at B and D and
maximum tensile stresses atA and D.
Therefore, maximum compressive stress at B,
Maximum tensile stress at C,
Deflection,
where K = 1/48, as the beam is simply supported and carries a concentrated load at
its centre.
Therefore,
Now, sin θ = 0.5, sin2 θ = 0.25, cos θ = 0.866, cos2 θ = 0.75
Position of the neutral axis
Exercise 18.7 A cantilever 2.8 m long having T-section with flange 12 cm × 2 cm
and web 13 cm × 2 cm carries a concentrated load W at its free end but inclined at
an angle of 45° to the vertical. Determine the maximum value of W if the
deflection at the free end in not to exceed 2 mm. Given that E = 200 kN/mm2, what
is the direction of neutral axis with respect to the vertical axis.
Shear Centre
We have studied about the distribution of shear stresses in the transverse section of
a beam subjected to a bending moment M and a shear force F. Summation of the
shear stresses over the section of the beam gives a set of forces which must be in
equilibrium with the applied shear force F. In case of symmetrical sections such as
rectangular and I-sections, the applied shear force is balanced by the set of shear
forces summed over the rectangular section or over the flanges and the web of Isection and the shear centre coincides with the centroid of the section. If the
applied load is not placed at the shear centre, the section twists about this point
and this point is also known as centre of twist. Therefore, the shear centre of a
section can be defined as a point about which the applied shear force is balanced
by the set of shear forces obtained by summing the shear stresses over the section.
For unsymmetrical sections such as angle section and channel section,
summation of shear stresses in each leg gives a set of forces which should be in
equilibrium with the applied shear force.
Figure 18.19(a) shows an equal angle section with principal axis UU. We have
learnt in previous examples that a principal axis of equal angle section passes
through the centroid of the section and the corner of the equal angle as shown in
the figure. Say this angle section is subjected to bending about a principal
axis UU with a shearing force F at right angles to this axis. The sum of the shear
stresses along the legs, gives a shear force in the direction of each leg as shown. It
is obvious that the resultant of these shear forces in legs passes through the corner
of the angle and unless the applied force F is applied through this point, there will
be twisting of the angle section in addition to bending. This point of the equal
angle section is called its shear centre or centre of twist.
For a beam of channel section subjected to loads parallel to the web, as shown
in Fig. 18.19(b), the total shearing force carried by the web must be equal to
applied shear force F, then in flanges there are two equal and opposite forces
say F1 each. Then, for equilibrium, F × e is equal to F1 × h and we can determine
the position of the shear centre along the axis of symmetry, that is, e = (F1 × h/F).
Figure 18.19
Similarly, Fig. 18.19(c) shows a T-section and its shear centre. Vertical force in
the web F is equal to the applied shear force F and horizontal forces F1 in two
portions of the flange balance each other at shear centre.
Example 18.8 Figure 18.20(a) shows a channel section, determine its shear centre.
Solution
Figure 18.20(a) shows a channel section with flanges b × t1 and web h × t2. XX is
the horizontal symmetric axis of the section. Say, F is the applied shear force,
vertically downwards. Then, shear force in the web will be F upwards. Say, the
shear force in the top flange = F1.
Shear stress in the flange at a distance of x from right hand edge
Figure 18.20 Shear stresses in channel sections (a) and (b)
where F = applied shear force
first moment of area about axis X–X
t = t1 (thickness of the flange)
Shear stress
Figure 18.20(b) shows, the variation of shear stress in flanges and web.
Shear force in elementary area (t1 dx = dA)
= τdA = τt1 dx
Total shear force in top flange
, where b = breadth of the flange
(say)
There will be equal and opposite shear force in the bottom flange.
Say shear centre is at a distance of e from web along the symmetric axis XX.
Then, for equilibrium
Moment of inertia,
in which, the expression (2bt13/12) is negligible when comparing to other terms
Substituting this in the expression for e
if we take,
bt1 = area of flange = Af
ht2 = area of web = Aw
Then,
Exercise 18.8 A channel section has flanges 6 mm × 1 mm and web 8 cm × 0.5
cm. Determine the position of its shear centre.
Problem 18.1 Find the product of inertia of a quadrant of a circle about
axes X and Y as shown inFig. 18.21.
Figure 18.21 Problem 18.1
Solution
Figure 18.21 shows the quadrant of a circle of radius R. Consider a small element
at radius r, radial thickness dr and subtending an angle dθ at the centre.
Area of the element,
dA = rdθdr
Co-ordinates of the element
x = r cos θ and y = r sin θ
Product of inertia,
Problem 18.2 A beam of angle section shown in Fig. 18.22 is simply supported
over a span of 1.6 m with 15 cm leg vertical. A uniformly distributed vertical load
of 10 k N/m is applied throughout the span. Determine (a) maximum bending
stress, (b) direction of neutral axis and (c) deflection at the centre. Given E = 210 k
N/mm2.
Solution
Figure 18.22 Problem 18.2
Let us first determine the position of the centroid
Moment of inertia
Co-ordinates of G2 and G1: [−1.875(8 − 4.875) and (5 − 2.375) − 4.375]
Ixy = 14(8 − 4.875)(− 1.875) + 10(5 − 2.375) (− 4.375)
= −82.031 − 114.843 = −196.874 cm4
(Note that parallel axes theorem for product of inertia is used here and product of
inertia of each rectangular strip about its own centroidal axis is zero.)
Directions of principal axes
Principal moments of inertia
(a) Maximum bending stress
w = rate of loading = 10 k N/m = 10,000 N/m = 100 N/cm
Components,
wu = w sin θ = 100 × 0.4065 = 40.65 N/cm
wv = w cos θ = 100 × 0.9138 = 91.38 N/cm.
The beam is simply supported and carries uniformly distributed load, the
maximum bending moment occurs at the centre of the beam.
Bending moment Mu = (wul2/8) = (40.65 × 160 × 160/8) = 130,080 N cm (as span
length I = 160 cm)
Bending moment Mu = (wul2/8) = (91.38 × 160 × 160/8) = 292,416 N cm
As it is obvious, maximum bending stress occurs at the point A with co-ordinates,
x = −2.375, y = 15 − 4.875 = 10.125.
Co-ordinates u = x cos θ + y sin θ = −2.375 × 0.9138 + 10.125 × 0.4065
= −2.170 + 4.116 = +1.946 cm
v = y cos θ − x sin θ = 10.125 × 0.9138 + 2.375 × 0.4065
= 9.252 + 0.965 = 10.217 cm.
Maximum bending stress,
Direction of neutral axis,
Deflection at the centre
where
w = rate of loading
Constant,
Span length,
l = 160 cm. E = 210 × 105 N/cm2
Deflection,
(in the direction perpendicular to the neutral axis)
Problem 18.3 A 3-m-long cantilever of I-section carries a load of 2 kN at the free
end and 3 kN at its middle. Line of load 2 kN is passing through the centroid of the
section and inclined at an angle of 30° to the vertical and the line of application of
a load of 3 kN is also passing through the centroid but inclined at 45° to the
vertical on the other side of a load of 2 kN as shown in Fig. 18.23. I-section has
two flanges 12 cm × 2 cm and web 16 cm × 1 cm. Determine the resultant bending
stress at the corners A, B, C and D.
Figure 18.23 Problem 18.3
Solution
Moment of inertia,
I-section shown is symmetrical about XX and YY axes, so principal
axes UU and VV passing through the centroid of the section are
along XX and YY axes.
Loads sved into components along U and V directions.
Components of 2 kN load,
Wu1 = 2,000 × sin 30° = 1,000 N
Wu1 = 2,000 × cos 30° = 1,732 N
Components of 3 kN load,
Wu2 = 3,000 × sin 45° = 3,000 × 0.707 = 2,121 N
Wu2 = 3,000 × cos 45° = 3,000 × 0.707 = 2,121 N
Bending moments at the fixed end,
Mu = Wu1 × 3 + Wu2 × 1.5 = −1,000 ×3 + 2,121 ×1.5 Nm = 181.5 Nm = 0.18 ×
10 N cm
5
Mv = Wv1 × 300 + Wv2 × 150 = 1,732 × 300 + 2,121 ×150 = 8.38 × 105 N cm
Resultant bending stresses
Due to Mu, there will be tensile stresses at points B and C and compressive stresses
at points D and A.
Due to Mv, there will tensile stress at points A and B, and compressive stress at
points C and D.
Stress,
Stress,
Problem 18.4 For an extruded beam, the cross-section is shown in Fig. 18.24.
Determine (a) location of the shear centre O and (b) the distribution of the shearing
stresses caused by vertical shearing force, F = 35 kN applied at O. Given Iz = 0.933
× 106 mm4.
Figure 18.24 Problem 18.4
Solution
Shear force,
F = 35,000 N
Moment of inertia,
Iz = 0.933 × 106 mm4
Shear stress at A,
Shear stress at D,
Shear stress at C,
(at the centre of web)
Shear force in the segment AB,
Shear force in the segment DE,
Taking moments about the centre of the web
Problem 18.5 Determine the location of the shear centre O of a beam of a uniform
thickness of 3 mm having the cross-section shown in Fig. 18.25.
Figure 18.25 Problem 18.5
Solution
Shear force in segment AB and EF will be equal and let us say it is equal to F1.
Shear force in segmentBC will pass through the point C. Therefore, we need not
calculate F2.
Shear force
Let us take dimensions in centimetre. Say, F is the vertical shear force on the
section applied at O.
Figure 18.26 Problem 18.5
Shear force,
We need not calculate the shear force F2 in segment BD as it is passing through the
point D.
Moment of inertia, Iz
Shear force,
Taking moments of the forces about the point D
or
e = 1.454 cm = 14.54 mm
Problem 18.6 Determine the location e of the shear centre, point O for the thinwalled member having the cross-section shown in Fig. 18.27. The member
segments have the same thickness t.
Figure 18.27 Problem 18.6
Solution
Moment of inertia of the section
Let us calculate the shear force in members AB or DE due to the vertical shear
force F at shear centreO, then
substituting the value of Ioz
Taking moments of the forces about the point C
Problem 18.7 A force P is applied to the web of the beam as shown in Fig. 18.28.
If e = 250 mm, determine the height h of the right side flange so that the beam will
deflect without any warping. The member segments have the same thickness t.
Figure 18.28 Problem 18.7
Solution
Moment of inertia about xx axis
neglecting web.
Shear force in the vertical segment CD (as in Fig.18.29),
Now,
e = 250 mm = 25 cm
Taking moment of the forces about the point E
Figure 18.29
Problem 18.8 Determine the location of the shear centre O of a thin-walled beam
of uniform thickness having the cross-section as shown in Fig. 18.30.
Solution
Take the dimensions in centimetre. Let us calculate the shear forces in vertical
segments DE and FG.
Figure 18.30 Problem 18.8
t = thickness of each segment
Say, the shear force on beam = F at O.
SF in vertical number DE,
where Iz = moment of inertia of the section about OZ axis
Similarly, shear force in vertical member FG,
Moment of inertia,
The moments of the shear forces about the point H is taken as shown in Fig. 18.31.
Figure 18.31
Problem 18.9 A beam with thin-walled semi-circular section is shown in Fig.
18.32. It is loaded in a principal plane xy, so as to produce simple bending in this
plane. Find the distance e for the shear centre O as shown in figure.
Solution
The shear stress τ at each section s = r·ø along the centre line of radius r will be in
the direction of the tangent to the line. τ is tangential to the centre line at s′ or
perpendicular to the line GS′ as shown. The magnitude of shear stress at s′,
We note that shear stress is maximum at (ø = π/2) and is zero at ø = 0 and π.
dF, elementary shear force = τb ds = τ br dø, dT, Moment of dF about centre G = δ
br2 dδ.
Total moment
Figure 18.32 Problem 18.9
The horizontal component of the elemental shear force, τbr dø above the neutral
axis OZ, cancel the horizontal components of those below the neutral axis, hence
the shear stress resultant is a vertical force equal to the shear force F at the section.
To produce twisting moment T, calculated above, this force must act through a
point O such that,
or,
where
Substituting the value of Iz,
Problem 18.10 Determine the location e of the shear centre point O for the thinwalled open pipe with the cross-section shown in Fig. 18.33.
Solution
The shear stress τ at any section, s = rφ along the centre line of radius r will be in a
direction tangential to centre line as shown.
Magnitude of shear stress
Figure 18.33 Problem 18.10
where,
Torque about the centre,
For the moment of inertia, consider an element at an angular displacement of θ,
Therefore, twisting moment T,
Problem 18.11 Determine the position of the shear centre of the section of a beam
shown in Fig. 18.34
Solution
Figure 18.34 shows the section for which the shear centre is to be determined. In
the diagram, direction of shear flow is given. Due to symmetry shear
forces, F1 = F5, shear forces, F2 = F4. The section is symmetrical about the axis XX;
therefore, shear centre will lie on this axis.
Let us determine shear force F1 or F5.
Shear stress in the vertical portion AB,
Figure 18.34 Problem 18.11
where F is the applied shear force on the section.
Now,
dA = t1dy
Shear force
Shear force in the horizontal portion BC,
Taking moments of the shear forces about the centre O of the vertical web,
Moment of inertia
Problem 18.12 For a section shown in Fig. 18.35, determine the position of the
shear centre. The thickness of the section is t throughout.
Figure 18.35 Problem 18.12
Solution
Due to symmetry
Shear force in portion AB, F1 = F4, shear force in portion CD
Shear force in portion BO, F2 = F3, shear force in portion OC
Shear force F1
Shear stress,
where F = applied shear force
Shear force,
where
a = z t (as shown) z is along AB
Therefore,
dA = t dz
Distance,
Shear force,
shear forces F2 and F3 are passing through the point O and will not produce any
moment about O.
Moment of inertia Ixx
Moment of inertia of AB, about their principal axes
(as shown in Fig. 18.35)
substituting cos θ = sin θ = 0.707
Figure 18.36 Problem 18.12
Similarly, the moment of inertia of BO
Now,
θ = 45°
as
Total moment of inertia of section,
Taking moments of the shear forces about the point O
or,
Key Points to Remember
o
o
o
o
Unsymmetrical bending occurs in a beam: (i) if the section is symmetrical
but load-line is inclined to the principal axes or (ii) if section itself is
unsymmetrical.
Product of inertia, Ixy = ∫ xydA
Product of inertia of a section about its principal axes is zero.
For a symmetrical section, principal axes are along the axes of symmetry.
Parallel axes theorem for product of inertia,
where
Ixy = Product of inertia about any co-ordinates axes X–Y
= Product of inertia about centroidal axes
= Coordinate of the centroid of the section about XY co-ordinates.
o
If Ixy, Iyy, Ixy are moments of inertia about any co-ordinates axes X–
Y passing through the centroid of the section. Inclination of principal axis
with respect to X–X axes.
Principal moments of inertia
o
If principal moments of inertia of a section are Iuu, Ivv, then moment of
inertia about an axis X–Xinclined at angle θ to U–V axis is
Ixx = Iuu cos2θ + Ivv sin2θ.
o
Stresses due to unsymmetrical bending, if u, v are the co-ordinates of a
point and M is the bending moment applied on the section and θ is the
angle of inclination of axis of M, with respect to the principal axes UU.
Resultant bending stress at the point
o
Angle of inclination of neutral axis with respect to principal axis UU
o
Deflection of a beam under load W causing unsymmetrical bending
where
K = Constant depending upon end conditions of the beam and position
of the load
θ = Angle of inclination of load W with respect to VV principal axis.
o
o
o
o
If the direction of the applied load on a beam passes through the shear
centre of the section, no twisting takes place in the beam.
For a section symmetrical about two axes, shear centre lies at the centroid
of the section.
For a section symmetrical about one axis only, shear centre lies along the
axis of symmetry.
About the shear centre, the moment due to the applied shear force is
balanced by the moment of the shear forces obtained by summing the
shear stresses over the various portions of the section.
Multiple Choice Questions
1. The product of inertia of a rectangular section of a breadth of 4 cm and a
depth of 6 cm about its centroidal axis is
1. 144 cm4
2. 72 cm4
3. 36 cm4
4. None of the above
2. The product of inertia of a rectangular section of a breadth of 3 cm and a
depth of 6 cm about the co-ordinate axes passing at one corner of the
section and parallel to the sides is
0. 81 cm4
1. 72 cm4
2. 54 cm4
3. None of these
3. For an equal angle section, co-ordinate axes XX and YY passing through
centroid are parallel to its length. The principal axes are inclined
to XY axes at an angle
0. 22.5°
1. 45.0°
2. 67.5°
3. None of the above
4. For an equal angle section, moment of inertia Ixx and Iyy are both equal to
120 cm4. If one principal moment of inertia is 180 cm4, the magnitude of
other principal moment of inertia is
0. 180 cm4
1. 120 cm4
2. 60 cm4
3. 30 cm4
5. For a section, principal moments of inertia are Iuu = 360 cm4 and Ivv = 160
cm4. Moment of inertia of the section about an axis inclined at 30° to
the U–U axis, is
0. 310 cm4
1. 260 cm4
2. 210 cm4
3. 120 cm4
6. For an equal angle section, Ixx = Iyy = 32 cm4 and Ixy = −20 cm4. The
magnitude of one principal moment of inertia is
0.
1.
2.
3.
52 cm4
42 cm4
32 cm4
16 cm4
7. For a T-section, shear centre is located at
0.
1.
2.
3.
Centre of the vertical web
Centre of the horizontal flange
At the centroid of the section
None of the above.
8. For an I-section (symmetrical about X–X and Y–Y axis) shear centre lies at
0.
1.
2.
3.
Centroid of top flange
Centroid of bottom flange
Centroid of the web
None of the above.
9. For a channel section symmetrical about X–X axis, shear centre lies at
0. The centroid of the section
1. The centre of the vertical web
2. The centre of the top flange
3. None of the above.
10.If the applied load passes through the shear centre of the section of the
beam, then there will be
0. No bending in the beam
1. No twisting in the beam
2. No deflection in the beam
3. None of these
Practice Problems
1. Figure 18.37 shows Z-section of a beam simply appalled over a span of 2 m. A
vertical load of 2 kN acts at the centre of the beam and passes through the centroid
of the section. Determine the resultant bending stress at points A and B.
2. Figure 18.38 shows a section of a beam subjected to shear force F. Locate the
position of the shear centre as defined by e.
Figure 18.37 Practice Problem 1
Figure 18.38 Practice Problem 2
3. Determine the location e of the shear centre point C for the thin-walled member
having the cross-section shows in Fig. 18.39, where b2 > b1, the member segments
have the same thickness t.
4. For an extruded beam having the cross-section shown in Fig. 18.40, determine (a)
location of shear centre and (b) distribution of shear stresses caused by vertical
shear force F = 12 kN.
Figure 18.39 Practice Problem 3
Figure 18.40 Practice Problem 4
5. Determine the location e of the shear centre for the thin-walled member having the
cross-section shown in Fig. 18.41. The member segments have the same thickness.
6. Determine the location of shear centre O of a thin-walled beam of uniform
thickness having the equilateral triangular section as shown in Fig. 18.42.
Figure 18.41 Practice Problem 5
Figure 18.42 Practice Problem 6
7. Determine the location of shear centre of a thin-walled beam of uniform thickness
having the cross-section shown in Fig. 18.43.
Figure 18.43 Practice Problem 7
Answers to Exercises
Exercise 18.2:
Exercise 18.3: 147.25 × 104 mm4, +32.30 × 104 mm4.
Exercise 18.4: 540 cm4, 540 cm4, −324 cm4
Exercise 18.5: 28° 31′, 118° 31′, 360 cm4, 38.33 cm4
Exercise 18.6: σA = –491.67 N/cm2, σB = +2,812.5 N/cm2
Exercise 18.7: 221.2 N, 15° 13′
Exercise 18.8: 2.7 cm
Answers to Multiple Choice Questions
1. (d)
2. (a)
3. (b)
4. (c)
5. (a)
6. (a)
7. (b)
8. (c)
9. (d)
10. (b)
Answers to Practice Problems
1. 147.44 MPa, 17.42 MPa
2.
3.
4. 23.2 mm, 12.36 MPa at B, 25.2 MPa at neutral axis
5.
6. e = 0.1443a
7. e = 2a
4b. Combined Bending and Direct Stresses
CHAPTER OBJECTIVES
The columns and struts are subjected to axial compressive load and if the load
becomes eccentric; then, in addition to direct compressive stress, there will be a
bending stress developed in the column/strut and these bending stresses are
compressive and tensile as well. However many columns made of brick/stone
masonry are incapable of withstanding tensile stress.
o
o
Students will learn, through the chapter, to calculate the resultant stresses
due to eccentric load and to mark the area of cross section, on which if the
load is applied at any point there will not be any tensile stress anywhere in
the section.
Similarly, frame of manufacturing machines are subjected to eccentric
load and section of frame has to be designed so as to withstand the
compressive and/or tensile stress.
Introduction
In this chapter, we will take various sections of columns/struts such as circular,
rectangular, I-section and hollow section subjected to eccentric loading causing a
bending moment in addition to a compressive force on struts/columns. For each
section, a core will be established, and if load is applied within the core, there will
not be any tensile stress developed anywhere in column section.
Wind loads on walls and chimneys cause bending moment on wall/chimney due
to which tensile and compressive stresses are developed in section of
wall/chimney. Depending upon the stress due to self-weight of the wall/chimney,
resultant stresses at the base of wall/chimney will be determined.
Eccentric Axial Thrust on a Column
A column of rectangular section B × D is shown in Fig. 10.1. G is the centroid of
the section abcd. A vertical load P is applied at G along xx axis such that GG′ = e,
eccentricity.
If this load acts at the CG of the section, then a direct compressive stress is
developed in section. Effect of the eccentric load is to produce bending
moment, M = Pe, on the section producing tensile and compressive stresses in the
section. Along the edge ad, there will be maximum tensile stress due to bending
and along edge bc, there will be maximum compressive stress due to bending.
Figure 10.1 Column under eccentric load
Say, Iyy = moment of inertia of section about yy axis
Direct compressive load = P
Direct stress,
Section modulus,
Maximum bending stresses developed in section,
Resultant stress at edge bc
Resultant stress at edge ad
If σb > σd, then resultant stress at edge ad will be tensile. Resultant stress
distribution along xx axis is shown in the Fig. 10.2.
Figure 10.2
Along edge bc
σb + σd = compressive stress
Along edge ad
Note that if
σb − σd = 0, no tensile stress is developed along edge ad
Therefore,
eccentricity must be less than
for no tensile stress to develop in
section, when load is applied along xx axis
Example 10.1 A cast iron column of a diameter of 250 mm carries a vertical load
of 600 kN at a distance of 50 mm from the centre of circular section. Determine the
maximum compressive and tensile stresses along the diameter passing through the
point of loading.
Solution
Compressive load,
P = 600 kN = 6,00,000 N
Diameter of column,
d = 250 mm
Area of cross section,
Section modulus,
= 1,533,980 mm3
Eccentricity,
e = 50 mm
Moment,
M = Pe = 6,00,000 × 50
= 30 × 106 N mm
Direct stress,
Bending stress,
= 12.223 N/mm2 (Compressive)
= ±19.557 N/mm2
Maximum compressive stress = 19.557 + 12.223 = 31.78 N/mm2
Maximum tensile stress = 19.557 − 12.223 = 7.334 N/mm2
Figure 10.3 shows the resultant stress distribution along xx axis
Figure 10.3
Exercise 10.1 A cast iron column of rectangular section 150 mm × 250 mm is
subjected to an eccentric vertical load of 750 kN. The vertical load is applied at a
point 40 mm away from centroid of the section along xx axis, which is parallel to
longer side. Determine maximum compressive and maximum tensile stresses in
section along the centroidal axis xx.
Load Eccentric to Both Axes (Rectangular Section)
Consider a column of rectangular section B × D as shown in Fig. 10.4. Centroid of
the section abcd is atG but a vertical load P is applied at G′ such that GG′ = e,
eccentricity. There are two components of this eccentricity along xx and yy axes,
that is, ex and ey as shown is the Fig. 10.4.
Figure 10.4 Load eccentric to both axes
Direct vertical load on column = P
Bending moment in xx plane,
Mx = Pex
Bending moment in yy plane,
My = Pey
Section modulus,
Section modulus,
Bending stresses along xx axis, edges ad and bc =
(Compressive along edge bc and tensile along edge ad)
Bending stresses along yy axis,
(Compressive along edge cd and tensile along edge ab)
Direct compressive stress,
Resultant stresses at corners
Note that we have taken compressive stress as positive stress in a column as the
tensile stress is undesirable.
Example 10.2 A cast iron column of a section of 200 mm × 250 mm is subjected
to a vertical load of 300 kN acting at a point 40 mm away (along the diagonal)
from the centre. Determine the resulting stress at the corners a, b, c and d of the
section (Fig. 10.5).
Solution
Sides B = 250 mm
D = 250 mm
Diagonal =
Figure 10.5 Example 10.2
= 320 mm
Eccentricity,
e = 40 mm
Eccentricity,
= 31.25 mm
= 25 mm
Bending moment,
Mx = Pex = 300,000 × 31.25 = 9.375 × 106 Nmm
My = Pey = 300,000 × 25 = 7.5 × 106 Nmm
Section modulus,
Direct stress,
Bending stress
Resultant stresses
σa = σd − σbx − σby = 6 − 4.5 − 4.5 = −3 N/mm2 (tensile)
σb = σd + σbx − σby = 6 + 4.5 − 4.5 = +6 N/mm2 (compressive)
σc = σd + σbx + σby = 6 + 4.5 + 4.5 = +15 N/mm2 (compressive)
σd = σd − σbx + σby = 6 − 4.5 + 4.5 = +6 N/mm2 (compressive)
Note that only at corner a, there is −3 N/mm2 (tensile stress) at all other corners
resultant stress is compressive.
Exercise 10.2 A cast iron column of square section 400 × 400 mm is subjected to a
compressive load of 500 kN acting at a point that is 60 mm away from xx axis and
80 mm away from yycentroidal axis. Determine the resultant stresses at the
extreme corners of the section.
[Hint: ex = 80 mm, ey = 60 mm]
Core of Rectangular Section
The core of a rectangular section is the part of the column section in which the load
can be applied without causing tensile stress anywhere in the section. Since the
columns are made of cast iron, brick work, concrete etc., which are not supposed to
withstand tensile stresses because tensile stress may develop cracks in column
leading to its failure.
Consider a rectangular section B × D of a column. Say a load P is applied at
point G′, such that GG′ = ex(Fig. 10.6).
Bending moment,
Mx = Pex
Section modules,
Stress due to bending,
Direct stress,
If the resultant stresses σd ± σbx has to be only compressive and no tensile stress
is permitted, then
Figure 10.6 Core of rectangular section
or
,
Similarly, load can be considered on the other side of G, and
In the figure, dimension
If the load is applied within the line ln along x axis, then there will not be any
tensile stress anywhere in the section.
Eccentricity along yy axis = ey
Bending moment,
My = Pey
Section modulus,
Bending stress,
Resultant stress,
If no tensile stress is permitted in the section, then
line mo shows that
If load is applied anywhere within line mo, no tensile stress is produced in the
section. Joining the endsl, m, n and o makes a rhombus of diagonals B/3
and D/3. This rhombus is termed as core of a rectangular section.
Core of Circular Section
Let us consider that the section of a column is circular of diameter d as shown
in Fig. 10.7, xx and yyare centroidal axes, with G as centroid of the section. Say
load P is applied along xx axis at G′, such thatGG′ = e, eccentricity
Bending moment,
Mx = Pe
Section modulus,
Direct stress,
Bending stress,
If the resultant stress is not to be tensile anywhere in the section, then:
or eccentricity,
Similarly, load can be considered along yy axis and on the other side of yy axis,
the
Figure 10.7 Core of a circular section
Core of the circular section is a circle of radius d/8 or diameter d/4 as shown in
figure. Area covered by a circle of diameter d/4 is called CORE or KERNEL of
circular section. If a load on column is applied within the core, then no tensile
stress will be developed anywhere in the section.
Core of Any Section
For any section, core can be marked if we know the area of section, Zx and Zy of
the section. Figure 10.8shows an I-section, D × B, with moment of
inertia Ixx and Iyy about xx and yy axes. Section modulus,
Maximum eccentricity along x axis,
where A = area of only section
Similarly, eccentricity along yy axis,
Core is a rhombus with diagonals 2ex and 2ey. A simple example will further
clarify the dimensions of core of an I-section.
Figure 10.8
Example 10.3 An ISLB 150 joist has following properties, area A = 1,808 mm2,
depth D = 150 mm, width, B = 80 mm, Ixx = 688.2 × 104 mm4 and Iyy = 55.2 ×
104 mm4. Mark the core of the I-section.
Figure 10.9
Solution
Allowable eccentricity,
Diagonals of rhombus of core are 108.14 and 15.26 mm as in fig 10.9.
Exercise 10.3 Mark the core of I-section with the following properties: Area =
4,320 mm2, depth = 203 mm, width = 106 mm, Ixx = 27 × 106 mm4, Zx = 266 ×
103 mm3, Iyy = 1.81 × 106 mm4, Zy = 34.2 × 103 mm3.
Example 10.4 A short hollow pier 1.6 m × 1.6 m outer sides and 1 m × 1 m inner
sides supports a vertical load of 3,000 kN at a point located on a diagonal 0.5 m
from vertical axis of the pier (Fig 10.10). Neglecting self-weight of the pier,
calculate the normal stress at four outside corners on a horizontal section of the
pier.
Figure 10.10 Example 10.4
Solution
Outer side = 1.6 m
Inner side = 1.0 m
Area,
A = 1.62 − 12
= 1.56 m2
Eccentricity about x axis,
Eccentricity about y axis,
Bending moment,
Mx = 3,000 × 0.353 = 1,059 kN m
My = 3,000 × 0.353 = 1,059 kN m
Direct stress,
Bending stress,
Resultant stresses
σa = 1,923 − 1,830.6 − 1,830.6 = −1,738.2 kN/m2 (tensile)
σb = 1,923 − 1,830.6 + 1,830.6 = +1,923 kN/m2 (compressive)
σc = 1,923 + 1,830.6 + 1,830.6 = +5,584.2 kN/m2 (compressive)
σd = 1,923 + 1,830.6 − 1,830.6 = +1,923 kN/m2 (compressive)
Exercise 10.4 A short cast iron column is of hollow circular section with an outer
diameter of 1.5 m and an inner diameter of 1.0 m as shown in Fig. 10.11. It is
subjected to a vertical load of 2,500 kN at G′, such that GG′ = 0.25 m. Determine
resultant stress at abcd edges of the section.
Figure 10.11
Wind Pressure on Walls
A masonry wall of breadth B, length L and height H is shown in Fig. 10.12.
Base of wall is abcd of area B × L. Horizontal wind pressure p acts on front face of
the wall.
Figure 10.12 Wind pressure on wall
Total wind pressure = p × H × L = Pw.
C.G of wind load Pw lies at H/2 from base.
Bending moment due to wind load,
Section modulus,
Bending stress due to M,
Say weight density of masonry = w
Direct compressive stress due to self-weight, σd = wH
Resultant stress along edge
(compressive)
Resultant stress along edge ad =
Note that wH should be
to avoid tension in base.
Example 10.5 A 10-m-high brick masonry wall of rectangular section 4 m × 0.5 m
is subjected to a horizontal wind pressure of 150 kN/m2 on the 4 m side. Find the
maximum and minimum stress intensities induced in the base, weight density of
masonry is 22 kN/m3.
Solution
Wall Height = 10 m
Length, L = hm
Width,
B = 0.5 m
w = 22 kN/m3, p = 150 kN/m2
Direct stress,
Bending stress,
σd = 22 × 10 = 220 kN/m2
= 180 kN/m2
σmax = 220 + 180
= 400 kN/m2 (compressive) = 0.4 N/mm2
σmin = 220 − 180 kN/m2
= 40 kN/m2 (compressive)
= 0.04 N/mm2
Exercise 10.5 A masonry brick wall of height 3 m, length 6 m and width 0.4 m is
subjected to a wind pressure p on the face of length 6 m and height 3 m. Weight
density of masonry structure is 21.8 kN/m3. What is the maximum value of p if
tensile stress is not to be developed in base due to wind ressure?
Wind Pressure on Chimney Shafts
High chimneys are subjected to wind pressures which may cause overturning of
the chimney at the base; therefore, these chimneys must be properly grouted in the
ground. Consider a chimney of heightH, internal diameter d and external
diameter D as shown in Fig. 10.13.
Figure 10.13 Wind pressure on chimney
Say weight density of chimney material = w
Direct compressive stress at the base due to self-weight, σd = wH.
Consider a small strip of width Rdθ, subtending arc angle dθ at the centre 0, and
making an angle θwith the axis xx as shown.
δP = wind force reaching the small strip.
δP = p Rdθ H cos θ = pHR cos θ dθ
Component of the force δP, normal to the strip, δPn = δP cos θ = pHR cos2θ dθ
Horizontal component of δP, δPHx = δPncos θ
= pHR cos3θ dθ
(10.4)
Another horizontal component of δPn, that
is, δpHy = δPsin θ = pHR cos2θ sin θ dθ is cancelled out when we consider a strip in
other quadrant (as shown)
Total force component in xx direction = 2pHR cos3θ dθ
Integrating over the whole exposed surface from θ = 0 to
Total wind force,
where k = coefficient of wind resistance, D = 2R
DH = projected area of curved surface of chimney
CG of the force lies at a distance of
From the base, in the case of uniform cylindrical chimney, M bending moment
due to wind force =
Section modulus,
Bending stress,
Generally the coefficient of wind resistance is taken as 0.6 for cylindrical
chimneys.
Example 10.6 A 20-m-high cylindrical chimney shaft is of hollow circular section,
2.4 m external diameter and 1.2 m internal diameter. The intensity of horizontal
wind pressure varies as y2/3, where yis the height above the ground. Determine the
maximum and minimum intensity of stress at the base (Fig 10.14). Given that:
1.
Density of masonry structure is 22.2 kN/m3
2. Coefficient of wind resistance, k = 0.6
3. Wind pressure at a height of 27 m is 1.8 kN/m2
Figure 10.14
Solution
Say intensity of wind pressure, p = Cy2/3
At y = 27 m and p = 1.8 kN/m2
Therefore,
1.8 = C × 272/3
9C = 1.8
C = 0.2
Pressure at any height,
Area,
y = 0.2y2/3
dA = 2.4dy
p = 0.2y2/3
dP = kP dA = k × 0.2y2/3 × 2.4 dy
= 0.48 ky2/3 dy
However,
k = 0.6
dP = 0.6 × 0.48 × y2/3dy = 0.288y2/3dy
Moment of the force from the base
dM = dPy = 0.288 y5/3dy
Total moment,
Section modulus,
Bending stress,
Height,
H = 20 m
Direction stress,
σd = wH = 22.2 × 20 = 444 kN/m2
Maximum stress at base = 444 + 248.50 = 692.5 kN/m2 (compressive)
Minimum stress at base = 444 − 248.5 = 195.5 kN/m2 (compressive)
Exercise 10.6 A 20-m-high masonry chimney of uniform circular section, 5 m
external diameter and 3 m internal diameter has to withstand a horizontal wind
pressure of 2 kN/m2 of the projected area. Find the maximum and minimum stress
intensity at the base. Density of masonry structure w= 21 kN/m3.
Problem 10.1 A short column of hollow circular section of internal diameter d and
external diameterD is loaded with a compressive load W. If D = 1.5d, determine
the diameter of the core (Fig 10.15).
Figure 10.15
Solution
Outer diameter = 1.5d
Inner diameter = d
Area of a cross section
Moment of inertia,
Section modulus,
Radius of core,
Eccentricity
2e = 0.271 × 2d = 0.542d
Diameter of the core
= 0.542d
Problem 10.2 The cross section of a short column is of the shape of an arrow. Find
the position of the vertical downward force on the line of symmetry of the section
so that at edge A stress is just zero (Fig. 10.16).
Solution
AB is line of symmetry or the axis yy.
Let us locate the position of G with respect to lower edge B,
Moment of inertia,
Direct stress,
Bending stress at A,
GG′ = e, as shown in Fig. 10.16.
Figure 10.16
Problem 10.3 To avoid interference, the cross-sectional area of a link in a machine
is reduced in section as shown is Fig. 10.17. The thickness of the link is 20 mm.
Determine the maximum force that can be applied if the maximum normal stress in
section AB is limited to 100 MPa.
Solution
Say P is the force on N.
Area,
A = 20 × 50
= 1,000 mm2
Figure 10.17
Eccentricity of load for section AB
Moment = Pe
= 25 P N mm
Section modulus,
Direct stress,
Bending stress,
Maximum normal stress = σd + σb
or
P + 3P = 1,00,000
4P = 1,00,000
P = 25,000 N = 25 kN
Problem 10.4 The cross section of a short column is shown in Fig. 10.18. Load of
160 kN is applied atP, 75 mm from edge AD. Section is symmetrical about xx axis,
determine the stresses at corners A, B, Cand D of the section.
Figure 10.18 Problem 10.4
Solution
Position of G along xx axis
Eccentricity,
e = 75 − 71.82 = 3.18 mm
Area, A = 11,000 mm2
P = 1,60,000 N
Direct compressive stress,
Moment of inertia,
Edge BC will be in compression, edge AD will be in tension
Resultant stresses
σAD = 14.54 − 1.322 = 13.218 N/mm2 (compressive)
σBC = 14.54 − 1.44 = 15.98 N/mm2 (compressive)
Problem 10.5 A steel rod with a diameter of 20 mm passes through a copper tube,
30 mm internal diameter and 40 mm external diameter. Rigid cover plates are
provided at each end of the tube and steel rod also passes through these cover
plates as shown in Fig.10.19. Nuts are screwed on the projecting ends of the rod, so
that the cover plates put pressure on the ends of the tube. Determine maximum and
minimum stresses in copper tube if the nut is tightened to produce a linear strain of
0.001 in the rod. The centre of the rod is 3 mm out of the centre of the tube.
Given ES = 2EC = 210 GPa.
Solution
Strain in steel rod = 0.001 tensile
Es of steel = 2,10,000 N/mm2
Stress in rod,
σs = 2,10,000 × 0.001
= 210 N/mm2 (tensile)
Area of cross section of steel rod
= 314.16 mm2
Area of cross section of the copper tube =
= 175p mm2 = 549.78 mm2
Figure 10.19 Problem 10.5
Pull in steel rod = push in copper tube
P = 210 × 314.16 = sc × 549.78 = 65973.6 N
Stress,
σc = 120 N/mm2 (compressive)
Now rod centre is 3 mm out of the centre of the tube (Fig. 10.19)
Moment of inertia,
Iyy =
= 8.59 × 104 mm2
e = eccentricity = 3 mm
Bending stress in tube,
Maximum stress in copper tube,
σmax = 120 + 460 8
= 166.08 N/mm2 (compressive)
σmin = 120 − 46.08
= 73.92 N/mm2 (compressive)
Problem 10.6 A large C clamp is shown in Fig. 10.20. As the screw is tightened
down, strain observed in vertical direction at point B is 400 μ strain. What is the
load on the screw if E = 200 GPa.
Figure 10.20 C clamp
Solution
Section is symmetrical about xx axis, G can be located along xx axis
Moment of Inertia,
Eccentricity,
e = 300 + x1 = 300 + 25 = 325 mm
Moment,
M = P e = P × 325 N mm.
Area,
A = 800 + 800 = 1,600 mm2
Bending stress at B,
At point B, direct stress and bending stress both are compressive so
Problem 10.7 A machine part is transmitting a pull of 3 kN. It is offset as shown in
the Fig. 10.21. Determine the largest normal stress in the offset position.
Solution
Eccentricity of the load
= 20 mm
Moment
= Pe
= 3,000 × 20
= 60,000 Nmm
Figure 10.21 Machine part
Edge A comes in compression, edge BC comes is tension.
Direct stress,
Bending stress,
Resultant stresses
σA = 53.33 − 6.667 = 46.666 N/mm2 (compressive)
σBC = −26.667 − 6.667 = −33.334 N/mm2 (tensile)
Maximum normal stress is 46.666 N/mm2 (compressive)
Problem 10.8 A tapering chimney of hollow circular section is 45 m high. Its
external diameter at the base is 3.6 m and at the top it is 2.4 m. It is subjected to a
wind pressure of 22 kN/m2 of the projected area. Calculate the overturning moment
at the base. If the weight of the chimney is 6,000 kN, and the internal diameter at
the base is 1.2 m, determine the maximum and minimum stress at the base.
Solution
Weight of chimney = 6,000 kN
Area at base
σd = direct camp.
Stress at base
Projected area
= 135 m2
Total wind load = 135 × 22 = 2,970 kN
CG of the load from the base
Overturning moment at base, M = 21 × 2,970 = 62,370 kN m
Figure 10.22 Tapering chimney
Section modulus, base section,
Bending stress,
σmax = 13,786.78 + 663.15 = +14,450 kN/m2 (compressive)
σmax = −13,786.7 + 663.15 = −13,123.6 kN/m2 (tensile)
or
σmax = 14.45 N/mm2 (compressive)
σmin = −13.1236 N/mm2 (tensile)
Problem 10.9 A masonry pillar of diameter D in m is subjected to a horizontal
wind pressure of intensity p kN/m2. If the coefficient of wind resistance is k, prove
that the maximum permissible heightH of the pillar so that no tension is induced at
the base is given by:
where w = weight density of masonry
Solution
Weight density of masonry = w kN/m2
Say permissible height = Hm
Direct stress due to self-weight, σd = wH (compressive)
Intensity of wind pressure = p kN/m2
Bending moment,
Section modulus,
Bending stress,
For no tension,
or,
Key Points to Remember
o
o
o
o
o
o
o
A short column of rectangular section,
load along x axis, with ex as eccentricity
, carries eccentric
Resultant stress,
A short column of circular cross section of diameter D, supports an
eccentric load P at an eccentricity e, Resultant stress,
The core or the kernel of a section is a small area located around the
centroid of the section of a column and if any vertical load is applied on
the column within this area of core, there will not be any tensile stress
developed anywhere in the section.
Core of a rectangular section B × D, is a rhombus of diagonals B/3
and D/3.
Core or kernel of a circular section of diameter D is a circular area of
diameter D/4.
For a wall of rectangular section B × D, wind pressure p acting on the face
of breadth B and heightH, stress due to bending moment created by the
wind pressure is
at the base of the wall.
For a chimney of outside diameter D, height H, coefficient of wind
resistance k, wind pressure p, and inner diameter d.
Total wind force,
Moment,
Section modulus,
Bending stress,
Pw = kp DH
Direct compressive stress,
Review Questions
1. What is the core or kernel of a section? What is its importance?
2. Mark the core of following sections.
1. A rectangular section
2. A circular section
3. A hollow circular section
4. I-section.
3. How the wind pressure on a wall produces an overturning moment on the
wall?
4. What do you mean by coefficient of wind resistance for a chimney?
5. How the overturning moment due to wind load on a chimney is
calculated?
6. Consider a C-clamp and show how the direct and bending stresses are
developed in the critical section of the frame.
7. A cast iron column of hollow circular section is made by casting. What is
the effect on stresses developed in column due to an axial compressive
load if the core is offset during casting?
Multiple Choice Questions
1. A cast iron column is of circular section of a diameter of 200 mm. What is
the diameter of core of the column?
1. 25 mm
2. 40 mm
3. 50 mm
4. None of these
2. A steel column has an outside diameter of 80 mm and an inside diameter
of 60 mm. What is the diameter of core of the column?
0. 62 mm
1. 31.25 mm
2. 30 mm
3. None of these
3. A column with I-section has following properties: Zx = 90 × 103 mm3, Zy =
54 × 103 mm3, A = 1,800 mm2. What is the allowable eccentricity ey for no
tension in column?
0. 30 mm
1. 40 mm
2. 50 mm
3. None of these
4. A short column is of hollow square section with outer side 2a and inner
side a. A load acts at a distance of 0.25a from CG of the section, along
one diagonal. The maximum and minimum stresses developed at corners
of the section are 4.8 and −1.2 N/mm2, respectively. The bending stress
introduced at the extreme corners of the section by the eccentric load are
0. ±3.6 N/mm2
1. ±2.4 N/mm2
2. ±1.2 N/mm2
3. None of these
5. For a cylindrical chimney of hollow circular section, subjected to wind
pressure, the coefficient of wind resistance is generally taken as
0. 0.3–0.5
1. 0.45–0.55
2. 0.6–0.7
3. None of these
6. A column is of hollow circular section. Due to an eccentric load,
maximum and minimum stresses are 7 and 1 N/mm2 (both compressive),
respectively. Now, the eccentricity is doubled, what will be the maximum
stress developed in column section?
0.
1.
2.
3.
14 N/mm2
10 N/mm2
8 N/mm2
None of these
7. A short masonry square section of 1 m side is 10 m high. Wind pressure of
intensity 2kN/m2 acts on a vertical face of column. Weight density of
masonry is 20 kN/m3 The maximum stress at the base of the column is
0. 800 kN/m3
1. 400 kN/m2
2. 200 kN/m2
3. None of these
8. A chimney is of brick masonry of a weight density of 22 kN/m3. Height of
chimney is 10 m. Outside diameter of chimney is 1 m while inside
diameter is 0.5 m. What is the compressive stress developed at base, due
to self-weight?
0. 0.275 N/mm2
1. 0.22 N/mm2
2. 0.20 N/mm2
3. None of these
9. A cast iron column of circular section carries an eccentric load due to
which maximum and minimum stresses developed in column section are
+420 kN/m2 and +120 kN/m2 (compressive), respectively. If Z = 1.2 m3,
what is the bending moment due to eccentric load?
0. 270 kN m
1. 225 kN m
2. 180 kN m
3. None of these
10.A cylindrical chimney of hollow circular cross section is subjected to wind
pressure p. Density of masonry work is 20 kN/m3. The maximum and
minimum stresses developed at the base of chimney are 650 and 150
kN/m2, respectively. If intensity of wind pressure is increased by 50 per
cent, what is the maximum stress developed in section of column?
0. 850 kN/m2
1. 775 kN/m2
2. 700 kN/m2
3. None of these
Practice Problems
1. A short column of hollow circular section of internal diameter d and external
diameter D is subjected to a vertical load (eccentric). If d = 0.8 D, determine the
diameter of the core.
2. A short block has cross-sectional area of a triangle as shown in Fig. 10.23.
Determine the range along the axis yy over which the downward vertical force
could be applied at the top of the block without causing any tension anywhere
in the base. Neglect the weight of the block.
3. A flat plate of section of a thickness of 20 mm and a width of 60 mm, which is
placed in a testing machine, is subjected to 60 kN of load acting along line AB as
shown in Fig. 10.24, an extensometer adjusted along the line of the load recorded
an extension of 0.078 mm on a gauge length of 150 mm. Determine (i) minimum
and maximum stresses set up in plate, and (ii) Young’s modulus of the plate.
Figure 10.23 Practice Problem 2
Figure 10.24 Practice Problem 3
4. The cross section of a short column is shown in Fig. 10.25. Load of 625 kN is
applied at P, 150 mm from edge AB, on yy axis. Determine the stresses developed
at corners A, B, C and D of the section.
5. A short cast iron column has an external diameter of 200 mm and an internal
diameter of 160 mm. The distance between the centre of outer and inner circles due
to the displacement of core during casting is 6 mm. A load of 400 kN acts through
a vertical centre line passing through the centre of the outer circle. Calculate the
values of greatest and least compressive stresses in a horizontal cross section of the
column.
[Hint: Locate G along x-axis, determine x1 and x2, calculate Iyy.]
Figure 10.25 Practice Problem 4
6. A C clamp is shown in the Fig. 10.26. As the screw is tightened on wooden blocks,
it was observed that a compressive force of 2 kN is applied on blocks. Section
of C clamp is given in Fig. 10.26. Determine the stresses at points A and B of the
section.
7. A bar of circular section with a diameter of 20 mm is subjected to a load P, 10 mm
off the centre as shown in Fig. 10.27. If the maximum stress in bar is not to exceed
80 MPa, calculate the value of P.
Figure 10.26 Practice Problem 6
Figure 10.27 Practice Problem 7
8. A tapering chimney of hollow circular section is 30 m high. Its external diameter
at the base is 3 m and at the top it is 2 m. It is subjected to a wind pressure of 2
kN/m2 of the projected area. Calculate the overturning moment at the base, if the
weight of the chimney is 5,000 kN and its internal diameter at the base is 1.2 m.
Determine the maximum and minimum stresses at the base.
9. A masonry pillar of diameter 0.8 m is subjected to a horizontal wind pressure of
1.2 kN/m2. If the coefficient of wind resistance is 0.6, determine the maximum
permissible height of the pillar so that no tension is induced at the base. Given w =
23.2 kN/m3
Special Problems
1. Mark the core of an I-section with the following properties: area = 12,700 mm2,
depth = 229 mm, width = 210 mm, Ixx = 113 × 106 mm4, Iyy = 36.6 × 106 mm4.
2. The cross section of a short column is as shown in Fig. 10.28. A vertical load
of W is applied at G′,GG′ = 10 mm. Determine (a) magnitude of W if maximum
stress set up in the cross section is not to exceed 75 MPa and (b) stress distribution
along the edge AD.
Figure 10.28 Special Problem 2
3. A short hollow cylindrical column carries a vertical load of 400 kN. Its external
diameter is 200 mm and internal diameter is 120 mm. Find the maximum
permissible eccentricity of the load if the allowable stresses are 60 MPa in
compression and 25 MPa in tension.
[Hint: Calculate σd, σd − σb = −25, σd + σb = 60]
4. An 80-mm (t = 6 mm) steel pipe is used as a support for a basket ball backboard as
shown in Fig. 10.29. It is securely fixed into the ground. Calculate the maximum
normal stress that would be developed if a 750-N player hangs on the base of the
rim of the basket.
[Hint: (e = 1.2 m) (80-mm pipe inside diameter)]
Figure 10.29 Special Problem 4
Answers to Exercises
Exercise 10.1: 39.2 N/mm2 (compressive), no tensile stress
Exercise 10.2: +4.063, 9.687, 2.187, −3.437 MPa (tensile)
Exercise 10.3: ey < 60.87 mm, ex < 7.83 mm, Rhombus with diagonals 15.66 mm
and 121.74 mm
Exercise 10.4: σa = 196.1 kN/m2 (compressive), σb = σd = 2546.6 kN/m2, σc =
4897.1 kN/m2(compressive)
Exercise 10.5: (0.3875 kN/m2)
Exercise 10.6: 607.26 kN/m2 (compressive), 232.74 kN/m2 (compressive)
Answers to Multiple Choice Questions
1. (c)
2. (b)
3. (c)
4. (a)
5. (c)
6. (b)
7. (a)
8. (b)
9. (c)
10. (b)
Answers to Practice Problems
1. (0.41 D)
2. yG = 100 mm, Gy = 200 mm, e1 = 25 mm, e2 = 50 m, range = 75 mm
below AB to 150 mm below edge AB
3. −40 MPa, 140 MPa, 200 GPa
4. σA = σB = +8.775 N/mm2 (compressive), σD = σC = 11.225
N/mm2 (compressive)
5. 46.01, 26.78 MPa
6. 61.62 N/mm2 at B, −31.36 N/mm2 at A (tensile)
7. 6.2832 kN
8. 2,100 kN m; 1653.1 kN/m2 (compressive), 29.1 kN/m2 (compressive)
9. 4.05 m
Answers to Special Problems
1.
2.
3.
4.
Rhombus of diagonals 54.88 and 155.4 mm
237.2 kN; 75 MPa (compressive), 17.4 MPa (compressive)
48.50 mm
27.952 N/mm2
5a. Thin Cylindrical and Spherical Shells
CHAPTER OBJECTIVES
Thin cylindrical and spherical shells are used mainly for storage of gas, petrol,
liquid, chemicals, grains and so on. Some are subjected to internal/external
pressures and the order of pressure is low (10–30 atmospheres). Their D/t ratio,
that is, the ratio of diameter to wall thickness, is large, that is, D/t is greater than
20. Because in comparison to diameter, thickness is very small, so the variation of
stresses along the thickness is taken negligible. In this chapter, we will learn about:
How to determine hoop stress and axial stress in a thin shell
subjected to internal pressure.
o
How to determine axial, circumferential and volumetric strains in
shell subjected to internal pressures.
o
The hoop stress developed is double the axial stress in thin
cylindrical shell. To reduce the effect of hoop stress; cylinder is wound
with a wire under tension, with the purpose of reducing the effect of hoop
stress.
o
A thin spherical shell is equally strong in two hoop directions;
therefore, generally the cylindrical shells are provided with hemispherical
ends.
o
Some pressure vessels have double curved wall, that is, at a
particular element radii of curvature in two directions are ρ1 and ρ2 (radius
of curvature). So pressure vessels with double curved wall will also be
analysed.
o
Conical tanks are available for storage of water. The hoop stress
developed is proportional to the square of the depth of element from free
water surface. Stress in conical water tanks will also be analyzed.
o
Introduction
Thin pressure vessels are subjected to internal uniform pressure or hydrostatic
pressure of low magnitude. Development of axial and hoop stresses in the wall
thickness of the pressure vessel provides useful data for the designer. Stresses in
thin shells are dependent on the D/t ratio, that is, the ratio of shell diameter to wall
thickness.
Derivation of axial and hoop stresses, strains in cylindrical pressure vessel and
derivation of hoop stresses, hoop strain and volumetric strain in both shells forms
the text of this chapter.
The effect of wire winding on the reduction of hoop stresses (which is greater
than axial stress) will be analysed.
Pressure vessels with the double curved surface of wall and the conical surface
of wall will also be analysed. In such shells, equations for hoop stresses will be
derived.
Cylindrical pressure vessel can be conveniently used to determine
experimentally E and v, that is, elastic constants of material of pressure vessel.
Thin Cylinder Subjected to Internal Pressure
Consider a thin cylindrical shell of inner diameter D, length L, with wall thickness
equal to t. The shell is full of liquid. The volume of the liquid inside the shell is
[π/4 (D2L)]and the pressure of the liquid is atmospheric. Now, the additional
volume of the liquid is pumped under pressure inside the shell; as a result, there is
overall expansion in the shell, that is, expansion in both diameter and length.
However, the expansion of liquid is partly prevented by the wall of the metallic
cylinder, which offers equal and opposite reaction and compresses the liquid inside
the cylinder (Fig 5.1).
Figure. 5.1 Small element of cylinder subjected to internal pressure
Mathematically
δV = additional volume of the liquid pumped inside.
= δv1 + δv2
= expansion in the volume of cylinder + contraction in the volume of the
liquid.
Since the liquid exerts pressure on cylinder wall, the length and diameter of the
cylinder get increased introducing axial and circumferential stresses, in the
cylinder.
Figure 5.2 shows a small element of the cylinder subjected to internal
pressure p; σc is the circumferential a hoop stress developed (along circumference)
and σa is the axial stress developed (along the axes); at outer surface pa is
atmospheric pressure, such that p >>pa.
Axial stress
Diameter of cylinder = D
Wall thickness = t
Axial stress = σa
Figure. 5.2 Small element under internal pressure
Figure. 5.3 Axial bursting force
Axial bursting force on cylinder in axial direction, [Pa = π/4(D2p)] acting on the
end plates as shown inFig. 5.3
Resisting area = πDt
Axial stress developed,
Circumferential stress (σc)
Consider a small portion along length δL and inner radius D/2, subjected to internal
pressure p, Fig 5.4Take a small element at an angle θ from x-axis, subtending an
angle dθ at inner surface.
Area of the small element
Force on the small element,
Vertical component of
Horizontal component of
The horizontal component of the force is cancelled out when the force is
integrated over the semicircular portion of the cylinder.
Figure. 5.4 Diametral bursting force
Therefore, total diametral bursting force,
(= p × projected area of the curved surface)
Area of cross-section resisting the diametral bursting force = 2t dl
Circumferential stress developed,
Strains
Consider Fig. 5.2, σc, σa and p stresses are perpendicular to each other. Note
that p acts as a compressive radial stress on inner surface of cylinder.
Stresses σc and σa are much larger than p; Therefore in calculation of strains, the
effect of internal pressure p is neglected. Say E is Young’s modulus and v is
Poisson’s ratio of the material.
Circumferential strain,
Axial strain,
Note that εc = circumferential strain
εa = axial strain
Change in the diameter of cylinder,
Change in the length of cylinder,
δD = εcD
δL = εaL
Volumetric strain
Volume of cylinder [v = π/4 (D2L)], taking partial derivative
or
= 2 × circumferential strain + axial strain
Change in volume
Change in volume of cylinder,
δV1 = ev V
Contraction in volume of liquid,
where K is bulk modulus of liquid.
Additional volume of liquid pumped inside the cylinder, δv = δv1 + δv2.
Note that if, in any problem, K for the liquid is not given that δv2, that is,
reduction in volume of liquid can be taken as negligible.
Example 5.1 A thin cylindrical shell made of 5-mm-thick steel plate is filled with
water under pressure of 3 N/mm2. The internal diameter of the cylinder is 200 mm
and its length is 1.0 m. Determine the additional volume of the water pumped
inside the cylinder to develop the required pressure. Given for steel E = 208
kN/mm2 and v = 0.3, and for water K = 2,200 N/mm2.
Solution
p = 3 N/mm2, E = 208 kN/mm2
D = 200 mm, v = 0.3
t = 5 mm
L = 1.0 m
Volume,
, putting the values
Volumetric strain,
= 0.548 × 10−3
Change in volume of the cylinder,
Change in volume of water,
δV1
=
εvV
=
0.548 × 10−3 × π × 107
=
1.72 × 104 mm3
, where P/K is volumetric strain on water
Additional volume of water pumped in shell, δV = δV1 + δV2
= 1.72 × 104 + 4.284 × 104 mm3 = 6.004 × 104 mm3
= 60 cm3 = 60 cc of water.
Exercise 5.1 A thin cylindrical shell made of 4-mm-thick copper plate is filled
with oil under a pressure of 2.4 N/mm2. The internal diameter of the cylinder is
200 mm and its length is 800 mm. Determine the additional volume of oil pumped
inside the cylinder so to develop the required pressure. Given E for copper = 104
kN/mm2, v for copper = 0.32, and K for oil = 2,800 N/mm2.
Thin Spherical Shell
Consider a thin spherical shell of diameter D and wall thickness t subjected to
internal fluid pressure pas shown in Fig. 5.5(a). The internal pressure p acts
throughout the inner curved surface and thediametral bursting force tends to break
the thin shell in two halves, that is, two hemispherical halves, as shown in Fig.
5.5(b).
Figure. 5.5 Thin spherical shell
Diametral bursting force, PD = p × projected area of curved surface
Say σc = hoop stress developed in the wall of the shell.
Area resisting the diametral bursting force = πDt
So
Strains
Consider a small element of thin spherical shell subjected to internal radial stress p,
due to which circumferential stress σc is developed in shell, as shown in Fig. 5.6.
Figure. 5.6 Small element of thin spherical shell
pa = atmosphere pressure on outer surface.
If E and v are Young’s modulus and Poisson’s ratio of the material, respectively,
then
Circumferential strain
=
, putting value of σc
Internal pressure p << σc; therefore, the effect of p is not considered in
determining the strain εc.
Volume of the shell,
as diametral strain = circumferential strain
Volumetric strain,
, putting the value of εc
Note that circumferential stress in thin spherical shell is only PD/4t; but in thin
cylindrical shell, circumferential stress is PD/2t, that is, two times, the stress which
is developed in a spherical shell, so it is advantageous to use a thin spherical shell,
but economically it is costlier to make a thin spherical shell than a thin cylindrical
shell. To make a thin spherical shell, two hemispherical portions are made and then
joined. To make a hemispherical portion, several segments are made and joined to
develop the hemispherical part.
Example 5.2 A thin spherical shell of a wall thickness of 2.5 mm and a diameter of
500 mm is subjected to an internal pressure p. What is the magnitude of p if
diametral strain in the shell is limited to 0.0005 only. E = 200 GPa, v = 0.3
Solution
Diameter strain, εD = εc circumferential strain
or
Exercise 5.2 A thin spherical shell of a diameter of 300 mm and wall thickness t is
subjected to internal pressure of 2.4 N/mm2. Determine the wall thickness t if the
hoop stress in the shell is limited to 60 MPa. What is the change in diameter of the
shell, if E = 105 kN/mm2 and v = 0.32.
Cylindrical Shell with Hemispherical Ends
For thin cylindrical shell subjected to internal pressure, we have derived
expressions for axial strain, εa= [(PD/4tE)(1 − 2v)] and we have assumed that
along the length of the shell, axial strain remains constant, meaning that there is no
distortion of end plates. But in actual practice there is distortion of end plates.
Moreover, we have learnt that hoop stress developed in spherical shell is only half
the hoop stress developed in cylindrical shell for same values of p, D and t.
Therefore, it is advantageous to use hemispherical ends for thin cylindrical shell as
shown in Fig. 5.7. Figure shows a cylindrical shell ofdiameter D, length L and wall
thickness t, with hemispherical ends of diameter D and wall thickness t2.
Obviously, the thickness of hemispherical portion will be less than the wall
thickness of cylindrical part.
Figure. 5.7
Say, σc′ = hoop stress in cylindrical part
σc″ = hoop stress in hemispherical part
Similarly hoop strain
εc′ = hoop strain in cylindrical part
εc″ = hoop strain hemispherical part
There can be two criteria for design in this case:
1. Hoop stress developed in both portions is the same.
or
t2 = 0.5t1
(5.4)
2. Hoop strain in both the portions is the same, so that there is no distortion
at the junction of two portions.
Example 5.3 A thin cylindrical steel shell of a diameter of 150 mm and a wall
thickness of 5 mm has hemispherical ends. Determine the thickness of
hemispherical ends, if there is no distortion of the junction under pressure. E = 208
GPa, v = 0.3
Solution
Thickness of the cylindrical portion, t1 = 5 mm
Poisson’s ratio, v = 0.3
For no distortion of junction under pressure,
or,
= thickness of the hemispherical end.
Exercise 5.3 A thin cylindrical copper shell of a diameter of 125 mm and a wall
thickness of 4 mm has hemispherical ends. Determine the thickness of
hemispherical ends, if hoop stress developed in both portions is the same. What is
the ratio of their hoop strains given v = 0.32?
Wire Winding of Thin Cylindrical Shells
We have learnt that due to internal pressure, hoop stress developed in the
cylindrical shell is double the axial stress. Under pressure, there are chances of
development of longitudinal crack in the cylinder, if circumferential stress exceeds
the safe limit. Moreover, due to high hoop stress, the pressure-bearing capacity of
cylindrical shell is reduced. Therefore, to reduce the intensity of hoop stress and to
strengthen the cylinder in longitudinal direction, a wire under tension is wound
around the circumference of the cylinder, as shown in Fig. 5.8 tensile stress in
winding wire puts pressure on cylinder from outside and a negative hoop stress is
developed in the cylinder wall.
Figure. 5.8
Consider a thin cylinder of diameter D, length L, and wall thickness t. It is
wound with a single layer of wire of diameter dw under an initial tension of σw.
Numbers of turns over the cylinder,
Tensile force exerted by the wire over cylinder
(Note that there are two sections)
Compressive force developed in cylinder
= σc 2Lt
For equilibrium
Putting the value of
This is the initial compressive hoop stress developed in cylinder wall due to wire
winding.
Figure 5.8 shows, the development of σc, in cylinder
Internal pressure
Now the wire-wound cylinder is subjected to internal pressure p. Due to this say
the stresses developed in cylinder are σa′, σc′ and σw′.
Axial bursting force in cylinder,
Axial stress developed,
Dimensional bursting force, PD = pDL
Say, the stresses due to internal pressure developed in cylinder and wire are,
respectively, σc′, σw′.
Then,
Putting the value of
or
There are two unknowns σc′ and σw′ but only one equation, so let us consider
compatibility condition, taking strain into account.
Circumferential strain in wire = circumferential strain in cylinder.
Let us take Ec and vc as Young’s modulus and Poisson’s ratio for cylinder,
respectively, and Ew and vw as Young’s modulus and Poisson’s ratio of wire,
respectively.
From Eqs (5.9) and Eqs (5.10), σc′ in cylinder and σw′ in wire can be determined.
Resultant stresses
In cylinder,
σcr = σc + σc′ (tensile)
In wire,
σwr = σw + σw′ (tensile)
Example 5.4 A thin bronze cylinder of 250 mm internal diameter and 6 mm thick
is wound with a single layer of 1.5-mm-thick steel tape under a tensile stress of
100 N/mm2. Find the maximum internal pressure if the hoop stress in the cylinder
is not to exceed 50 N/mm2. Determine also the tensile stress in steel tape. For
bronze E = 117 N/mm2, v = 0.33 and for steel E = 208 kN/mm2.
Solution
Let us consider only a unit length of cylinder
σct × 2 × 1 = σt t′ 2
where t = thickness of cylinder = 6 mm
t′ = thickness of tape = 1.5 mm
Say internal pressure is p, then
p · D · 1 = σ′c × 2t × 1 + σ′t × 2t′ × 1
p × 250 = σ′c × 2 × 6 + σ′t × 2 × 1.5
250 p = 12 σ′c + 3σ′t
(5.8)
where σc′ and σt′ are the additional stresses in cylinder and tape, respectively due to
internal pressure in cylinder.
Axial stress in cylinder,
using strain compatibility
Putting the values
Putting the value σt′ of from Eq. (5.9) into Eq. (5.8)
250p
17.333 σ′c
σ′c
=
12σ′c + 3(1.77 σ′c − 6.11p)
=
12σ′c + 5.333 σ′c − 18.33p
=
268.33p
=
15.480p
Resultant stress in cylinder = 50 N/mm2
−σc + σ′c = 50 N/mm2
−25 + σ′c = 50
σ′c = 75 = 15.480p
or internal pressure,
σ′t
=
1.77σ′c − 6.11 × p
=
1.777 × 15.480p − 6.11 × p
=
(27.5 − 6.11)p = 21.39p
=
21.39 × 4.845 = 103.63 N/mm2
(putting the value of p)
(5.10)
σtr = σt + σ′t
Resultant stress in tape,
=
100 + 103.63
=
203.63 N/mm2
Exercise 5.4 A thin cylindrical shell of an internal diameter of 400 mm and a wall
thickness of 10 mm is closely wound around its circumference by a 3-mmdiameter steel wire under an initial tension of 10 N/mm2. The cylinder is further
subjected to an internal pressure of 2.4 N/mm2. Determine the resultant hoop stress
developed in the cylinder and wire. Cylinder is also made of steel. E = 208
Gpa, v = 0.3.
Pressure Vessel with a Double Curved Wall
Sometimes, the cylinder heads of a pressure vessel have double curved wall, that
is, a particular element of wall has two radii of curvatures in two planes.
Let us consider a pressure vessel, the form of whose wall is a surface of
revolution as shown in Fig. 5.9(a). Thickness of the wall is very small in
comparison to principal radii of curvatures r1, r2 as shown in Fig. 5.9(b), so
the wall can be treated as a membrane. The stresses are assumed to act on the
middle surface of the membrane. An elementary part shown in Fig. 5.9(a) is
enlarged in Fig. 5.9(b), element bounded by two circumferential arcs and two
meridional arcs. Say
σ1
=
stress in circumferential direction
σ2
=
stress along the meridional direction
r1
=
radius of curvature of circumferential arc
r2
=
radius of curvature of meridional arc
δθ1
=
angle subtended by circumferential arc
δθ2
=
angle subtended by meridional arc
Figure. 5.9 (a) Element of double curved wall (b) Stresses on a small element
Forces acting along edges of element are
σ1ds2t and σ2ds1t (refer to Fig. 5.9(a,b))
Let us resolve these forces along the normal to the element abcd as shown in Fig.
5.10.
Figure. 5.10
But
, putting these values
Simplifying the expression
For a thin spherical shell,
For a thin cylindrical shell,
Circumferential stress
Conical Water Tank
Figure 5.11 shows an open conical tank uniformly suspended in brackets around its
curved conical surface. Say, H is the depth of water in tank, from bottom. Take a
section X–X at a distance of y from the bottom apex point.
Radius of curvature,
Pressure,
Using the equation
p = w(H − y)
, for double curved surface
or
where w = weight density of water
or
Figure. 5.11
For the hoop stress to be maximum
Meridional stress σ2 can be calculated by considering the total weight of the
water (shown by dotted lines) which is equal to (w (H –
y)π y2 tan2α + wπy tan2α(1/3)) The component of force due to σ2 in the vertical
direction is equal to (σ2 (2πy tan α) t cosα).
So for equilibrium
For meridional stress, σ2 to be maximum
Example 5.5 A conical water storage tank has an included angle of 60° and water
is filled up to a vertical depth of 3 m. If the wall thickness is 5 mm, determine the
circumferential and meridional stresses at a depth of 2 m from top.
Solution
Water density, w = 10 kN/m3.
Then, α = 30°, semicone angle
t = 5 mm, wall thickness = 0.005 m
y=3−2=1m
H=3m
w = 10 kN/m3
Circumferential stress,
tan α = 0.577
cos α = 0.866
= 2.665 N/mm2
Meridional stress,
Putting the values
= 1,554,657.4 N/m2
= 1.55 N/mm2
Exercise 5.5 A conical water storage tank has an included angle of 60° and a
vertical depth of 4 m. If the wall thickness is 6 mm, determine the location and
magnitude of maximum hoop and maximum meridional stresses.
Water density, w = 10 kN/m3.
Problem 5.1 A stream boiler of an internal diameter of 1.6 m is subjected to an
internal pressure of 1.5 N/mm2. What is the tension per linear metre of the
longitudinal joint in the boiler shell. Calculate the thickness of the plate if the
maximum tensile stress in the plate section is not to exceed 100 N/mm2, taking an
efficiency of the longitudinal riveted joint equal to 75 per cent.
Solution
Diameter
D = 1,600 mm
Internal pressure,
p = 1.5 N/mm2
Circumferential stress,
Length = 1,000 mm (to calculate tension per linear metre)
Thickness = t mm
Area = 1,000t mm2
Tension per linear metre = σc × 1,000t
Plate thickness
Efficiency of longitudinal joint, ηl = 0.75
σc = 100 N/mm2 as given in problem.
Problem 5.2 A closed pressure vessel of a length of 400 mm, a thickness of 5 mm
and an internal diameter of 120 mm is subjected to an internal pressure of 8
N/mm2. Determine the normal and shear stresses on an element of the cylindrical
wall on a plane at 30° to the longitudinal axis of cylinder (Fig 5.12).
Solution
O–O—longitudinal axis of cylinder = AC
B–A—inclined plane
A–C—reference plane (Fig. 5.12)
Figure. 5.12
Problem 5.3 The ends of a thin cylindrical shell are closed by flat plates. It is
subjected to an internal fluid pressure, but the ends of the cylinder are rigidly
stayed and no axial movement is permitted. Determine the increase in the volume
of the shell. Take Poisson’s ratio as 0.30.
Solution
Due to internal pressure, the cylinder tries to expand longitudinally. However, in
this case, the ends are rigidly stayed, which means that the ends exert axial
compressive force on cylinder, so that the axial strain in cylinder becomes zero.
Due to internal pressure
Hoop stress
Axial stress
Due to resistance from end plate
Say σa′ is the axial stress on cylinder
Resultant axial stress,
σar = σa − σa′
Axial strain,
or
or
or
Volumetric strain,
Where
But
v = 0.3
εv = 2εc + εa but εa = 0
Volume
δV = change in volume
Problem 5.4 A thin-walled aluminium alloy pressure vessel of a diameter of 200
mm and a wall thickness of 3 mm is subjected to an internal pressure of 4 MPa.
Strain gauges that are bonded on the outer surface in the hoop and axial directions
give readings of 1,620 and 380 μ strain respectively at full pressure respectively.
Determine E and v for the material.
Solution
p = 4 N/mm2
D = 200 mm
t = 3 mm
Circumferential strain,
(5.13)
Axial strain,
(5.14)
Dividing Eq. (5.13) by Eq. (5.14)
From Eq. (5.13)
Problem 5.5 A 15-cm-long bronze tube with closed ends has a diameter of 75 mm
and a wall thickness of 3 mm. With no internal pressure, the tube just fits between
the two rigid end walls. Calculate the longitudinal and tangential stresses in the
cylinder for an internal pressure of 2.5 N/mm2 (Fig 5.13).
E = 100 kN/mm2, v = 0.32
Solution
Internal pressure
The rigid wall exerts axial pressure σa′ to keep axial strain zero as shown in Fig.
5.13. Net axial stress σa– σa′ = 15.625 – σa′
Axial strain
or
σ′a
=
15.625 − 10 = 5.625 N/mm2
Tangential stress in cylinder,
σc
=
31.25 N/mm2
Axial stress,
σar
=
σa − σ′a = 15.625 − 5 625
=
10 N/mm2
Figure. 5.13
Problem 5.6 A gun metal tube of 50-mm bore and a wall thickness of 1.5 mm is
closely wound externally by a steel wire 1.0 mm diameter. Determine the tension
under which the wire must be wound on the tube if an internal pressure of 1.8
N/mm2 is required before the tube is subjected to tensile stress in the
circumferential direction. For gun metal E = 105 GPa and v = 0.34 and for E = 210
GPa.
Solution
Let us first determine the stresses σa′, σc′ and σw′ developed in cylinder and the wire
due to the internal pressure.
Take the unit length of cylinder, ℓ = 1 mm
Number of wires per millimetre length of tube,
Axial stress,
Compatibility equation
1.68 (57.325 − 1.91σ′c)
=
2σ′c
96.308
=
(2 + 3.2088)σ′c = 5.2088σ′c
σ′c
=
18.5 N/mm2
σ′w
=
57.325 − 1.91σ′c
=
57.325 − 1.91(18.5)
=
57.325 − 35.335
=
21.99 N/mm2
Now, σc – σc′ = 0 (as given in the problem)
σc = σc′ = 18.5N/mm2 (the initial compressive stress in the cylinder in hoop
direction)
But
Problem 5.7 A 200-mm-long copper tube closed at its ends has an outer diameter
of 90 mm and a thickness of 4 mm. It fits without clearance in a 90-mm hole in a
rigid block. The tube is then subjected to internal pressure of 5 N/mm2. Taking v =
0.34 and E = 105 kN/mm2. Determine the final hoop stress in the tube (Fig 5.14).
Solution
The block is rigid so diametral strain due to internal pressure p will be zero.
Do = outer diameter = 90 mm.
Di = inner diameter = 90 − 2 × 4 = 82 mm.
The rigid block will offer radial pressure p′ to prevent diametral strain.
Internal pressure
Stresses
Figure. 5.14
Due to external pressure, p′
Diametral strain
Final hoop stress,
σcr = 51.25 − 3.78 × 11.25 = 8.71 N/mm2
Problem 5.8 A thin spherical shell of steel has an internal diameter of 3 m and a
wall thickness of 6 mm and is just filled with water at 20°C and at atmospheric
pressure. Find the rise in water pressure if the temperature of the water and the
shell rises to 60°C, then determine the volume of water that would escape if small
leak is developed at the top of the vessel. For steel E = 200 GPa, αs = 11 ×
10−6/°C,v = 0.3 and for water K = 2.2 × 103 N/mm2, α = 0.207 × 10−3/°C
(coefficient of volumetric expansion).
Solution
For thin spherical shell
Say, p = pressure developed
Because α for water >>α for steel
Hoop stress in shell,
Volumetric strain in sphere, εvs = volumetric strain in water, εvw
Equating
εvs = εvw
1312.5 p × 10−6 + 1320 × 10−6 = −0.4545 p × 10−3 + 8.20 × 10×3
1.3125 p × 10−3 + 1.32 × 10−3 = −0.4545 p × 10−3 + 8.28 × 10×3
Pressure,
The volume of water which escapes through the leak is simply the difference of
free expansion of the water and the vessel, as after leakage there is no pressure
remaining in the shell.
Problem 5.9 A brass hoop of an inside diameter of 40 cm and a thickness of 1 cm
fits snugly at 180°C over a steel hoop which is 1.5 cm thick. Both the hoops are 5
cm wide. If the temperature drops to 20°C, determine the circumferential stress in
each hoop and the radial pressure at common radius. For steelE = 200 GPa, α = 12
× 10−6/°C and for brass E = 100 GPa, α = 20 × 10−6/°C (Fig 5.15).
Figure. 5.15
Solution
Brass hoop initial temperature = 180°C
Final temperature = 20°C
Reduction in temperature = 160°C
As the brass hoop cools down, brass will try to contract, exerting pressure on steel
hoop, say the radial pressure developed at common radius of 200 mm is p′ as
shown.
Due to p′, compressive hoop stress in steel hoop will be developed.
At the same time steel hoop exerts outer pressure p′ on brass hoop and tensile
hoop stress will be developed in brass hoop.
In brass
In steel
Total strain
Problem 5.10 A bronze sleeve of an internal diameter of 200 mm and a thickness
of 5 mm is pressed over a steel liner of an external diameter of 200 mm and a
thickness of 10 mm with a force fit allowance of 0.06 mm on diameter.
Considering both the bronze sleeve and the steel liner as thin cylinder, determine
(a) radial pressure at the common radius, (b) hoop stresses in sleeve and liner, and
(c) percentage of fit allowance met by the sleeve.
Given for bronze E = 120 kN/mm2, vb = 0.33 and for steel E = 208 kN/mm2, vs =
0.3.
Solution
Figure 5.16 shows the bronze sleeve under an initial pressure p and the sleeve liner
under an external pressure p. The internal pressure p is developed due to force fit
between the bronze sleeve and the steel liner.
Figure. 5.16
(a) Stresses
Bronze,
Steel,
Radial strain in bronze sleeve
where σcb is tensile and p is compressive
Radial strain in steel liner
Total radial clearance,
δR
=
0.06mm
Common radius,
R
=
100 mm
So,
Radial pressure,
= 2.78 N/mm2 at common surface
(b) Hoop stresses
σcb = 20 p = 2.78 × 20 = 55.6 N/mm2
σcs = −10 p = −2.78 × 10 = −27.8 N/mm2
(c) Radial clearance
On bronze sleeve
On steel liner
Percentage of fit allowance of sleeve
Problem 5.11 A steel tyre of a thickness of 10 mm, a width of 80 mm and an
inside diameter of 1,500 mm is heated and shrunk on to a steel wheel of a diameter
of 1,501 mm. If the coefficient of static friction between the tyre and the wheel is
0.3, what twisting moment is required to twist the tyre relative to the wheel?
Neglect the deformation of the wheel. E = 200 GPa.
Solution
Width of tyre,
b = 80 mm
Thickness of tyre,
t = 10 mm
Tyre diameter,
D = 1,500 mm
Wheel diameter,
D′ = 1,501 mm
Change in the inner diameter of tyre = 1,501 − 1,500 = 1 mm
Circumferential or diametral strain,
E = 2,00,000 N/mm2
Hoop stress in tyre,
where p is junction pressure between the tyre and the wheel
or,
Twisting moment
Total radial force on tyre,
R
=
b × πDp
=
80 × π × 1,500 × 1.77
=
6,67,274 N
Coefficient of friction,
μ
=
0.3
Force of friction,
F
=
μR = 0.3 × 667,274
=
2,00,182 N
=
F × radius of wheel
=
2,00,182 × 750
=
1,50,136,500 N mm
=
150.136 kN m
Twisting moment,
T
Key Points to Remember
o
o
For a shell, D/t ratio is greater than 20, it is classified as a thin shell.
For a cylindrical thin shell is subjected to internal pressure, p.
Hoop stress,
Axial stress,
Hoop strain,
Axial strain,
Volumetric strain,
o
For a thin spherical shell subjected to internal pressure, p.
Hoop stress,
Diametral strain,
Volumetric strain,
o
o
δV, addition of volume of liquid pumped inside the cylinder δV1 (increase
in the volume of cylinder) + δV2 (decrease in the volume of liquid)
Due to wire winding, initial compressive hoop stress is developed in
cylinder. Pressure bearing capacity of cylinder is increased.
o
In double curved wall of a shell under pressure
o
o
o
In a cylindrical shell, σ1 = σc, σ2 = σa, r2 = ∝ (infinity)
In a conical water tank, stresses
Where α is semi-cone angle of tank, t is wall thickness, w is specific
weight and H is depth of liquid.
Review Questions
1. What is a thin cylindrical/spherical shell? On what D/t ratio it is classified
as a thin shell?
2. Derive the expressions for hoop and axial stresses developed in a thin
cylindrical shell subjected to internal pressure p.
3. Take a small element of a thin spherical shell and show the stresses acting
on this element.
4. Derive the expression for the volumetric strain of a thin spherical shell
subjected to internal pressure p.
5. A cylinder of diameter D and wall thickness t is wound with a single layer
of wire under tensionσw. Derive the expression for hoop stress developed
in cylindrical shell.
6. Derive the following expression for a double curved surface under
pressure p.
7. Derive the expression for maximum hoop stress developed in a conical
water tank.
8. How the strain gauges mounted on an external surface of a thin shell
subjected to an internal pressure can be used to measure E and v of the
material of the shell.
Multiple Choice Questions
1. A thin cylindrical shell with D/t = 30 is subjected to an internal pressure of
3 N/mm2. What is the hoop stress developed in shell?
1. 90 MPa
2. 45 MPa
3. 22.5 MPa
4. None of these
2. A thin spherical shell of an inner diameter of 400 mm is subjected to an
internal pressure of 2.5 N/mm2. If the hoop stress is not to exceed 100
MPa, what is the thickness of shell?
0. 2.5 mm
1. 5 mm
2. 10 mm
3. None of these
3. A thin cylindrical shell is made of steel with v = 0.3. It is subjected to an
internal pressure p. What is the ratio of hoop strain to axial strain?
0. 1.0
1. 2.0
2. 3.0
3. 4.25
4. A thin spherical shell is subjected to an internal pressure p. What is the
ratio of volumetric strain to circumferential strain in shell?
0.
1.
2.
3.
1.0
2.0
3.0
None of these
5. A cylindrical tank of an inside diameter of 1 m and a height of 20 m is
filled with water of a specific weight of 10 kN/m3. If the thickness of tank
is 25 mm, then the maximum stress developed in wall of the tank is
0.
1.
2.
3.
4 N/mm2
2 N/mm2
1 N/mm2
None of these
6. A thin cylindrical shell is made of steel with v = 0.30. It is subjected to an
internal pressure. What is the ratio of volumetric strain to circumferential
strain?
0.
1.
2.
3.
2.0
2.235
9.5
None of these
7. A closed pressure vessel of a length of 400 mm, a wall thickness of 5 mm
and an internal diameter of 100 mm is subjected to an internal pressure of
8 N/mm2. The normal stress on an element of the cylinder on a plane 30°
to the longitudinal axis will be
0.
1.
2.
3.
140 MPa
70 MPa
77.32 MPa
None of these
8. A steam boiler of an internal diameter of 1.5 m is subjected to an internal
pressure of 2 N/mm2. If the efficiency of the longitudinal joint is 80 per
cent and the maximum tensile stress in the plate section is not to exceed
125 MPa, what is the thickness of the plate?
0.
1.
2.
3.
6 mm
3 mm
15 mm
None of these
Practice Problems
1. A steam boiler of an internal diameter of 1.5 m is subjected to an internal pressure
of 1.2 N/mm2. What is the tension per linear metre of the longitudinal joint in the
boiler shell? Calculate the thickness of the plate if the maximum tensile stress in
the plate section is not to exceed 90 N/mm2, taking efficiency of longitudinal joint
as 70 per cent.
2. A single strain gauge making an angle of 15° with the horizontal plane is used to
determine the gauge pressure in a cylindrical tank with its axis vertical as shown
in Fig. 5.17. The tank is 6 mm in thickness and 500 mm in diameter. It is made of
steel with E = 200 GPa and v = 0.29. The strain gauge reading is 350 m strain.
Determine the pressure inside the tank.
3. The ends of a thin cylindrical shell are closed by flat plates. It is subjected to an
internal fluid pressure of 3 N/mm2, but the ends of the cylinder are rigidly stayed
and no axial movement is permitted. The diameter of cylinder is 200 mm and its
length is 800 mm, while the wall thickness is 5 mm. Determine the change in the
volume of cylinder, if E = 200 GPa and v = 0.3.
Figure. 5.17
4. A thin-walled copper alloy pressure vessel of a diameter of 250 mm and a wall
thickness of 5 mm is subjected to an internal pressure of 5 MPa. The strain gauges
that are bonded on the surface in the hoop and axial directions give readings of 988
and 191 μs at full pressure, respectively. Determine Eand n for the material.
5. A 20-cm-long copper tube with closed ends is 100 mm in diameter and has a wall
thickness of 4 mm with no internal pressure. The tube just fits between the two
rigid end walls. Calculate the axial and hoop stresses in the cylinder for an internal
pressure of 3 N/mm2. E = 105 kN/mm2 and n = 0.33.
6. A 50-mm bore gun metal tube of a wall thickness of 1.25 mm is closely wound
externally by a steel wire of a diameter of 0.5 mm. Determine the tension under
which the wire must be wound on the tube if an internal pressure of 1.5 N/mm2 is
required before the tube is subjected to a tensile stress in the circumferential
direction. For gun metal E = 102 GPa and v = 0.35 and for steel E = 210 GPa
7. A 150-mm-long bronze tube closed at its ends is 80 mm in outer diameter and has
a thickness of 3 mm. It fits without clearance in an 80-mm hole in a rigid block.
The tube is then subjected to an internal pressure of 4 N/mm2. Take v = 1/3 and E =
83 kN/mm2, determine the tangential stress in the tube.
8. A thin spherical shell of copper has an internal diameter of 2.5 m and a wall
thickness of 5 mm and is just filled with water at 25°C and at atmospheric
pressure. If the temperature of water and shell rises to 60°C, determine the volume
of water that would escape if a small leak is developed at the top of the vessel. For
copper E = 105 GPa, αcu = 18 × 10−6/°C and v = 0.34 and for water K = 2,200
N/mm2, α = 0.207 × 10−3/°C (for volumetric expansion).
9. A steel tube of an internal diameter of 150 mm and a wall thickness of 8 mm in a
chemical plant is lined internally with a well-fitting copper sleeve of a wall
thickness of 2 mm. If the composite tube is initially unstressed, calculate the hoop
stresses set up assumed to be uniform throughout the wall thickness, in a unit
length of each part of the tube due to increase in temperature of 100°C. Neglect the
temperature effect in axial direction.
[Hint: compressive stress in copper, αc > tensile stress in steel, αs]
For steel,
E = 208 GPa, α = 11 × 10−6/′C
For copper,
E = 104 GPa, α = 18 × 10−6/′C
10. A copper sleeve of an internal diameter of 150 mm and a thickness of 2.5 mm is
pressed over a steel liner of an external diameter of 150 mm and a thickness of 2.5
mm with a force fit allowance of 0.04 mm on diameter. Considering both the
copper sleeve and the steel liner as thin cylinders, determine (a) radial pressure at
common radius, (b) hoop stresses in sleeve and liner (c) percentage of fit
allowance met by the sleeve.
For copper, E = 105 kN/mm2, vc = 0.34
For steel, E = 210 kN/mm2, vs = 0.3
11. A thin tyre is to be shrunk onto a rigid wheel of a diameter of 1 m. Determine the
necessary internal diameter of the tyre if the maximum hoop stress in the tyre is 10
N/mm2. Also determine the least temperature to which the tyre must be heated
above that of wheel before it could be slipped on.
α for tyre = 11 × 10−6/°C
E = 204 GPa
Special Problems
1. A closed cylindrical vessel made of steel plate of a thickness of 5 mm with plane
ends carries fluid under a pressure of 4 N/mm2. The diameter of the cylinder is 250
mm and its length is 750 mm. Calculate the longitudinal and hoop stresses
developed in the cylinder. What are the changes in the diameter, length and volume
of the cylinder?
E = 210 GPa, v = 0.3
2. One method of determining Poisson’s ratio for a material is to subject a cylinder to
internal pressure and to measure axial strain εa and hoop strain εc on the outer
surface of cylinder show that
3. A cylindrical tank of an inside diameter of 2 m and a height of 20 m is filled with
water of a specific weight of 10,000 N/mm3. The material of the tank is structural
steel with yield strength of 250 N/mm2. What is the minimum thickness required at
the bottom of the steel tank if the efficiency of the longitudinal seams is 80 per
cent. Take factor of safety (FOS) as 4.
4. A thin spherical shell made of copper alloy is 300 mm in diameter and 1.5 mm in
wall thickness. It is full of water at an atmospheric pressure. Find by how much the
internal pressure will increase if 20 cc of water is pumped inside the shell. E = 100
GPa, v = 0.29
For water, K = 2,200 N/mm2
5. A pressurized steel cylinder tank has an inner radius of 600 mm and a thickness of
16 mm. The tank is subjected to a pressure p = 1,750 kPa and an axial force P =
125 kN. The butt weld seam forms an angle of 54° with the longitudinal axis of the
tank. Determine (a) the normal stress perpendicular to weld and (b) the in-plane
shear stress parallel to weld.
, net stress in axial direction
σa = 54° with the plane of σc
6. A spherical steel vessel is made of two hemispherical portions fitted together at
flanges. The inner diameter of the sphere is 600 mm and the wall thickness is 6
mm. Assuming that the vessel is a homogeneous sphere, what is the maximum
working pressure for an allowable tensile stress in a shell of 150 MPa. If 20 bolts
of a diameter of 16 mm are used to hold flanges together, what is the tensile stress
in bolts when the sphere is under full pressure?
7. For a hydraulic test, a steel tube of an internal diameter of 80 mm, a wall thickness
of 2 mm and a length of 1.2 m is fitted with end plugs and filled with oil at a
pressure of 2 MPa. Determine the volume of oil leakage which would cause the
pressure to fall to 1.5 MPa.
For oil, K for oil = 2.8 GN/m2,
For steel, E = 208 GPa, v for steel = 0.29
Answers to Exercises
Exercise 5.1: 48.51 cc
Exercise 5.2: 3mm, ec = 0.3888 × 10−3, dD = 0.1166 mm
Exercise 5.3: t2 = 2 mm, 1.235
Exercise 5.4: σcr = +37.864 N/mm2
σwr = 43.02 N/mm2
Exercise 5.5:
Answers to Multiple Choice Questions
1. (b)
2. (a)
3. (d)
6. (b)
7. (b)
8. (c)
4. (c)
5. (a)
Answers to Practice Problems
1.
2.
3.
4.
5.
6.
7.
8.
9.
400 kN/linear metre, t = 14.3 mm
p = 2.07 N/mm2
14.17 cc
E = 105 kN/mm2, n = 0.34
37.5 N/mm2, 12.335 N/mm2
64.52 N/mm2
8.72 N/mm2
43,810 cm3
σcs = 16.18 N/mm2; (tensile)
= 57.325 – 1.91(18.5)
10.(a) 1.24 N/mm2; (b) 37.2 N/mm2 (tensile), 37.2 N/mm2 (compressive); (c)
67.14%
11.44.56°C, 999.51 mm
Answers to Special Problems
1.
2.
3.
4.
5.
6.
50 MPa, 100 MPa, 0.101 mm, 0.0714 mm, 33.3 × 103 mm3
4 mm
0.93 MPa
42.46 MPa, 14.51 MPa
p = 12 N/mm2; 211 Mpa
δv = 1.63 ccd
5b. Thick Shells
CHAPTER OBJECTIVES
For low pressures, thin pressure vessels are used where D/t ratio is greater than 20.
For high internal pressure as in the case of tubes used for high pressure gauge and
bulk compression gauge, tubes with considerable thickness are used where D/t <<
20. The objectives of this chapter are to make the students learn about:
Derivation of Lame’s equations for radial and circumferential
stresses.
o
Variation of circumferential and radial stresses along the thickness
of the shell.
o
There is large variation of hoop stress along the thickness of the
shell. The maximum hoop stress cannot be more than the allowable stress;
therefore, two cylinders are compounded.
o
How compressive hoop stress in inner cylinder and tensile hoop
stress in outer cylinder are developed in a compound cylinder.
o
Calculation of proper shrinkage/force fit allowance between two
compounded cylinders.
o
Calculation of hoop and radial stresses in a hub and shaft assembly.
o
Derivation of radial and hoop stresses in a thick spherical shell.
o
Introduction
In a thin pressure vessel, the hoop and axial stresses are much greater than the
intensity of fluid pressure and in many applications such as tube (for high-pressure
gauge) it is subjected to high internal pressure. The tube section can be designed so
that the ratio of hoop stress/internal pressure is not large.
In this chapter we will deal with Lame’s equations for hoop and axial stresses in
terms of internal pressure. To increase the pressure-bearing capacity of a thick
cylinder, it is compounded with another cylinder which introduces external
junction pressure on inner cylinder.
We will study about stresses developed due to shrinkage and shrinkage
allowance in compound cylinders.
Stresses developed in thick spherical shell will also be briefly discussed.
Lame’s Equations
Stresses in a thick cylinder are determined on the assumption that plane sections
(perpendicular to the axes) of the cylinder remain plane after the application of
internal pressure. In other words, axial strain in cylinder remains constant along its
lengths and there is no distortion of end plates.
Figure 6.1 shows a section of a thick cylinder of which thickness t = R2 − R1 is
considerable. It is subjected to internal pressure p such that axial and circumstantial
stresses develop in the cylinder. Radial stress σr varies from p (compressive) at
radius R1 to zero stress at radius R2, a nonlinear variation ofσr. Consider a thin circular
cylinder of inner radius r and outer radius r + dr, that is, radial thickness
drsubjected to radial stress σr at inner radius r and σr + dσr at outer radius r + dr.
Consider the equilibrium of half of the cylinder, that is, XAX as shown in Fig.
6.1, σc is the circumstantial stress (taking force equal to stress × projected area with
cylinder length equal to unity).
2(σr + dσr) (r + dr) + 2σc dr = 2rσr
2σcdr + 2rdσr + 2σrdr + 2dσrdr = 0
But, 2dσrdr is a negligible quantity,
So,
σcdr + σrdr + rdσr = 0
or,
Say E and v are elastic constants
Axial stress developed in cylinder,
remains constant.
Figure 6.1 Radial and hoop stresses on thin cylinder
Figure 6.2 Stresses on a small element
Stresses on a small element of thick cylinder are shown in Fig. 6.2, that
is, σr, σc and σa in radial, circumstantial and axial directions. Note that σr is
compressive stress, while σa and σc are tensile stresses.
Axial strain,
But stress σa is constant axial stress and E and v are elastic constants; therefore,
σc − σr = constant
= 2A (say)
From Eqs (6.3) and (6.4)
σc − σr = 2A
(6.4)
From Eqs (6.1)
Subtracting Eq. (6.4) from Eq. (6.5)
or
or
or ln (A + σr) = −2 ln r + ln B, where ln B is another constant
or
or
Radial stress,
From Eq. (6.4)
Circumferential stress,
Boundary conditions [Refer to Fig. 6.1(a)]
Radial stress at
Radial stress at
or
putting the value of A in Eq. (6.9)
or constant
constant
putting the values of A and B in equation for σc we get,
at
Figure 6.3 Variation of radial and circumferential stresses
at
Figure 6.3 shows the variation of radial and circumstantial stresses in a thick
cylinder subjected to internal pressure p.
Example 6.1 A thick cylinder of inner radius 150 mm and outer radius 210 mm is
subjected to internal premiere p such that the maximum hoop stress developed in
cylinder is 154.16 N/mm2. Draw the hoop stress and radial stress distribution along
the thickness of cylinder. If E = 200 GPa, what is circumstantial strain in cylinder
at the outer surface? Given v = 0.3.
Solution
Internal radius,
R1 = 150 mm
Outer radius,
R2 = 210 mm
Internal pressure = p
Lame’s constants
Radial stress σr
At
r = R1, σr = p = +50 N/mm2 (compressive)
r = 180 mm,
= 70.891 − 52.083
= 18.80 N/mm2
r = 210 mm, σr = 0
Hoop stress
at r = R1,
σcmax = 154.16 N/mm2 (given)
r = 180 mm,
= 70.891 + 52.083
= 122.974 N/mm2
At r = 210 mm,
= 52.083 + 52.083
= 104.166 N/mm2
Figure 6.4 shows the distribution of σc and σr stresses along the thickness (R2 − R1)
of cylinder.
Axial stress
Figure 6.4
At the outer surface of the cylinder, there are only two principal stresses, that
is, σcmin and σa, and the radial stress is zero.
Circumstantial strain at outer surface of cylinder
Exercise 6.1 A cylindrical shell of inner radius 60 mm and outer radius 100 mm is
subjected to internal fluid pressure of 64 N/mm2. Draw the distribution
of σc and σr along the thickness of cylinder. If E = 105 kN/mm2 and v = 0.34, what
are the axial and circumstantial strains at the outer surface of cylinder?
Thick Cylindrical Shell Subjected to External Pressure
In the previous chapter, we have studied about the thick cylinder subjected to an
internal pressure which causes tensile hoop and axial stresses in the cylinder. Yet,
in many applications, a cylinder may be subjected to an external pressure as in the
case of the submarine tank subjected to external pressure and the inner cylinder of
a compounded cylinder subjected to external pressure which causes compressive
hoop and axial stresses in the cylinder.
Figure 6.5 shows a thick cylinder of inner radius R1, outer radius R2 and
length L subjected to an external pressure p.
σr = p at r = R2
σr = 0 = at r R1
Using Lame’s equations,
or,
A = B/R12, putting the value in Eq.(6.12)
Figure 6.5 Thick cylinder subjected to external pressure
or constant,
Radial stress,
at
r = R1, σr = 0
at
Hoop stress
at
Note that
2R22 > (R22 + R12)
Therefore, the maximum hoop stress occurs at inner radius.
Axial stress,
as the pressure acts on the outer surface.
Stresses on an element of a cylinder subjected to pressure p are shown in Fig.
6.6. Note that all the stresses σc, σr and σa are compressive. The maximum hoop
stress (though compressive in this case) occurs at inner radius.
Figure 6.6
Example 6.2 A thick cylinder of inner radius 85 mm and outer radius 120 mm is
subjected to an external pressure of 50 N/mm2. Determine σcmax and σcmin and show
variation of σc and σr along the thickness of the cylinder.
Solution
Figure 6.7
σr = p, at outer radius
σr = 0, at inner radius
Figure 6.7 shows the variation of σc and σr in the cylinder due to external pressure.
Exercise 6.2 A thick cylinder with an external diameter of 240 mm and an internal
diameter of Dis subjected to an external pressure of 50 MPa. Determine the
diameter D of the maximum hoop stress in the cylinder that is not to exceed 200
MPa.
Compound Cylinder
To increase the pressure-bearing capacity of a cylinder and to decrease the
variation in hoop stress developed in cylinder, another cylinder is force fitted or
shrink fitted over the cylinder. This process is called compounding of cylinders.
Therefore, for compounding, the inner diameter of the outer cylinder is slightly
less than the outer diameter of the inner cylinder so as to provide the force fit
allowance or shrinkage allowance.
In shrinkage fitting, the outer cylinder is heated such that its inner radius R3″
expands to R3′ and then the outer cylinder is slipped over the inner cylinder. When
the outer cylinder cools, it exerts a radial pressure p′ on the outer surface of the
inner cylinder. In reaction, the inner cylinder exerts a pressure p′ on the inner
surface of the outer cylinder as shown in Fig. 6.8(c). The outer pressure p′ on the
inner cylinder develops compressive hoop stresses in the inner cylinder. Similarly
the inner pressure p′ on the outer cylinder develops tensile hoop stresses in the
outer cylinder.
The final junction radius between two cylinders is R3 such that R3 < R3′
but R3 > R3″
The difference R3″ ‒ R3′ is termed as shrinkage allowance.
Stresses due to junction pressure
Inner cylinder
(expression derived earlier in case of cylinder with external pressure)
Outer cylinder
Figure 6.8
Figure 6.9 shows the variation of hoop stresses developed in the inner and outer
cylinders due to junction pressure.
Figure 6.9
Compound cylinder
Say the compound cylinder is now subjected to an internal pressure p.
Say A and B are Lame’s constants, thus
or
constant,
Stresses at R1, R2 and R3 due to internal pressure
Resultant stresses
Inner cylinder
Outer cylinder
Figure 6.10 shows the variation of resultant hoop stresses developed due to
junction pressure p′ and internal pressure p on the inner and outer cylinders.
Figure 6.10 Resultant hoop stress in the inner and outer cylinders of a compound
cylinder
Example 6.3 A compound cylinder is made by shrinking one cylinder over another
such that the outer diameter is 200 mm, the inner diameter is 100 mm and the
junction diameter is 150 mm. If the junction pressure developed between two
cylinders is 10 N/mm2 and the internal pressure is 50 N/mm2, what are the hoop
stresses at inner and outer radii of both the cylinders?
Solution
Radii R1
=
50 mm
R2
=
100 mm
R3
=
75 mm
=
10 N/mm2
Pressure p′
p
Resultant hoop stresses
Inner cylinder
Putting the values
Outer cylinder
=
50 N/mm2
Exercise 6.3 A steel cylinder of an outer diameter of 180 mm is shrunk on another
cylinder of an inner diameter of 120 mm. The common diameter is 150 mm. If
after shrinkage the radial pressure at the common surface is 6 N/mm2, determine
the magnitude of internal pressure to which the compound cylinder can be
subjected so that the maximum hoop tension in the inner and outer cylinders is
equal.
Then,
Shrinkage Allowance
In a compound cylinder at the junction radius due to junction pressure, tensile hoop
stress in the outer cylinder and compressive hoop stress in the inner cylinder are
developed. At common radius R3
Inner cylinder
Junction pressure, p′ (compressive)
Radial or circumferential strain,
When vi and Ei are the Poisson’s ratio and the elastic modulus of the inner
cylinder, respectively
Outer cylinder
Junctions pressure, p′ (compressive)
where v0 and E0 are the Poisson’s ratio and the elastic modulus of the outer
cylinder.
Shrinkage allowance (R′3 − R′′3 from Eqs (16) and (18)
If both the cylinders are of different materials, then the shrinkage allowance is
calculated from Eq. (6.19).
If Ei = E0 = E,vi = v0 = v, that is, both the cylinders are of same material.
The shrinkage allowance on common radius R3 is
where R1, R3 and R2 are inner radius, junction radius and outer radius of
compound cylinder.
Remember that
(numerical sum of hoop stresses developed at the
common radius of two cylinders).
Example 6.4 A compound cylinder is formed by shrinking one outer steel cylinder
over bronze cylinder. The final dimensions are internal diameter 100 mm, external
diameter 200 mm and junction diameter 160 mm, and the shrinkage pressure at the
common surface is 12 N/mm2. Calculate the necessary difference in radii of two
cylinders at the common surface.
For steel,
E0 = 200 GPa, ni = 0.3
For bronze,
Ei = 100 GPa, n0 = 0.32
What is the maximum temperature through which the outer cylinder should be
heated before it can be slipped on?
For steel α = 11 × 10−6/°C
Solution
Junction pressure,
p′ = 12 N/mm2
R1 = 50 mm, R2 = 100 mm, R3 = 80 mm
Putting the values
The outer cylinder should be heated by 47.84°C so that it can be slipped over inner
cylinder.
Exercise 6.4 A steel tube of an outside diameter of 220 mm is shrunk on another
steel tube of an inside diameter of 140 mm. The diameter at junction is 180 mm
after shrinking on. The shrinkage allowance provided on the radius of the inner
tube is 0.08 mm. Determine (a) the junction pressure, (b) the hoop stresses at the
inner and outer radii of the inner tube and (c) the hoop stress at the inner and outer
radii of the outer tube if E = 210 kN/mm2.
Hub and Shaft Assembly
In a power transmission system, the shaft is connected to the hub with the help of a
key inserted in keyways cut on the shaft and in the hub. These keys severely
reduce the strength of shaft and hub due to the keyways having sharp corners. To
avoid keyways, the hub can be shrunk fitted on the shaft as in the case of
compound cylinders. Due to shrink fitting the junction pressure p′ is developed
between the shaft and the hub. Let as consider a shaft of radius R1 fitted in a hub of
outer radius R2 and inner radiusR1 as shown in Fig. 6.11.
Figure 6.11
Shaft
Let as take Lame’s constants A and B for shaft.
Radial stress,
Circumferential stress,
But material exists at centre, that is, at r = 0, and the stress cannot be infinite;
therefore, constant B = 0.
In shaft σr = − A = p′ (junction pressure) or A = −p′
σc = + A = − p′
While deriving Lame’s equations, we have considered σr as negative and σc as
positive. So now σc and σrboth are compressive and equal to junction pressure p′.
Radial strain in shaft
at radius R1
negative strian
Hub
Due to junction pressure p′, the hoop stress developed
Radial strain in hub,
positive strain
Because p′ is compressive
Shrinkage allowance at radius R1
δR1 = ɛcsR1 + ɛchR1
If both shaft and hub are of same material such that Es = Eh = E; Vs = Vh = V then
Shrinkage allowance on radius
Example 6.5 A steel shaft of a diameter of 100 mm is driven into a bronze hub.
The driving allowance provided is on diameter 1/1,000 of the shaft diameter.
Determine the thickness of the hub, if the maximum bursting stress in the hub is
limited to 80 N/mm2.
Given
Es = 200 GPa, Vs = 0.3
EB = 100GPa, VB = 0.34
Solution
Say junction pressure is p′
Hub
Shaft
But total strain = εch + εcs
As given in the problem
putting the values of ES and EB
Multiplying throughout by 1,00,000, we get
80 + 0.34 p‴ + 0.5 p ′ − 0.15 p′ = 100
0.69 p′ = 20
Junction pressure,
Outer radius of hub
Thickness of hub = R2 − R1 = 73.08 − 50 = 23.08 mm
Exercise 6.5 A bronze liner of an outside diameter of 60 mm and an inside
diameter of 39.94 mm is forced over a steel shaft of 40 mm in diameter. Determine
(a) the radial pressure between the shaft and the liner and (b) the maximum hoop
stress in the liner.
For steel,
E = 208 GPa, v = 0.29
For steel,
E = 125 GPa, v = 0.33
Thick Spherical Shell
To determine radial and hoop stresses in a thick spherical shell subjected to an
internal pressure, let us first learn about the radial and circumferential strains in
such axi-symmetric cases. Consider a thin disc of inner radius R1 and outer
radius R2 subjected to an internal pressure p. A small element abcd is subtending
an angle δθ at the centre is considered as shown in Fig. 6.12.
Figure 6.12
Radius oa = r
Radius od = r + δr before the application of internal pressure.
After the application of internal pressure, say
r → changes to r + u
δr → changes to δr + δu
Circumferential strain in element,
Radial strain in element,
These strains are tensile if positive.
On the element of a spherical shell, say stresses are σr + δσr at radius r + δr
Circumferential stress on element = σc
Bursting force P on the elementary shell (taking the projected area of
hemispherical portion)
πr2σr − π (r + δr)2 (σr + δσr)
Resisting force = σc × 2 πrδr , where δr is the radial thickness.
For equilibrium
πr2σr − π (r + δr)2 (σr + δσr) = σc × 2πrδr
2σrδr − rδσr = 2σcδr (neglecting the small quantities of higher order terms)
Three principal stresses at any point in the elementary shell are:
1. radial pressure, σr (compressive),
2. hoop stress, σc (tensile), and
3. hoop stress, σc on another plane (at right angle to the first plane).
Radial strain at any point
or
Circumferential strain at the point
From initial Eqs. (22) and (23)
Substituting for εr and εc from Eqs. (25) and (26)
After simplification we get
From Eq. (6.24)
Differentiating Eq. (6.29) we get
Substituting for σc and dσc/dr in Eq. (6.24) and after multiplication we get
Let us put dσr/dr = k, then we have
So
ln k = −4 ln r + ln C1
where C1 is a constant of integration
or
Integrating Eq. (6.30),
σr =
Now
σc =
or
σc =
, where C2 is another constant of integration.
(from Eq. 6.29)
(using Eq. 6.30)
σc =
Let us put constants C1 = 6B and C2 = ‒A
Then,
Radial stress,
Hoop stress,
(compressive)
(tensile)
For a thick spherical shell subjected to an internal pressure p, the boundary
conditions are
Constants,
Example 6.6 Calculate the thickness of a spherical shell of an inside diameter of
120 mm to withstand an internal pressure of 50 MPa, if the maximum permissible
tensile stress in the shell is 120 MPa.
Solution
Putting the values
or
120 (R23 − 1203)
95 R23
=
25R23 + 50 × 603
=
206.905 × 104
=
2 06905 × 106
R2
=
127.42mm
Thickness of shell = 127.42 − 60 = 67.42 mm
Exercise 6.6 Determine the maximum shear stress at the inner surface of a
spherical shell, having a diameter ratio of 1.5 for an internal pressure of 7 N/mm2.
Problem 6.1 Two thick cylinders are of the same dimensions. The external
diameter is 1.5 times the internal diameter. Cylinder A is subjected to an internal
pressure p1, while cylinder B is subjected to an external pressure p2. Find the ratio
of p1/p2 if the greatest hoop strain in both cylinders is the same.
Poisson’s ratio = 0.3
Solution
Cylinder A
Inner radius = R1
Outer radius, R2 = 1.5 R1
Internal pressure = P1
At inner radius,
Axial stress,
σr = p1 (compressive)
Hoop strain,
Cylinder B
External pressure = p2
At inner radius,
Axial stress,
Hoop strain,
However,
ɛc′ = ɛc″
Problem 6.2 A steel cylinder of an inside diameter of 800 mm and a length of 6 m
is subjected to an internal pressure of 40 MPa. Determine the thickness of the
cylinder if the maximum shear stress of cylinder is not to exceed 65 MPa. What is
the increase in the volume of the cylinder? E = 200 GPa, v = 0.3.
Solution
Inner radius,
p
τmax
R1
=
400 mm
=
40 MPa
=
65MPa
where
So
or
Thickness of cylinder,
At the inner radius
t = R2 − R1 = 645 − 400 = 245 mm
At the inner surface
σc
=
90MPa
σa
=
25MPa
p
=
−40MPa
E
=
200,000 N/mm2, v = 0 3
Strains
Volumetric strain,
Volume of cylinder,
Change in volume,
Problem 6.3 Strain gauge are mounted on the outer surface of a thick cylinder with
a diameter ratio of 2.5. The cylinder is subjected to an internal pressure of 150
MPa. The recorded strains are
1. longitudinal strain = 60 × 10‒6 and
2. circumferential strains = 241 × 10‒6
Determine E and v of the material.
Solution
Internal radius = R1
Outer radius = R2
Outer surface
Strains
or
57.14 − v × 28057 = 57 = 241 × 10−6 × E
28.57 − v × 57.14 = 60 × 10−6E
(6.31)
(6.32)
Dividing Eq. (6.31) by Eq. (6.32)
Poisson’s ratio,
Putting the value of v is Eq. (6.31)
Young’s modulus,
= 203,000 N/mm2
= 203 kN/mm2
Problem 6.4 A thick cylinder is subjected to both internal and external pressures.
The internal diameter of the cylinder is 120 mm and the external diameter is 200
mm. If the maximum permissible stress in the cylinder is 25 N/mm2 and the
external radial pressure is 5N/mm2, determine the intensity of internal fluid
pressure (Fig. 6.13).
Figure 6.13
Solution
R1 = 60mm
R2 = 100mm
Radial stress
σr = 5 N/mm2
at
r = R2
σr = p (say) at r = R1 as shown in Fig. 6.13
Using Lame’s equation,
From Eqs. (6.33) and (6.34)
from Eq. (6.33)
= (p − 5) × 0.5625 −5
(6.36)
Circumferential stress
Putting the value of σcmax, 25 = (p − 5) (2.125) − 5
Internal pressure
= 19.117 N/mm2
Problem 6.5 A compound cylinder is formed by shrinking one cylinder over
another. The outer diameter of the compound cylinder is 240 mm, the inner
diameter is 100 mm and the diameter at the common surface is 170 mm. The
junction pressure due to shrinkage is 5 N/mm2. If the compound cylinder is now
subjected to an internal fluid pressure of 60 N/mm2, determine the maximum hoop
tension in the cylinder. How much heavier a single cylinder of an internal diameter
of 100 mm would be if it is subjected to the same internal pressure in order to
withstand the same maximum hoop tension?
Solution
R1 = 50 mm, R3 = 85mm, R2 = 120 mm
p = 60 N/mm2, p′ = 5 N/mm2,
Inner cylinder
Resultant at R1,
Outer cylinder
Resultant at R3,
At inner radius of inner cylinder, σc is maximum.
Single cylinder
p
= 60 N/mm2
R1
= 50 mm
R′2
=?
σcmax
= 69.91
69.91
1.165
1.165 R′22 − 1.165 × 502
= R22 + 502
0.165 R22
= 5412.5
R′2
= 181.11 mm
Single cylinder, A′ π (R′22 − 502)
= π (181.112 − 502)
= π × 30303.5
Compound cylinder, A = π (1202 − 502) = π (11, 900)
Problem 6.6 A compound cylinder has a bore of 120 mm, an outer diameter of 240
mm and the diameter at common surface is 180 mm. Determine the radial pressure
at the common surface which must be provided by shrinkage fitting if the resultant
maximum hoop stress in the inner cylinder under the superimposed internal
pressure of 60 N/mm2 is to be 60 per cent of the value if the maximum hoop
tension at the inner radius of overall cylinder subjected to an internal pressure of
50 N/mm2 (Fig. 6.14).
Figure 6.14
Solution
Inner radius,
R1 = 60 mm
Junction radius,
R3 = 90 mm
Outer radius,
R2 = 120 mm
Internal pressure,
p = 60 N/mm2
Considering overall cylinder
Compound cylinder
Inner cylinder subjected to junction pressure p′ and internal pressure p,
with σcmax equal to 100 × 0.6 = 60 N/mm2 as given in the problem. This stress
occurs at inner radius R1 (Fig. 6.14).
σr = 50at r R1 = 60 mm
σr = p ′ at r = R3 = 90 mm
Moreover
at r = R1, σc = 60 N/mm2 as given
From Eqs. (6.37) and (39)
2A = 60 − 50 = 10
A = 5 N/mm2
From Eq. (6.37)
Problem 6.7 A thick cylinder of an internal diameter of 100 mm is subjected to an
internal pressure of 20 N/mm2. If the allowable stress for cylinder is 110 N/mm2,
determine the wall thickness of the cylinder. The cylinder is now strengthened by
wire winding so that it can be safely subjected to an internal pressure of 25 N/mm2.
Find the radial pressure exerted by wire winding.
Solution
Radius R1 = 50mm
R2 = to be calculated
P = 20 N/mm2
Internal pressure,
Allowable maximum stress,
or
5.5 (R22 − 502)
= R22 + 502
4.5 R22
= 6.5 × 502
R2
= 1.2018 × 50 = 60.1 mm
Internal pressure
σ″c, reduction in σ′cmax, due to wire winding.
= 137.5 − 110
= 27.5 N/ mm2
Pressure exerted by wire winding, p′ = 4.233 N/mm2.
Problem 6.8 A compound cylinder consists of a steel cylinder of an internal
diameter of 120 mm and an external diameter of 180 mm and a bronze liner of a
thickness of 10 mm with an external diameter of 120 mm as shown in Fig. 6.15.
Assuming liner to be a thin cylinder and that there is no stress in the compound
cylinder due to shrink fitting, determine the maximum hoop stress in the liner due
to an internal pressure of 40 N/mm2. Ignore the longitudinal stress and strain.
Solution
For compound cylinder
R1 = 50 mm
R2 = 90 mm
p = 40N/mm2
Figure 6.15
Liner
Problem 6.9 A steel cylinder has an internal diameter of 100 mm and an external
diameter of 200 mm. Another cylinder of same steel is to be shrunk over the first
cylinder so that shrinkage stresses just produce a condition of yield at the inner
surface of each cylinder. Determine the necessary difference in diameters of the
cylinder at the mating surface before shrinking and the required external diameter
of the outer cylinder, assuming that yielding occurs according to the maximum
shear stress criteria’s and that no axial stresses are set up due to shrinking. Yield
stress in simple tension or compression = 270 N/mm2 and E = 200 GPa.
Solution
In a simple tensile test on a bar, at yield point stress σyp, the maximum shear stress
occurs at ±45° to the axis of the bar and it is equal to 0.5 σyp. This can be obtained
by drawing a Mohr’s stress circle.
Maximum shear stress in simple tension or compression
Say, junction pressure is p′
R1 = inner radius of liner cylinder = 50mm
R3 = junction radius = 100mm
R2 = outer radius of outer cylinder = ?
Inner cylinder
σcmax occurs at inner radius due to p′ at outer radius
σr = 0, at inner radius of inner cylinder.
So,
1.333 p′ = 135
p ′ 101.25 N/mm2
Outer cylinder
Junction pressure or σr = p′ (compression)
or
R22 + 1002
= 1.667 R22 − 1.667 × 1002
0.667 R22
= 2.667 × 1002
R22
= 2.667 × 1002
Shrinkage allowance
Shrinkage allowance on diameter, δD3 = 2δR3 = 0.337 mm
Problem 6.10 A bronze sleeve is pressed onto a steel shaft of 60 mm in diameter.
The radial pressure between steel shaft and bronze sleeve is 14 N/mm2 and the
hoop stress at the inner surface of the sleeve is 50 N/mm2. If an axial compressive
force of 60 kN is now applied to the shaft, determine the change in radial pressure.
For steel,
E = 200 Gpa, v = 0.3
For bronze,
E = 100 Gpa, v = 0.33
Solution
Bronze sleeve
Junction pressure, p′
= 14 N/mm2
Hoop stress at inner surface
= +50 N/mm2
Diameter of shaft
= 60mm
Radius, R1
= 30mm
When compressive force is applied onto the shaft, axial compressive stress will be
developed in the shaft, which will cause positive radial strains (lateral strain in
shaft) putting internal pressure p″ on sleeve. In reaction, sleeve will exert p″ radial
pressure on the shaft.
Additional junction pressure p″
Additional hoop stress in sleeve,
Additional circumferential strain in sleeve
Shaft
Axial compressive stress,
Additional stresses in shaft are
p″, radial compressive
p″, circumferential compressive
σα axial compressive
Circumferential strain in shaft
Strain compatibility,
ε″cb = ε″cs
Problem 6.11 A steel rod of a diameter of 50 mm is forced into a bronze sleeve of
an outside diameter of 80 mm, thereby producing a tension of 40 N/mm2 at the
outer surface of sleeve. Determine (a) the radial pressure between the bronze
sleeve and the steel rod and (b) the rise in temperature which would eliminate the
force fit.
For steel,
E = 210 Gpa, v = 0.25, a = 11.2 × 10−6/°C
For bronze,
E114 Gpa, v = 0.33, α = 18 × 10−6/°C
Solution
Bronze sleeve
R1 = 25 mm
R2 = 40 mm
Tension on outer surface,
, where p′ is junction pressure
Radial pressure between bronze sleeve and rod
At junction
Circumferential strain,
Shaft
σ″c = − p′ (compressive) both are compressive
σ″c = − p′
Circumferential stress,
Total numerical sum of circumferential strains
αb > αs(on heating bronze sleeve will expand more than steel shaft.)
(αb − as) ΔT = ɛc
(18 − 11.2)× 10−6 × ΔT = 82.631 × 10−6
Rise in temperature
Problem 6.12 A steel plug of a diameter of 50 mm is forced into a copper ring of
an external diameter of 80 mm and a width of 30 mm. From a strain gauge fixed on
the outer surface of the ring in the circumferential direction, the strain is found to
be 64 × 10‒6. Considering that the coefficient of friction between the mating
surfaces is 0.25, determine the axial force required to push the plug out of the ring.
For bronze,
E = 100 × 103 N/mm2
For steel,
E = 200 × 103 N/mm2
Solution
Bronze ring
Bronze ring
R1 = 25 mm
R2 = 40 mm
At outer surface,
ɛc = 64 × 10−6
At outer surface,
σc = 64 × 10−6 × EB
= 64 × 10−6 × 100,000
= 6.4 N/mm2
Say p′ = junction pressure
Normal force on shaft, N = 2πR1 × B × p′,
where B is the width of ring
N = 2π × 25 × 30 × 4.99
= 23,515 N
Tangential force, F = μN = 0.25 × 23515 = 5878.7 N (along the length of shaft)
Axial force required to push the plug out of the ring = 5878.7 N
Key Points to Remember
o
o
o
In a thick shell, the wall thickness is significant and stresses vary along the
thickness of the cylinder and cannot be assumed to be uniform.
In a thick cylindrical shell subjected to internal pressure, Lame’s equations
can be used to determine radial and hoop stresses.
Constants A and B are determined by using boundary conditions.
In a thick cylindrical shell with inner radius (R1) and outer radius (R2)
subjected to internal pressurep
At inner radius,
o
Axial stress,
In a thick cylindrical shell with inner radius (R1) and outer radius (R2)
subjected to external pressurep
At inner radius,
o
Constant axial stress,
In a compound cylinder, the outer cylinder is shrink fitted over the inner
cylinder. The junction pressure is developed which causes tensile hoop
stress in the outer cylinder but compressive hoop stress in the inner
cylinder.
R1 = inner radius, R2 = outer radius and R3 = junction radius.
If both cylinders are made of same material, that is, same E
δR3 = shrinkage allowance
o
o
In a hub and shaft assembly, junction pressure introduces tensile hoop
stress in the hub but compressive hoop stress in the shaft.
In shaft, σc = −p′, σr = p′ (junction pressure)
In a thick spherical shell subjected to internal pressure,
With the help of boundary conditions, the values of Lame’s constants can
be determined.
Review Questions
1. In a thick cylinder subjected to internal pressure, explain the assumption
that plane sections remain plane after the application of internal pressure.
2. Show the variation of σc and σr in a thick cylinder along its thickness due
to external pressure p.
3. In a compound cylinder, in which one cylinder is shrink fitted over
another cylinder, show the variation of hoop stresses in outer and inner
cylinders along the wall thickness.
4. Derive expressions of shrinkage allowance in a hub and shaft assembly.
5. Explain the purpose of compounding two cylinders.
6. In an assembly of bronze sleeve and steel shaft, if the temperature of the
assembly is raised, at a particular temperature, the junction pressure
becomes zero, why?
Multiple Choice Questions
1. In a thick cylindrical shell with R2 = 2R1 subjected to external pressure of
45 N/mm2, what is the maximum hoop stress developed in the cylinder?
1. 120 N/mm2
2. 75 N/mm2
3. 60 N/mm2
4. None of these
2. In a compound cylinder at junction, the sum of circumferential strains in
outer and inner cylinders is 120 × 10‒6. If the junction’s diameter is 200
mm, what is the shrinkage allowance?
0. 0.048 mm
1. 0.024mm
2. 0.0024 mm
3. None of these
3. In a thick cylindrical shell, σcmax = 1.25p, what is the ratio of R2/R1?
0. 1.5
1. 2.0
2. 3.0
3. None of these
4. In a shaft and hub assembly with shaft diameter 50 mm, hub diameter 100
mm and junction pressure 30 N/mm2, what is the hoop stress in the shaft?
0.
1.
2.
3.
+50 N/mm2
+30 N/mm2
−30 N/mm2
None of these
5. In a thick cylinder, R2/R1 = 2, the internal pressure is 60 N/mm2. What is
the maximum shear stress at inner radius?
0.
1.
2.
3.
20 N/mm2
30 N/mm2
80 N/mm2
None of these
6. In a thick cylindrical shell, σcmax = 120 p = 50 N/mm2, what is the value of
Lame’s constant A?
0.
1.
2.
3.
30 MPa
35 MPa
70 MPa
None of these
7. In a compound cylinder, the hoop stresses developed at the junction in
outer and inner cylinders are +84 MPa and −66 MPa. If E = 200 GPa, the
junction radius is 100 mm, what is the shrinkage allowance as diameter?
0.
1.
2.
3.
0.3 mm
0.15 mm
168 mm
None of these
8. The variation of hoop stress across the thickness of a thick cylinder is
0. Linear
1. Uniform
2. Parabolic
3. None of these
9. Purpose of compounding cylinder is
0.
1.
2.
3.
To increase pressure-bearing capacity of a single cylinder
To reduce the variation in hoop stress distribution
To increase the strength of the cylinder
All the above
10.A bronze sleeve of an outer diameter of 100 mm is forced over a solid
steel shaft of a diameter of 60 mm. If the junction pressure is 32N/mm2,
the hoop strain at outer radius of sleeve is given by (if E = 100 GPa)
0. 320 mstrain
1. 360 mstrain
2. 720 mstrain
3. None of these
Practice Problems
1. Two thick cylinders A and B are of the same dimensions. The external diameter is
double the internal diameter. A is subjected to an internal pressure p1, while B is
subjected to an external pressure p2. Find the ratio of p1 / p2 if the greatest hoop
strain developed in both cylinders is the same. The Poisson’s ratio of the material
is 0.3.
2. A steel cylinder of an inside diameter of 1 m and a length of 4 m is subjected to an
internal pressure of 60 MPa. Determine the thickness of the cylinder if the
maximum shear stress in the cylinder is not to exceed 80 MPa. What is the change
in the volume of the cylinder if E = 210 GPa, and v = 0.3?
3. A pressure vessel of an internal diameter of 200 mm, an external diameter of 250
mm and a length of 1 m is tested under a hydraulic pressure of 20 N/mm2.
Determine the change in its internal and external diameters if E = 208
kN/mm2 and v = 0.3.
If a strain gauge is mounted on the surface of the pressure vessel in an axial
direction, what will be the strain reading?
4. A thick cylinder is subjected to both internal and external pressures. The internal
diameter of the cylinder is 150 mm and the external diameter is 200 mm. If the
maximum permissible stress in the cylinder is 20 N/mm2 and the external radial
pressure is 4 N/mm2, determine the intensity of internal radial pressure.
5. A compound cylinder is formed by shrinking one cylinder over another. The outer
diameter of the compound cylinder is 200 mm, the inner diameter is 120 mm and
the diameter at the common surface is 160 mm. The junction pressure developed
due to shrinkage is 4 N/mm2.
The compound cylinder is now subjected to an internal pressure of 50
N/mm2, determine themaximum hoop tension in the cylinder. How much
heavier a single cylinder of an internal diameter of 120 mm would be if it is
subjected to the same internal pressure in order to withstand the same
maximum hoop stress.
6. A compound cylinder has a bore of 120 mm and an outer diameter of 200 mm and
the diameter at the common surface is 160 mm. Determine the radial pressure at
the common surface which must be provided by the shrinkage fitting, if the
resultant maximum hoop stress in the inner cylinder under a superimposed internal
pressure of 50 N/mm2 is to be half the value of the maximum hoop tension in the
inner cylinder if this cylinder alone is subjected to an internal pressure of 50
N/mm2.
7. A thick cylinder of an internal diameter of 160 mm is subjected to an internal
pressure of 5 N/mm2. If the allowable stress for cylinder is 25 N/mm2, determine
the wall thickness of the cylinder. The cylinder is now strengthened by wire
winding so that it can be safely subjected to an internal pressure of 8 N/mm2. Find
the radial pressure exerted by wire winding.
8. A compound cylinder consists of a steel cylinder of an internal diameter of 180
mm and an external diameter of 250 mm and bronze liner of an external diameter
of 180 mm and an internal diameter of 170 mm as shown in Figure 6.16. Assuming
the liner to be thin cylinder and that there is no shrinkage stress in compound
cylinder, determine the hoop stress developed in the liner due to an internal
pressure of 50 N/mm2. Ignore the longitudinal stress and strain.
For steel,
E = 208 kN/mm2, v = 0.28
For bronze, E = 112 kN/mm2, v = 0.30
9. A steel tube has an internal diameter of 25 mm and an external diameter of 50 mm.
Another tube of the same steel is to be shrunk over the first tube so that shrinkage
stresses just produce a condition of yield at the inner surface of each tube.
Determine the necessary difference in the diameter of the tubes at the mating
surface before shrinking and the required external diameter of the outer tube.
Assuming that yielding occurs according to the maximum shear stress condition
and that no axial stresses are set up due to shrinkage. The yield stress in simple
tension or compression is 414 N/mm2, E = 207 GPa.
Figure 6.16 Compound cylinder
10. A steel sleeve is pressed onto a steel shaft of a diameter of 50 mm. The radial
pressure between the steel shaft and the steel sleeve is 20 N/mm2 and the hoop
stress at the inner surface of sleeve is 56 N/mm2. If a compressive force of 50 kN is
now applied to the shaft, determine the change in radial pressure. E = 210 Gpa, v =
0.3.
11. A steel rod of a diameter of 60 mm is forced into an aluminium sleeve of an
outside diameter of 80 mm, thereby producing a tension of 54 N/mm2 at the outer
surface of sleeve. Determine (a) the radial pressure between the bronze sleeve and
the steel rod and (b) the rise in temperature which would eliminate the force fit.
For steel, E = 200 Gpa, v = 0.3, α = 11.2 × 10−6/°C
For aluminum, E = 68 Gpa, v = 0.33, α = 22 × 10−6/°C
12. A steel plug of a diameter of 80 mm is forced into a steel ring of an external
diameter of 120 mm and a width of 50 mm. From a strain gauge fixed on the outer
surface of the ring in the circumferential direction, the strain is found to be 41 ×
10‒6. Considering that the coefficient of friction between the mating surface is 0.2,
determine the axial force required to push the plug out of the ring. E = 210 GPa.
[Hint: Normal force, N = p′ × 20 × 40 × 50, F = μN]
Special Problems
1. A compound cylinder is made by shrinking a cylinder of an outer diameter of 200
mm over another cylinder of an inner diameter of 100 mm. If the numerical value
of maximum hoop stress developed due to shrinkage fitting in both the cylinders is
the same, find the junction diameter.
2. A thick cylinder of internal diameter D and wall thickness t is subjected to an
internal pressure ofp. If the maximum hoop stress developed in the cylinder is
2.5p, determine the ratio of t/D.
3. A steel shaft of a diameter of 120 mm is force fitted in a steel hub of an external
diameter of 200 mm, so that the radial pressure developed at the common surface
is 12 MPa. If E = 200 GPa, determine the force fit allowance on the diameter.
What is the maximum hoop stress developed in the hub?
4. A steel ring of internal radius r and external radius R is shrunk onto a solid steel
shaft of radius r + dr. Prove that the intensity of pressure at the mating surface is
equal to
, where E is the modulus of elasticity of steel.
[Hint: use shrinkage formula for shaft + hub]
5. Two thick cylinders A and B are of same dimensions. The external diameter is
double the internal diameter. Cylinder A is subjected to an internal pressure p1,
while cylinder B is subjected to an external pressure p2. Find the ratio of
pressure p1 to p2 if the greatest circumferential stress developed in both the
cylinders is the same.
6. A steel cylinder of an internal diameter of 100 mm and an external diameter of 150
mm is strengthened by shrinking another cylinder onto it, the internal diameter of
which before heating is 149.92 mm. Determine the outer diameter of the outer
cylinder at the junction if the pressure at the junction after shrinking is 20 N/mm2.
E = 210 GPa.
[Hint: use expressions of shrinkage allowance are taking R1 = 50 mm
and R3 = 75 mm]
7. A compound cylinder is made by shrinking a tube of an outer diameter of 150 mm
over another tube of an inner diameter of 100 mm. Find the common diameter if
the greatest hoop stress in the inner tube is numerically 0.7 times that of outer tube.
8. A thick cylinder of an internal diameter of 120 mm and an external diameter of
180 mm is used for a working pressure of 15 N/mm2. Because of external
corrosion, the outer diameter of the cylinder is machined to 178 mm. Determine by
how much the internal pressure is to be reduced so that the maximum hoop stress
in the cylinder remains the same as before machining.
Answers to Exercises
Ecercise 6.1: 109.7, 56.9 ms
Exercise 6.2: D = 169.7 mm
Exercise 6.3: p = 102.78 N/mm2
Exercise 6.4: 20.485, −103.70, −83.22, +103.45, + 82.96 MPa.
Exercise 6.5: (a) 55.576 N/mm2, (b) 144.5 N/mm2
Exercise 6.6: 7.46 MPa
Answers to Multiple Choice Questions
1. (a)
2. (b)
3. (c)
4. (c)
5. (c)
6. (b)
7. (b)
8. (c)
9. (d)
10. (b)
Answers to Practice Problems
1.
2. t = 0.5m, δv = 5,206cc
3. 0.1707mm, 0.0727 mm, 68.375 μ strain
= 68.375 × 10−6μ strain
4. 10.72 N/mm2
5. σcmax = 100.71 N/mm2 at inner radius, single cylinder is 41.5 per cent
heavier than the compound cylinder
6. p′ = 3.71 N/mm2 due to shrinkage
7. p′ = 2 5 . N/mm2
8. 130.74 N/mm2
9. R2 = 50 mm, D2 = 100 mm, δD3 = 0.125 mm
10.2.01 N/mm2
11.(a) 21 N/mm2, (b) Rise in temperature 118.4°C
12.13.52 kN
Answers to Special Problems
1.
2.
3.
4.
5.
6.
7.
R3 = 75.12 mm D3 = 150.24 mm
0.2635
0.0225 mm, 25.5 MPa
p1 = 1.6p2
212 mm
129.92 mm
0.373 N/mm2