9111-SRM MCET Mechanical Engineering

Unit- III

TORSION

Part – A (Q & A)

1. Compute the torsional rigidity of a 100 mm diameter, 4 m length shaft c = 80

kN/mm2 (Nov/Dec’23)
Torsional rigidity (Ct) = Modulus of rigidity (G) x Polar moment of inertia (J)
4
πd
J= = (π x 0.14) / 32 = 9.82 x 10-6 m4
32
Ct = (80 x 109) x (9.82 x 10-6) = 785600 N m/ radian = 785.6 kN.m / radian.

2. What is meant by spring constant (Nov/Dec’23)

The spring constant, also known as the force constant or stiffness, is a measure of a
spring's resistance to being stretched or compressed. It is a physical property of the spring
and is denoted by the symbol "k."

3. List any four assumptions made in the theory of torsion. (Apr/May’23)

Material is Homogeneous and Isotropic, Circular Cross-Section, Pure Torsion, Small
Deformations

4. A solid circular shaft of 60 mm diameter transmits a torque of 1600 N.m. Determine

the value of maximum shear stress developed. (Apr/May’23)
16 T
Maximum Shear stress (ζ max ¿ = 3
3 = (16 x 1600) / (π x 0.06 ) = 37.761629.5 MPa
πd
5. Compare shafts in series and shafts in parallel connection (Apr/May’23)
Shafts in Series: Torque Transmission: In a series configuration, the same torque is
transmitted through each shaft, meaning T 1 = T 2 = T
Shafts in Parallel: Torque Distribution: In a parallel configuration, the total torque
applied is shared between the shafts, T = T 1 + T 2

6. Differentiate closed and open coiled helical springs (Apr/May’23, Apr/May’18)

Closed Coiled Helical Springs: These springs are tightly wound with no noticeable gaps
between adjacent coils. They are designed to resist stretching and twisting, making them
suitable for applications involving axial tensile loads.

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Open Coiled Helical Springs: These springs have sufficient gaps between adjacent
coils, allowing for significant compression and extension. They are primarily used for
applications that require flexibility under axial loads.

7. Draw the shear force diagram for a cantilever beam of span 2 m carrying a point
load of 5 kN at mid span. (Apr/May’24)
From A to B (0 m to 1 m)
At point A (0 m)
Shear Force (V) = + 5 kN
Just before point load (at 0.99 m):
Shear Force (V) = + 5 kN
From B to C (1 m to 2 m):
After the point load (from 1 m to 2 m):
V = 0 kN

Figure 3.1: Shear Force Diagram

8. Define stiffness of a spring (Apr/May’24)

The stiffness of a spring, also known as the spring constant, quantifies the spring's
resistance to deformation when subjected to an external force. It is defined as the amount
of force required to produce a unit displacement in the spring.
K=F/δ

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9. What is called twisting moment? (Apr/May’19)
A twisting moment, commonly referred to as torque, is a measure of the rotational force
applied to an object that causes it to twist around its axis. It is defined as the product of
the force applied and the perpendicular distance from the line of action of the force to the
axis of rotation. Mathematically, it can be expressed as: T = F.d.

10. Give any two functions of spring (Apr/May’19, Nov/Dec’19)

Energy Storage: Springs store mechanical energy when they are compressed or
stretched. This stored energy can be released to perform work, such as in mechanisms
like shock absorbers, where the spring absorbs energy from impacts and then releases it
gradually.
Force Transmission: Springs can transmit forces in mechanical systems. They provide a
restoring force that acts to return a system to its equilibrium position, which is essential in
applications like vehicle suspensions, where they help maintain stability and comfort by
absorbing road shocks.

11. Draw the diagram showing the shear stress distribution along the thickness of a
hollow shaft subjected to torsion. (Apr/May’22)

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Figure 3.2: Shear stress distribution along with thickness of the hollow shaft diagram

12. What is hollow shaft preferred over a solid shaft for transmitting power
(Apr/May’22)
Weight reduction, Higher Torsional strength, Material savings, Design flexibility,
Dynamic balancing.

13. Define torsional rigidity (Nov/Dec’19)

Torsional rigidity is a measure of a material's resistance to twisting under the application
of torque. It quantifies how much an object will deform (twist) when subjected to a
twisting force. Torsional rigidity is defined mathematically as the torque required to
produce a unit angle of twist per unit length of the shaft or structural member.
GJ = T/ (θ/L)

14. Write short notes on shaft (Nov/Dec’20)

A shaft is a crucial mechanical component, typically circular in cross-section, designed to
transmit power and rotational motion from one part of a machine to another. Shafts are
integral to various machines, supporting rotating elements such as gears, pulleys, and
wheels while also bearing the loads imposed by these components.

15. Distinguish the springs in parallel and series (Nov/Dec’20)

beam.
Springs in series: Springs are connected end-to-end, meaning the end of one spring is
attached to the beginning of the next. F= F1 = F2 = …
Springs in parallel: Springs are connected side-by-side, meaning they share both ends
and act simultaneously on the same load. Ftotal = F1 + F2 +

16. What is polar moment of inertia for solid shaft? (Apr/May’18)

The polar moment of inertia for a solid shaft is a measure of its ability to resist torsion
(twisting) when subjected to an applied torque. It is denoted as J.
4
πd
J=
32

17. Write down the expression for torque transmitted by hollow shaft (Apr/May’21)
The expression for the torque transmitted by a hollow shaft can be derived using the
polar moment of inertia and shear stress concepts. The maximum torque (T) that can be
transmitted by a hollow circular shaft is given by the formula:
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π
T= ζ (D4 – d4)
16 max
18. What are the various types of springs? (Apr/May’21)
Compression spring, Extension Spring, Torsion Spring, Constant force spring, Helical
Spring, Leaf Spring, Spiral Spring.

19. Sketch the shear stress distribution on a solid circular shaft due to torsion
(Apr/May’19)

Figure 3.3: Shear stress distribution diagram

20. What are the two types of shear stresses induced in a helical spring?
(Apr/May’19)
a) Direct shear stress
Cause: This is induced due to the axial force or load applied to the spring.
Nature: The applied load exerts a direct shearing action on the cross-section of the
spring wire.
Distribution: Uniform across the cross-section.
b) Torsional shear stress
Cause: This arises because the applied load causes the spring wire to twist along its
length.
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Nature: The twisting action induces torsional shear stress in the wire.
Distribution: Varies linearly across the cross-section, being zero at the center and
maximum at the outer surface.
Part – B (Q & A)

1. A hollow shaft of external diameter 120 mm transmits 300 W power at 200 r.p.m.
Determine the maximum internal diameter if the maximum stress in the shaft is not
to exceed 60 N/mm2. (Nov/Dec’23)
Given Data
External diameter (D) = 120 mm
Power (P) = 300 W
Speed (N) = 200 RPM
Maximum Shear Stress (ζ max ¿ = 60 N/mm2
To find:
maximum internal diameter
Solution:
Step 1: Calculate the Torque (T)
P = 2πNT/60
T = 60P/ 2πN
T= 60 x 300 / 2π x 200 = 14.32 Nm.
Step 2: Calculate the Polar moment of inertia (J)
π
J= (D4 – d4)
32
Step 3: Relate Torque to shear stress
T = ζ max J / R
Substituting for J:
π
T = (ζ max / R) x ( (D4 – d4))
32
R = D/2 = 60 mm = 0.06 m
Step 4: Substitute known values
π
14.32 = (60/ 1000) x ( (0.124 – d4))
32
Step 5: Solve for Inner Diameter (d)
π
14.32 = (60/ 1000) x ( (0.00020736 – d4))
32
Answer: d4 = 7.6333 (Since this value cannot be negative)

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2. A close-coiled helical spring is to have a stiffness of 900 N/m in compression, with a
maximum load of 45 N and a maximum shearing stress of 120 N/mm 2. The solid
length of the spring (i.e., coils touching) is 45 mm. Find
(i) The wire diameter
(ii) The mean coil radius and
(iii) The number of coils
Take modulus of rigidity of material of the spring = 0.4 x 10 5 N/mm2 (Nov/Dec’23,
Apr/May’19)
Given Data: Stiffness (k) = 900 N/m
Maximum load (F) = 45 N
Maximum Shearing stress (ζ max) = 120 N/ mm2
Solid length of the spring = 45 mm
Modulus of rigidity (G) = 0.4 x 105 N/mm2
To find:
(i) The wire diameter
(ii) The mean coil radius and
(iii) The number of coils
Solution:
Step 1: Spring stiffness
K = Gd4 / 64R3n
n = number of active coils

𝜻 = 16 WR / πd3
Step 2: Shearing stress

Step 3: Solid Lenth and Number of coils

Ls = nd
n = Ls / d
Step 4: Substitute n in the stiffness formula
K = Gd4 / 64R3n = Gd5 / 64R3 Ls
Rearranging for R:
R3 = Gd5 / 64kLs
R = (Gd5 / 64kLs )1/3

𝜻 = 16 WR / πd3
Step 5: Solve for d using Shearing stress

This equation will be solved numerically for 𝑑. Once d is found, we can determine
= [16 W(Gd5 / 64kLs )1/3] / πd3

𝑅 and 𝑛.

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Answer:
The wire diameter (d) = 3.22 mm.
Mean coil radius (R) = (Gd5 / 64kLs )1/3 = 17.48 mm
Number of coils (n) = Ls / d = 13.98

3. A solid steel shaft is subjected to a torque of 50 Nm. If the angle of twist is 0.5 0 per
metre length of the shaft and the shear stress is not to be allowed to exceed 90
MN/m2. Take C= 80 GN/m2. Find the following
(i) Suitable diameter for the shaft
(ii) Final Angle of twist
(iii) Maximum shear strain in the shaft (Apr/May’23)
Given Data
Torque T = 50 Nm
ζ max = 90 MN m2 = 90 x 106 N m2
G = 80 GN m2 = 80 x 109 N m2
θ
Angle of twist per meter length, = 0.05 rad m.
L
To find:
(i) Suitable diameter for the shaft
(ii) Final of twist
(iii) Angle of twist
(iv) Maximum shear strain in the shaft
Solution:
(i) Suitable Diameter for the shaft
16 T
ζ max = 3
πd
Rearranging for d:
16T
d3 = = (16 x 5) / ( π x 90 x 106) = 14 m
π ζ max
(ii) Final Angle of Twist
θ = TL / JG
4
πd
J= = 1.0367 x 10-9 m4
32
θ = [50 N m (1 m)] / [(1.0367 x 10-9 m4) (80 x 109 N m2)]
= 6.032 rad
(iii) Angle of Twist per Unit Length = 0.50 rad / m
(iv) Maximum Shear strain in the shaft
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ζ max= G . γmax
γmax = ζ max / G = 90 x 106 / 80 x 109 = 0.001125 (or) 1.125 %.
Answer:
(i) d3 =14 m
(ii) θ = 6.032 rad
(iii) Angle of Twist per Unit Length = 0.50 rad / m
(iv) ζ max= 1.125 %.

4. For the close-coiled helical spring subjected to an axial load of 350 N having 12 coils
of wire diameter of 16 mm and made with coil diameter of 250 mm. Take C= 80
GN/m2. Find the following:
(i) Axial Deflection
(ii) Strain energy
(iii) Maximum torsional shear stress in the wire
(iv) Maximum shear stress using wahls correction factor
(Apr/May’23)
Given data for the close-coiled helical spring:
Axial Load (F) = 350 N
Number of Coils (n) = 12
Wire Diameter (d) = 16 mm = 0.016 m
Coil Diameter (D) = 250 mm = 0.25 m
Modulus of Rigidity (C) = 80 GN/m² = 80 × 1 0 9 80×10 9 N/m²
To find:
(i) Axial Deflection
(ii) Strain energy
(iii) Maximum torsional shear stress in the wire
(iv) Maximum shear stress using wahls correction factor
Solution:

𝛿 = (64 𝐹 𝐷3𝑛) / (𝐺 𝑑4)

(i) Axial Deflection (𝛿)

Where: F = axial load, D = coil diameter, n = number of coils, G = modulus of rigidity

(shear modulus), d = wire diameter.
Substitute the given values:
δ = (64 x 350 x (0.25)3 x 12) / ((80 x 109) (0.016)4)
δ = 0.0818 m = 81.8 mm
(ii) Strain Energy

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𝑈=𝐹𝛿/2
The strain energy U stored in the spring under an axial load is given by:

U= 350 x 0.0818 / 2 = 14.315 J

(iii) Maximum Torsional Shear Stress in the Wire
ζ max = 8 F D / (πd3)
ζ max = (8 x 350 x 0.25) / (π (0.016)3) = 54.47 MPa
(iv) Maximum Shear Stress Using Wahl's Correction Factor
Kw = 1 + [(d / 4D) x (1 + (d/D)]
Kw = 1 + [(0.016 / 4 x 0.25) x (1 + (0.016/0.25)]
Kw = 1.017
Now, the corrected maximum shear stress using Wahl's factor is:
ζ max corrected = Kw × ζ max
ζ max corrected = 1.0107 x 54.47 MPa = 55.36 MPa
Summary of Results
Axial Deflection: 81.8 mm
Strain Energy: 14.315 J
Maximum Torsional Shear Stress: 54.47 MPa
Maximum Shear Stress with Wahl's Correction Factor: 55.36 MPa

5. A solid shaft of 200 mm diameter has the same cross-sectional area as a hollow shaft
of the same material with an inside diameter of 150 mm. Find the ratio of Power
transmitted by both the shafts at the same angular velocity, Angles of twist in equal
lengths of the shafts, when stresses to the same intensity. (Apr/May’23)

Solid shaft diameter, 𝑑 = 200 mm

Given:

Hollow shaft inner diameter, di =150 mm

To find:
Power Transmitted Ratio:
Angle of Twist Ratio:
Solution:
Step 1: Cross-Sectional Areas
a) Solid shaft area:
2 2
πd π 200
Asolid = = = 31415.93 mm2
4 4
b) Hollow shaft area:

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π
Ahollow = (d02 – di2)
4
Substitute di =150mm and the area A solid :
Ahollow = Asolid = 31415.93mm2
π
(d02 – 1502) = 31415.93mm2
4
d0 = 250 mm

Step 2: Polar Moment of Inertia

a) Solid shaft: The polar moment of inertia for a solid shaft is:
4 4
πd π 200
Jsolid = = = 157079632.7 mm4
32 32
b) Hollow shaft: The polar moment of inertia for a hollow shaft is:
π π
Jhollow = (d04 – di4) = (2504 – 1504) = 335103273.2 mm4
32 32
Step 3: Power Transmitted
P = τ⋅ ω⋅ J
Where: τ is the shear stress, ω is the angular velocity, J is the polar moment of inertia.
Since both shafts are transmitting power at the same angular velocity (ω) and under the
same shear stress (τ), the ratio of the power transmitted is proportional to the ratio of the
polar moments of inertia.
Power Ratio = Phollow / Psolid = Jhollow / Jsolid
= 157079632.7 / 33510323.2 = 0.469
Step 4: Angle of Twist
Θ = TL/JG
Where: T is the torque, L is the length of the shaft, G is the shear modulus (same for both
shafts since they are of the same material), J is the polar moment of inertia.
Since the stresses are the same and the material is the same, the ratio of the angles of
twist will be the same as the ratio of the polar moments of inertia.
Angle of Twist Ratio= Jsolid / Jhollow = 157079632.7 / 335103273.2 =0.469
Final Results:
Power Transmitted Ratio: The solid shaft transmits approximately 46.9% of the power
transmitted by the hollow shaft.
Angle of Twist Ratio: The angle of twist for the solid shaft is approximately 46.9% of the
angle of twist for the hollow shaft

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6. A closely coiled helical spring is made of 6 mm wire. The maximum shear stress and
deflection under a 200 N load is not to exceed 80 MPa and 11 mm respectively.
Determine the number of coils and their mean diameter. Take C = 84 MPa.
(Apr/May’23)
To determine the number of coils and the mean diameter of the helical spring, we will use
the formulas for the shear stress and deflection of a closely coiled helical spring under a
load.
Given:
Wire diameter dw = 6 mm = 0.006 m
Load applied, P = 200 N
Maximum shear stress ζ max = 80 MPa = 80 x 106 N/m2 (or) 80 x 106 Pa
Maximum Defelection, Δ = 11 mm = 0.011 m
Modulus of rigidity, G = 84 MPa = 84 x 109 Pa
Applied Load (P) = 200 N
To find:

Number of coils 𝑛
Mean diameter D

Solution:
Step 1: Calculate Mean Diameter using Shear Stress formula
ζ max = 16 P R/ (πd3)
R = ζ max πd3 / (16 P)
R = 80 x 106 x π x (0.006)3 / (16 x 200)

Thus, the mean radius 𝑅 R is approximately 0.00427 m 0.00427m, and the mean
= 0.00427 m

diameter 𝐷 = 2 𝑅 D=2R is:

D = 2 x 0.00427 = 0.00854 m = 8.54 mm
Step 2: Calculate the Number of Coils using the Deflection Formula
Δ= 64PR3n / (Gd4)
Rearranging the formula to solve for n:
n=ΔGd4 / 64PR3
= (0.011 x 84 x 109 x 1.269 x 10-9) / (64 x 200 x 7.783 x 10-8)
n = 9.49 coils
Final Answer: The mean diameter D is approximately 8.54 mm. The number of coils n
is approximately 9.5 coils.

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7. A hollow circular shaft is to have an inside diameter one half the outside diameter.
It is to be designed to transmit 50 kW at a speed of 450 rpm and shear stress is not
to exceed 85 N/mm2. Calculate
(i) External diameter of the hollow shaft,
(ii) The angle of relative twist in degrees between two sections 1.5 m apart,
and.
(iii) Percentage difference in the weight of the hollow circular shaft as
compared with a solid shaft designed for the same condition. Consider G
= 8.4 x 104 N/mm2. (Apr/May’24)

Given Data:
Power transmitted (P) = 50 kW = 50,000 W
Speed of shaft (N) = 450 rpm
Shear stress (τ max) = 85 N/mm²
Length of the shaft section (L) = 1.5 m

The inside diameter (di) is half the outside diameter (do), so 𝑑𝑖 = (1/ 2) 𝑑𝑜
Modulus of rigidity (G) = 8.4 × 10⁴ N/mm²

To find:
(i) External diameter of the hollow shaft,
(ii) The angle of relative twist in degrees between two sections 1.5 m apart, and.
(iii) Percentage difference in the weight of the hollow circular shaft as compared
with a solid shaft designed for the same condition.
Solution:
Step 1: Calculate the External Diameter of the Hollow Shaft
P= ζ max Jω / L
Where: ζ maxis the maximum shear stress, J is the polar moment of inertia for the shaft's
cross-section, ω is the angular velocity in rad/s, L is the length of the shaft.
Step 1.1: Angular velocity (ω)
ω = 2πN/ 60 = 2π x 450 / 60 = 47.123 rad / s
Step 1.2: Polar Moment of Inertia (J) for Hollow Shaft
π
J= (d04 – di4)
32
π
J= (d04 – (d0/2)4) = (15 π d04 ) / 512
32
Step 1.3: Relate Power to Shear Stress and Moment of Inertia
P= ζ max Jω / L
Rearranging to solve for do :
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do4 = (512 P L) / (15 π ζ maxω)
do4 = (512 x 50,000 x 1.5) / (15 π x 85 x 106 x 47.123)
do4 = 0.1905
do = 655 mm
Step 2: Calculate the Angle of Relative Twist
θ = TL / JG
We need to first calculate the torque (T) using the formula for power:
P = Tω / 1000
T= ω P × 1000 = (50,000 x 1000) / 47.123 = 1061032.8 N.mm
θ = TL / JG = (15 π x (655)4) / 512 = 6.023 109 mm4
θ = (1,061,032.8 x 1500) / (6.023 x 109 x 8.4 104) = 3.15 x 10-6 radians
Convert to degrees:
θ = 3.15 x 10-6 x (180 / π) = 0.00018 degrees
Step 3: Calculate the Percentage Difference in Weight

Weight of solid shaft ∝ do4

For a solid shaft with diameter do:

Weight of hollow shaft ∝ do4 −di4

For the hollow shaft:

Given that 𝑑𝑖 = 1/2 do, we have:

di4 = (1/ 2 𝑑𝑜)4 = (1/16) 𝑑𝑜4

Weight of hollow shaft / Weight of solid shaft = [d04 - (1/16) 𝑑𝑜4] / d04 = 1- (1/16) =
Thus, the weight ratio between the hollow and solid shafts is:

15/16
The percentage difference in weight is:
Percentage difference = (1- 15/16) 100 = 6.25 %
Summary:
(i) The external diameter of the hollow shaft is approximately 655 mm.
(ii) The angle of relative twist between two sections 1.5 m apart is approximately
0.00018 degrees.
(iii) The percentage difference in weight between the hollow shaft and a solid shaft is
approximately 6.25%.

8. A solid shaft transmits 250 KW at 100 rpm. If the shear stress is not to exceed 75
MPa, what should be the diameter of the shaft? If this shaft is to be replaced by a
hollow one, whose diameter ratio is 0.6, determine the size and percentage of savin
in weight, the maximum shear stress being the same. (Apr/May’24)

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Given Data:
Power transmitted (P) = 250 kW =2, 50,000 W
Speed of shaft (N) = 100 rpm
Shear stress (τmax ) = 75 MPa = 75 x 106 Pa
For the hollow shaft, the diameter ratio 𝑑𝑖 / 𝑑𝑜 = 0.6, where 𝑑𝑖 is the inner diameter and
do is the outer diameter.
We need to determine:
a) The diameter of the solid shaft.
b) The size of the hollow shaft.
c) The percentage saving in weight when replacing the solid shaft with the hollow shaft.
Solution:
Step 1: Calculate the diameter of the solid shaft
P = T ω / 1000
Step 1.1: Calculate Angular Velocity (ω)
ω = 2πN / 60
Substitute N=100 rpm:
ω = 2π x 100 / 60 = 10.47 rad /s
Step 1.2: Calculate Torque (T)
T= P×1000 / ω = 2,50,000 x 1000 / 10.47 = 23, 900, 000 N/ mm = 23.9 kN/mm
Step 1.3: Relating Torque and Shear Stress for Solid Shaft
T = (π d3 τmax) / 16
d = (16 T / (π τmax))1/3 = (16 x 23,900,000 / (π x 75))1/3 = 116.8 mm
Step 2: Diameter of the Hollow Shaft
π
J= (d04 – di4)
32

T = ((π d03 τmax) / 16) x (1- (di4 / 𝑑𝑜4))

The relationship between torque and shear stress for the hollow shaft is:

𝑑𝑖 = 0.6 𝑑𝑜
T = ((π d0 τmax) / 16) x (1- (0.64))
3

d03 = (23.9 x 106 x 16) / (75 π x 0.8704) = 1.9 x 106

d0 = 125.6 mm
Step 3: Percentage Saving in Weight
The weight of a shaft is proportional to its volume, which is in turn proportional to the
cross-sectional area.
Step 3.1: Volume of the Solid Shaft
cross-sectional area A = πd2 / 4.
Asolid = π x 116.82 / 4 = 10,775 mm2
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Step 3.2: Volume of the Hollow Shaft A = π / 4 (d02 - di2)
Ahollow = π / 4 (125.62 – 75.362) = 7,946 mm2
Step 3.3: Calculate the Percentage Saving in Weight
Percentage saving = ((Asolid – Ahollow) / Asolid) x 100 = 26.4 %
Summary:
The diameter of the solid shaft is approximately 116.8 mm.
The outer diameter of the hollow shaft is approximately 125.6 mm, and the inner
diameter is approximately 75.36 mm.
The percentage saving in weight when replacing the solid shaft with the hollow shaft is
approximately 26.4%.

9. A hollow shaft of diameter ratio 3/5 is required to transmit 400 kW at 120 rpm, the
maximum torque being 30 % greater than the mean. The shear stress is not to
exceed 50 MPA and the twist in a length of 4 m is not to exceed 1.6 o. Calculate the
minimum external diameter satisfying these conditions. Take shear modulus as 80
GPa. (Apr/May’22)
Given Data
Power transmitted (P) = 400 kW = 400,000 W
Speed of shaft (N) = 120 rpm
Maximum shear stress (τ max) = 50 MPa = 50 x 106 N/m²

𝜃 = 1. 6∘ = (1.6 × 𝜋) / 180 radians

The twist in a length of 4 meters is not to exceed 1.6 degrees, so the twist angle,

Shear Modulus (G) = 80 GPa = 80 × 103 N/mm²

Diameter ratio: di / d0 = 3/5, The inside diameter (di) is half the outside diameter (do) of
the hollow shaft
We need to determine the minimum external diameter do of the hollow shaft that
satisfies these conditions.
Solution:
Step 1: Calculate the Mean Torque
The power transmitted by a shaft is given by:
P = Tmean ω/ 1000
Step 1.1: Calculate the Angular Velocity ω
ω= 2πN/ 60
ω= 2π x 120 / 60 = 12.566 rad/ s
Step 1.2: Calculate the Mean Torque: Tmean
Tmean = P x 1000 / ω = 31,831 Nm
Step 1.3: Calculate the Maximum Torque: Tmax
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Tmax = 1.3 x Tmean =41,380.3 Nm
Step 2: Relate Torque to Shear Stress
T = (π / 16) x τmax (do3 – di 3)
di = (3/5) do = (27/125) do3
Substitute this into the torque equation:
T = (π / 16) x τmax (do3 – (27/125) do3)
T = (98 x π x τmax do3) / 2000
τmax = 50 x 106 N/m2
T = 7.7 x 106 do3 ; T = 41,380 Nm.
41,380 = 7.7 x 106 do3
do = 17.25 cm = 172.5 mm
Step 3: Calculate the Twist in the Shaft
θ= TL/JG
Step 3.1: Polar Moment of Inertia J for Hollow Shaft
π
J= (d04 – di4)
32
di4 = ((3/5) do) 4
J = (544 π / 20000) d04
Step 3.2: Relate Twist to Known Values
θ= TL/JG = 1.60 = 0.029 radians

Step 3.3: Solve for 𝑑𝑜

J = (31,831 x 4) / (0.0279 x 8 x 107) = 0.057 m4

0.057 = (544 π / 20000) d04

𝑑𝑜 = 0.179 m = 179 mm

𝑑𝑜 = 179 mm
Answer:

10. A carriage spring 1 m long is made up with steel plates with width equal to six
times, the thickness, Design the spring for a load of 15 kN such that the bending
stress does not exceed 160 MPa and the deflection does not exceed 16 mm. Tae
modulus of elasticity as 200 GPa (Apr/May’22)

Length of the spring, 𝐿 = 1 m = 1000 mm

Given Data

Load on the spring, 𝑊 = 15 kN = 15 , 000 N

Bending stress, 𝜎𝑏 = 160 MPa = 160 × 1 06 N/m2
Maximum deflection, 𝛿 = 16 mm

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CE3491 STRENGTH OF MATERIALS Page | 17
9111-SRM MCET Mechanical Engineering
Modulus of elasticity, 𝐸 = 200 GPa = 200 × 103 N/mm2
Plate width, 𝑏 = 6 𝑡, where t is the thickness of the plate.
We need to determine:
a) Number of plates (n)
b) Dimensions of each plate (b and t).
Solution:
Step 1: Formula for Deflection in a Leaf Spring
The deflection (δ) of a semi-elliptic leaf spring is given by:
δ= (6WL3) / (nbt3E)
nbt3 = (6WL3) / (δ E) = (6 x 15,000 x 10003) / (16 x 200 x 103) = 28,125
Step 2: Formula for Bending Stress
The bending stress (σb) in a leaf spring is given by:
σb = 3WL / 2nbt2
nbt2 = 3WL / 2 σb = 140,625

From the problem, b=6t. Substituting b=6t into the expressions for nbt3 and 𝑛𝑏𝑡2:
Step 3: Relate b and t

Substituting into nbt3= 28,125:

n(6t)t3= 28 , 125:

Substituting into 𝑛𝑏𝑡2 = 140,625

nt4 = 4,687.5

𝑛(6t)𝑡2 = 140 , 625

nt3 = 23,437.5
Step 4: Solve for t and n
4 3
nt / nt = 4687.5/ 23,437.5; t = 0.2 mm
nt3 = 23,437.5
n 0.23 = 234337.5
n = 2,929,687.5
Answer:
n = 2,929,687.5

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