0

A Strong Partner for Sustainable Development

Module
In
ES 107A

MECHANICS OF DEFORMABLE
BODIES

College of Engineering and Technology

WPU-QSF-ACAD-82A Rev. 00 (09.15.20)

Bachelor of Science in Civil Engineering

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Module No. 3

Torsion and Helical Spring

Second Semester AY 2022 - 2023

Ryan A. Limco
Instructor II

WPU-QSF-ACAD-82A Rev. 00 (09.15.20)

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Table of Contents

Content Page

Title Page 1
Table of Contents 2
Instruction to User 3
Introduction 4
Chapter 3
Torsion 5
Overview 5
Learning Outcomes 5
Pretest 6
Lesson 1: Torsion 7
Specific Learning Outcome 7
Time Allotment 7
Discussion 7
1.1 Torsion 7
1.2 Torsional Shearing Stress 8
1.3 Angle if Twist 9
1.4 Power Transmitted by the Shaft 9
Lesson 2: Helical Spring 13
Specific Learning Outcome 13
Time Allotment 13
Discussion 13
2.1 Helical Spring 13
2.2 Spring in Series 14
2.3 Spring in Parallel 15
Activities/Exercises 20
References 21

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INSTRUCTION TO THE USER

This module would provide you an educational experience while

independently accomplishing the task at your own pace or time. It aims as
well to ensure that learning is unhampered by health and other challenges. It
covers the topic about the Torsion.

Reminders in using this module:

1. Keep this material neat and intact.

2. Answer the pretest first to measure what you know and what to be
learned about the topic discussed in this module.
3. Accomplish the activities and exercises as aids and reinforcement for
better understanding of the lessons.
4. Answer the post-test to evaluate your learning.
5. Do not take pictures in any parts of this module nor post it to social
media platforms.
6. Value this module for your own learning by heartily and honestly
answering and doing the exercises and activities. Time and effort were
spent in the preparation in order that learning will still continue amidst
this Covid-19 pandemic.
7. Observe health protocols: wear mask, sanitize and maintain physical
distancing.

Hi! I’m Blue Bee, your WPU

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Welcome to Western Philippines

University!
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experience.

STAY SAFE AND HEALTHY!

WPU-QSF-ACAD-82A Rev. 00 (09.15.20)

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INTRODUCTION

Mechanics of Deformable Bodies is a basic engineering science course dealing

with axial stress and strain, stresses for torsion and bending, combined stresses, beam
deflections, indeterminate beams and elastic instability.
The module includes discussions as input about the topic. Problems with
detailed solutions are also included. Enhancement activities and exercise problems are
also provided to reinforce the understanding of the topics.
As user, you are expected to read the discussions carefully, solve the sample
problems and follow instructions provided while performing the activities and
exercises. You may use books, internet and other references to further your knowledge
on the topics.

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Chapter 3

Torsion

A. Overview

Module 3 covers Torsion. The module includes one lesson: Lesson 1 – Torsion,
and Lesson 2 – Helical Springs.

B. Learning Outcomes

At the end of the module, you can:

1. Introduce the concept of torsion in structural members and machine parts.

2. Define shearing stresses and strains in a circular shaft.
3. Define angle of twist in terms of the applied torque, geometry of the shaft,
and material.
4. Use torsional deformation to solve indeterminate problems.
5. Design shaft for power transmission.
6. Review stress concentrations and how they are included in torsion
problems.
7. Analyze torsion for noncircular member.
8. Describe the elastic-perfectly plastic response of circular shafts.
9. Define the behavior of thin-walled hollow shafts.
10. Compute stresses in helical spring.
11. Compute elongation and spring constant.
12. Compute problems in spring in series.
13. Compute problems in spring in parallel.

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Lesson 1

Torsion

A. Specific Learning Outcomes

At the end of the lesson, you can:

1. Introduce the concept of torsion in structural members and machine parts.

2. Define shearing stresses and strains in a circular shaft.
3. Define angle of twist in terms of the applied torque, geometry of the shaft,
and material.
4. Use torsional deformation to solve indeterminate problems.
5. Design shaft for power transmission.
6. Review stress concentrations and how they are included in torsion
problems.
7. Analyze torsion for noncircular member.
8. Describe the elastic-perfectly plastic response of circular shafts.
9. Define the behavior of thin-walled hollow shafts.

B. Time allotment: 4 hours

C. Discussion

1.1 Torsion

Consider a bar to be rigidly attached at one end and twisted at the other end by a torque
or twisting moment T equivalent to F × d, which is applied perpendicular to the axis
of the bar, as shown in the figure. Such a bar is said to be in torsion.
𝑇 = 𝐹𝑑

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1.2 Torsional Shearing Stress

For a solid or hollow circular shaft subject to a twisting moment T, the torsional
shearing stress τ at a distance ρ from the center of the shaft is

𝜏= and 𝜏 =
where J is the polar moment of inertia of the section and r is the outer radius.
For solid circular shaft:
𝜏𝑟
𝜏 =
𝐽
Since 𝐽 = and 𝑟 =

Substituting J and D in the equation, 𝜏 = 𝜋𝐷4

32

𝟏𝟔𝑻
𝝉𝒎𝒂𝒙 =
𝝅𝑫𝟑

For solid circular shaft:

𝜏𝑟
𝜏 =
𝐽
Since 𝐽 = (𝐷 − 𝑑 ) and 𝑟 =

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Substituting J and D in the equation, 𝜏 = 𝜋

(𝐷4 −𝑑4 )
32

𝟏𝟔𝑻𝑫
𝝉𝒎𝒂𝒙 =
𝝅(𝑫𝟒 − 𝒅𝟒 )
1.3 Angle of Twist
The angle θ through which the bar length L will twist is

𝜃= in radians
Where: T is the torque in N·mm, L is the length of shaft in mm, G is shear modulus in
MPa, J is the polar moment of inertia in mm4, D and d are diameter in mm, and r is
the radius in mm.

Watch discussion at this link: https://youtu.be/vZi9BQvZ9Eg

1.4 Power Transmitted by the Shaft

A shaft rotating with a constant angular velocity ω (in radians per second) is being
acted by a twisting moment T. The power transmitted by the shaft is
𝑃 = 𝑇𝜔 = 2𝜋𝑇𝑓
where :T is Torque in 𝑁 ∙ 𝑚, 𝑓 is the number of revolutions per second, and 𝑃 is the
power in Watts.
Problem 1. A steel shaft 3 ft long that has a diameter of 4 in is subjected to a torque
of 15 kip·ft. Determine the maximum shearing stress and the angle of twist. Use G =
12 × 106 psi.
Solution:

answer

Watch discussion at this link:

https://youtu.be/G3EwD95unS0

answer

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Problem 2. What is the minimum diameter of a solid steel shaft that will not twist
through more than 3° in a 6-m length when subjected to a torque of 12 kN·m? What
maximum shearing stress is developed? Use G = 83 GPa.
Solution:

answer

answer

Watch discussion at this link: https://youtu.be/bN5YuGWWEvc

Problem 3. A compound shaft consisting of a steel segment and an aluminum

segment is acted upon by two torques as shown in Fig. P-316. Determine the maximum
permissible value of T subject to the following conditions: τst ≤ 83 MPa, τal ≤ 55 MPa,
and the angle of rotation of the free end is limited to 6°. For steel, G = 83 GPa and for
aluminum, G = 28 GPa.

Solution:
Based on maximum shearing stress, τmax = 16T / πd3:
Steel

Aluminum

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Based on maximum angle of twist, θmax = 6°:

Use the least value of T. Thus, T = 679.04 N·m answer

Watch discussion at this link: https://youtu.be/qlfu_pECaXY

Problem 4. A steel propeller shaft is to transmit 4.5 MW at 3 Hz without exceeding

a shearing stress of 50 MPa or twisting more than 1◦ in a length of 26 diameters.
Compute the proper diameter if G = 83 GPa.
Solution:

𝑃 4.5 𝑥 10 𝑊
𝑇= =
2𝜋𝑓 2𝜋(3 𝐻𝑧)

𝑇 = 238,732.41 𝑁 ∙ 𝑚

Based on maximum allowable shearing stress:

16𝑇
𝑇 =
𝜋𝑑
16 (238,732.41𝑥10 𝑁 ∙ 𝑚𝑚)
50 𝑀𝑃𝑎 =
𝜋𝑑

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𝑑 = 289.71 𝑚𝑚

Based on maximum allowable angle of twist.

𝑇𝐿
𝜃=
𝐽𝐺

𝜋 (238,732.41𝑥10 𝑁 ∙ 𝑚𝑚)(26𝑑)
1° =
180° 1
𝜋𝑑 (83𝑥10 𝑀𝑃𝑎)
32
𝑑 = 352.08 𝑚𝑚

Use the larger diameter, thus, d = 352 mm

Watch discussion at this link: https://youtu.be/AbyFsxhvbXg

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Lesson 2

Helical Spring

A. Specific Learning Outcomes

At the end of the lesson, you can:

1. Compute stresses in helical spring.

2. Compute elongation and spring constant.
3. Compute problems in spring in series.
4. Compute problems in spring in parallel.

B. Time allotment: 4 hours

C. Discussion

2.1 Helical Springs

When close-coiled helical spring, composed of a wire of round rod of
diameter d wound into a helix of mean radius R with n number of turns, is subjected
to an axial load P produces the following stresses and elongation:

The maximum shearing stress is the sum of the direct shearing stress 𝜏 = 𝑃/𝐴 and
the torsional shearing stress 𝜏 = 𝑇𝑟/𝐽, with 𝑇 = 𝑃𝑅.

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𝜏=𝜏 +𝜏
𝑃 𝑇𝑟
𝜏= +
𝐴 𝐽
𝑃 16𝑃𝑅
𝜏= +
𝜋𝑑 /4 𝜋𝑑
𝟏𝟔𝑷𝑹 𝒅
𝝉= 𝟏 +
𝝅𝒅𝟑 𝟒𝑹
Note: This formula neglects the curvature of the spring. This is used for light spring
where the ratio is small.

For heavy springs and considering the curvature of the spring, R. A. M. Wahl
formula, a which is more precise, is used. It is given by:
𝟏𝟔𝑷𝑹 𝟒𝒎 − 𝟏 𝟎. 𝟔𝟏𝟓
𝜏= +
𝝅𝒅𝟑 𝟒𝒎 − 𝟒 𝒎

Where m is called the spring index and is the Wahl Factor.

𝐷
𝑚=
𝑑
The elongation is
𝟔𝟒𝑷𝑹𝟑 𝒏
𝜹=
𝑮𝒅𝟒
Notice that the deformation 𝛿 is directly proportional to the applied load P. The ratio
P to 𝛿 is called the spring constant, k and is equal to
𝑷 𝑮𝒅𝟒
𝒌= = 𝑵/𝒎𝒎
𝜹 𝟔𝟒𝑹𝟑 𝒏
2.2 Spring in Series
For two or more springs with spring laid in series, the resulting spring constant k is
given by

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𝟏 𝟏 𝟏
= + +⋯
𝒌 𝒌𝟏 𝒌𝟐
Where 𝑘 , 𝑘 , … are the spring constants for different spring.
2.3 Spring in Parallel
For two or more springs in parallel, the resulting spring constant is

𝒌 = 𝒌𝟏 + 𝒌𝟐 + ⋯
Where 𝑘 , 𝑘 , … are the spring constants for different spring.

Watch discussion at this link: https://youtu.be/q-JMz68fl0A

Problem 2.1. A spring having a length of 500 mm

deform by 50 mm when loaded by 50 kN. a). If the
spring is cut by half of its length and is loaded by the
same load, compute the deformation of the 250 mm
spring. b.) Give the spring constant of the 500 mm
spring. C.) Give the spring constant of the second
spring.
Solution:
a. Deformation of the second spring.

𝑃𝐿
𝛿 =
𝐴𝐸
50,000 𝑁(500𝑚𝑚)
50𝑚𝑚 =
𝐴𝐸
𝐴𝐸 = 500,000 𝑁

𝑃𝐿
𝛿 =
𝐴𝐸

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50,000 𝑁(250𝑚𝑚)
𝛿 =
500,000 𝑁
𝜹𝟐 = 𝟐𝟓 𝒎𝒎

b. Spring constant of first spring.

𝑃
𝐾=
𝛿
50,000 𝑁
𝐾= = 𝟏𝟎𝟎𝟎 𝑵/𝒎𝒎
50 𝑚𝑚
c. Spring constant of first spring.
𝑃
𝐾=
𝛿
50,000 𝑁
𝐾= = 𝟐𝟎𝟎𝟎 𝑵/𝒎𝒎
25 𝑚𝑚

Watch discussion at this link: https://youtu.be/ZHHMln8OKJU

Problem 2.2. A load P is supported by two

concentric steel springs arranged as shown. The
inner spring consists of 30 turns of 20 mm wire on
the main diameter of 150 mm, the outer spring has
20 turns of 30 mm wire on a mean diameter of 200
mm. Use G=83,000 MPa.
a. Give the maximum load that the inner spring
could carry so as not to exceed a shearing
stress of 140 MPa.
b. Give the maximum load that the outer spring
could carry so as not to exceed shearing stress
of 140 MPa.
c. Give the maximum load that the spring could
carry.

a. Load carried by inner spring:

16𝑃𝑅 𝑑
𝜏= 1+
𝜋𝑑 4𝑅
16𝑃 (75𝑚𝑚) 20 𝑚𝑚
140 = 1+
𝜋(20𝑚𝑚) 4(75𝑚𝑚)
𝑷𝟏 = 𝟐𝟕𝟒𝟗 𝑵
b. Load carried by outer spring:

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16𝑃𝑅 𝑑
𝜏= 1+
𝜋𝑑 4𝑅
16𝑃 (100𝑚𝑚) 30 𝑚𝑚
140 = 1+
𝜋(30𝑚𝑚) 4(100𝑚𝑚)
𝑷𝟏 = 𝟔𝟗𝟎𝟒 𝑵
c. Maximum load that the spring could carry.
64𝑃𝑅 𝑛
𝛿=
𝐺𝑑
𝛿 =𝛿

64𝑃 (75𝑚𝑚) (30𝑡𝑢𝑟𝑛𝑠) 64𝑃 (100𝑚𝑚) (20𝑡𝑢𝑟𝑛𝑠)

=
83(20𝑚𝑚) 83(30𝑚𝑚)
𝑃 = 3.2 𝑃
If 𝑃 = 2749 𝑁 from solution A, then
𝑃 = 3.2(2749𝑁)
𝑃 = 8,746.8 𝑁 > 6904 𝑁 (outer spring fails)
Therefore,
𝑃 = 6904 𝑁
6949 = 3.2𝑃
𝑃 = 2157.5 𝑁 > 2794 𝑁(safe)
𝑃 =𝑃 +𝑃
𝑃 = 2157.5 + 6904
𝑷 = 𝟗𝟎𝟔𝟏. 𝟓 𝑵

Watch discussion at this link: https://youtu.be/xs_yBzr9ZZ8

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Problem 2.3 Two steel springs arranged in series as shown

in the figure support a load P. The upper spring has 12 turns
of 25 mm diameter wire on a mean radius of 100 mm. The
lower spring consists of 10 turns of 20 mm diameter wire on
a mean radius of 75 mm. If the maximum shearing stress in
either spring must not exceed 200 MN/m2. Use Wahl
Formula and G=83 GPa.
a. Compute the maximum value of P.
b. Compute the total elongation of the assembly.
c. Compute the equivalent spring constant by dividing
the total elongation.

Solution:
a. Maximum value of P.

16𝑃𝑅 4𝑚 − 1 0.615
𝜏= +
𝜋𝑑 4𝑚 − 4 𝑚
For the upper spring:
𝐷 200
𝑚= = =8
𝑑 25
16𝑃 (100) 4(8) 0.615
200 = +
𝜋(25) 4(8) − 4 8
𝑃 = 5182.29 𝑁
For the lower spring:
𝐷 150
𝑚= = = 7.5
𝑑 20
16𝑃 (75) 4(7.5) 0.615
200 = +
𝜋(20) 4(7.5) − 4 8
𝑃 = 3498.28 𝑁
Use 𝑷 = 𝟑𝟒𝟗𝟖. 𝟐𝟖 𝑵

b. Total Elongation

64𝑃𝑅 𝑛
𝛿=
𝐺𝑑
For the upper spring:
64𝑃𝑅 𝑛 64(3498.28)(100) (12)
𝛿 = = = 82.87 𝑚𝑚
𝐺𝑑 (83000)(25)

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For the lower spring:

64𝑃𝑅 𝑛 64(3498.28)(75) (10)
𝛿 = = = 71.12 𝑚𝑚
𝐺𝑑 (83000)(20)
𝛿 = 𝛿 + 𝛿 = 82.87 + 71.12
𝜹 = 𝟏𝟓𝟑. 𝟗𝟗 𝒎𝒎 𝒔𝒂𝒚 𝟏𝟓𝟒 𝒎𝒎

c. Equivalent spring constant

3498.28
𝐾=
154
𝑲 = 𝟐𝟐. 𝟕𝟐 𝒌𝑵/𝒎

Watch discussion at this link: https://youtu.be/CzNgKLEHJhA

Problem 2.4 A Rigid plate of negligible mass rests on a central spring which is 20 mm
higher than the symmetrically loaded outer springs. Each of the outer springs consists
of 18 turns of 10 mm wire on a mean diameter of 100 mm. The central spring has 24
turns of 20 mm wire on a mean diameter of 150 mm. If a load P = 5 kN is now applied
to the plate, use G=83 GN/m2.
a. Determine the load carried by the central spring.
b. Determine the shearing stress of the central spring.
c. Determine the shearing stress of the outer spring.

Solution:
a. Load carried by the central spring:

𝛿 = 𝛿 + 20

64𝑃 𝑅 𝑛 64𝑃 𝑅 𝑛
= + 20
𝐺𝑑 𝐺𝑑

64𝑃 (75) (24) 64𝑃 (50) (18)

= + 20
(83,000)(24) (83,000)(10)

0.0488𝑃 = 0.1735𝑃 + 20

𝑃 = 3.5552𝑃 + 409.8 ------- eq. 1

𝑃 + 2𝑃 = 5000 ------ eq. 2

Substitute eq. 1 to eq. 2:

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(3.5552𝑃 + 409.8) + 2𝑃 = 5000

𝑃 = 826 𝑁

𝑃 + 2(826) = 5000

𝑷𝟏 = 𝟑𝟑𝟒𝟖 𝑵

b. Shearing stress of the central spring:

16𝑃𝑅 𝑑
𝜏= 1+
𝜋𝑑 4𝑅
16(3348)(75) 30
140 = 1+
𝜋(20) 4(75)
𝑷𝟏 = 𝟏𝟕𝟎. 𝟓 𝑵

c. Shearing stress of the outer spring:

16𝑃𝑅 𝑑
𝜏= 1+
𝜋𝑑 4𝑅
16(826)(50) 10
140 = 1+
𝜋(10) 4(50)
𝑷𝟏 = 𝟐𝟐𝟎. 𝟗 𝑵

Watch discussion at this link: https://youtu.be/e-FeQLgSAL0

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D. Activities/Exercises

Exercise 1:
Direction: Design a compound shaft attached to rigid support using bronze with G=35
GPa and allowable shear stress of 60 MPa, and Steel with G = 200 GPa and allowable
shear stress of 80 MPa. Assume different values of diameter for Bronze and Steel.
Compute for the ratio of the length of Steel to the length of Bronze so that each
material will be stressed to its maximum permissible limit. Give the reaction at the
supports, maximum torque carried by each material, and the maximum angle of twist.

Exercise 2:
Using 2 springs of different spring constants, wire diameter, mean radius, Modulus of
Rigidity, and turns, compute the maximum value of load and total deformation using
the following conditions: Concentric (Activity 2A), Series (Activity 2B) and Parallel
(Activity 2C). Note that each material has Constant values for Modulus of Rigidity.

F. References

Beer, F. P., Johnston, E. R. Jr., DeWolf, J. T. & Mazurek, D. F. (2015) Mechanics of

Material (7th Ed). McGraw-Hill Companies, Inc.
Besavilla, V.I. Jr. (2009). Simplified Structural Design. VIB Publisher
Hibbeler, R. C. (2018). Mechanics of Materials (10th Ed. in Si Units). Pearson
Education, Inc.
Singer, F. L. & Pytel, A. (1980). Strength of Materials (3rd Ed.). Harper & Row,
Publishers, Inc., 10 East 53rd Street, New York, N.Y. 10022
Vertera, R. T. F. & Gillesania, D. I. T. (2013). Solutions to Problems in Strength of
Materials. Diego Inocencio Tapang Gillesania Publishing

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WPU-QSF-ACAD-82A Rev.
WPU-QSF-ACAD-82A Rev. 00
00 (09.15.20)
(09.15.20)