Mechanics of Deformable Bodies 17

UNIT III. TORSION

Overview
In this unit, we will consider the derivation and application of the twisting or torsion
problem only in connection with circular shafts. The twisting of noncircular shafts is so
complex that we shall only state the formulas that are used.

Learning Objectives:

At the end of this unit, I am able to:

1. Calculate the diameter of a shafting base on the maximum allowable shearing
stress
2. Calculate the diameter of a shafting base on the maximum allowable angle of twist
3. Understand the concept and principle of keys
4. Understand the concept and principle of a flanged bolt coupling
5. Understand the concept and principle of a helical spring
 Mechanics of Deformable Bodies 18

Lesson Proper

Introduction
In deriving the torsion formulas, we make the following assumptions. These
assumptions may be proved mathematically, and some may be demonstrated experimentally.
The first two only apply to shafts of circular section.
1. Circular sections remain circular.
2. Plane sections remain plane and do not warp.
3. The projection upon transverse section of straight radial lines in the section remains
straight.
4. Shaft is loaded by twisting couples in planes that are perpendicular to the axis of the
shaft.
5. Stresses do not exceed the proportional limit.

Torsion Formulas
The figure below shows views of a solid circular shaft. If a torque T is applied at the
ends of the shaft, a fiber AB on the outside surface, which is originally straight, will be twisted
into a helix AC as the shaft is twisted through the angle θ. This helix is formed as follows:

From the above figure, the shearing deformation δs is equal to arc DE. The length of
this deformation is the arc of a circle whose radius is ρ and which subtended angle of θ
radians; the length is given by
 Mechanics of Deformable Bodies 19

δs = DE = ρθ
The unit deformation (shearing strain) will be equal to

𝛿𝑠 ρθ
γ= =
L L
Then the shearing stress at this typical fiber is determined from Hooke’s Law to be

ρθ
τ = Gγ = G ( )
L
The figure below shows the free body diagram of the left side of a haft which divided
into two by cutting plane M-N. A differential area of section M-N at a radial distance ρ from
the axis of the shaft carries the differential resisting load dP = τdA.

The applied torque T is equal to the resisting torque Tr. The resisting torque Tr is the sum of
the resisting torques developed by all differential loads dP.

T = Tr = ∫ ρ dP = ∫ ρ (τdA)
Replacing τ by its value from the previous equation gives

ρθ
T = ∫ ρ [G ( ) dA]
L
Gθ
T=( ) ∫ ρ2 dA
L
Since ∫ ρ2 dA , the polar moment of inertia (J) of the cross section,
 Mechanics of Deformable Bodies 20

Gθ
T=( )J
L
𝐓𝐋
𝛉=
𝐉𝐆
where: θ = angular deformation or angle of twist (radian)
T = torque (N-mm, lbf-in)
L = length (mm, in)
J = polar moment of inertia (mm4, in4)
G = modulus of rigidity (MPa, psi)

By replacing Gθ/L by its equivalent T/J, we can obtain the torsion formula:

Gθ Tρ
τ = ρ( )=
L J
Now, to obtain the maximum shearing stress, we have to replace the ρ by the radius of the
shaft:
𝐓𝐫
𝐌𝐚𝐱. 𝛕 =
𝐉
where: Max. τ = maximum shearing stress (MPa, psi)
T = torque (N-mm, lbf-in)
r = radius (mm, in)
J = polar moment of inertia (mm4, in4)

*Note: Since Hooke’s law was used in deriving these equations, the stresses must not exceed
the shearing proportional limit; also, these formulas are applicable only to circular shafts,
either solid or hollow.
 Mechanics of Deformable Bodies 21

The values of polar moments of inertia for circular shafts are given with the figure below.

Using the values above, we can obtain the formula for maximum shearing stress both
for solid and hollow shafts.
𝟐𝐓 𝟏𝟔𝐓
𝐌𝐚𝐱. 𝛕 = 𝛑𝐫𝟑 = 𝛑𝐝𝟑 : For Solid Shaft

𝟐𝐓𝐑 𝟏𝟔𝐓𝐃
𝐌𝐚𝐱. 𝛕 = 𝛑(𝐑𝟒−𝐫𝟒) = 𝛑(𝐃𝟒−𝐝𝟒) : For Hollow Shaft

In many practical applications, shafts are used to transmit power. From dynamics, it
is known that the power P transmitted by a constant torque T rotating at a constant angular
speed ω is given by,

𝐏 = 𝐓𝛚
If the shaft is rotating with a frequency of n revolution per unit time, ω = 2πn, thus
the torque can be expressed as
P = T(2πn)
𝐏 = 𝟐𝛑𝐓𝐧
where: P = power (kW, hp)
T = torque (N-m, lbf-in)
ω = angular speed (rad/sec)
n = frequency (Hz, rev/sec)
 Mechanics of Deformable Bodies 22

To simplify the solving in terms of units, we can use the formulas below

𝐓𝐧
𝐏=
𝟔𝟑, 𝟎𝟎𝟎
where: P = power (hp)
T = torque (lbf-in)
n = frequency (rev/min)

or we can use

𝐓𝐧
𝐏=
𝟗𝟓𝟓𝟎
where: P = power (kW)
T = torque (N-m)
n = frequency (rev/min)

Example:
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 × 10 6 psi.
Solution:
Solving for the maximum shearing stress that can resist by the steel shaft.
2T 16T
Max. τ = πr3 = πd3
in
16(15 kip−ft)(12 ft )
Max. τ = π(4 in)3

𝐌𝐚𝐱. 𝛕 = 𝟏𝟒. 𝟑𝟐 𝐤𝐬𝐢

Solving for the angle of twist.

TL
θ= JG
TL
θ = π(d)4
G
32
in lb in
(15 kip−ft)(12 )(1000 )(3 ft)(12 )
ft kip ft
θ= π(4 in) 4 lb
(12 x106 2)
32 in
 Mechanics of Deformable Bodies 23

360ᵒ
θ = 0.0215 rad (2πrad)
𝛉 = 𝟏. 𝟐𝟑ᵒ

2. A solid steel shaft 5 m long is stressed at 80 MPa when twisted through 4°. Using G = 83
GPa, compute the shaft diameter. What power can be transmitted by the shaft at 20 Hz?

Solution:
First, solve for the relation between the torque and the shaft diameter.
TL
θ= JG
mm
2πrad T(5 m)(1000 m )
4ᵒ ( 360ᵒ ) = π(d)4 N
mm4(83,000 )
32 mm2

T = 0.1138(d)4 , N − mm

Solving for the shaft diameter.

16T
Max. τ = πd3
16[0.1138(d)4]
80 MPa =
πd3
d = 138.03 mm (5.43 in. ) use 𝟏𝟓𝟐. 𝟒 𝐦𝐦 (𝟔. 𝟎 𝐢𝐧. )

Solving for the torque

T = 0.1138(d)4 , N − mm
T = 0.1138(152.4 mm)4
T = 61,387,808.56 N − mm
T = 61,387.81 N − m

Solving for the power transmitted.

 Mechanics of Deformable Bodies 24

Tn
P = 9550
rev sec
(61,387.81 N−m)(20 )(60 )
sec min
P= 9550

𝐏 = 𝟕𝟕𝟏𝟑. 𝟔𝟓 𝐤𝐖

3. A steel propeller shaft is to transmit 4.5 MW at 3 Hz without exceeding a shearing stress of

50 MPa or twisting through more than 1° in a length of 26 diameters. Compute the proper
diameter if G = 83 GPa.

Solution:
Solving first for the torque.
9550 P
T= n
9550 (4500 kW)
T= rev sec
(3 )(60 )
sec min

T = 238,750 N − m

Solving for the diameter based on the maximum allowable shearing stress.
16T
Max. τ = πd3
mm
16(238,750 N−m)(1000 m )
50 MPa = πd3

𝐝 = 𝟐𝟖𝟗. 𝟕𝟐 𝐦𝐦

Solving for the diameter based on the maximum allowable angle of twist.
TL
θ= JG
mm
2π (238,750 N−m)[26(d)](1000 )
m
1ᵒ (360ᵒ) = π(d)4 N
mm4(83,000 )
32 mm2
 Mechanics of Deformable Bodies 25

𝐝 = 𝟑𝟓𝟐. 𝟎𝟗 𝐦𝐦

Therefore, we must use d = 352.09 mm for safe operation.

4. An aluminum shaft with a constant diameter of 50 mm is loaded by torques applied to gears

attached to it as shown in the figure below. Using G = 28 GPa, determine the relative angle of
twist of gear D relative to gear A.

Solution:
TL
θ= JG

Since the shaft is a

homogenous material and
have a constant diameter,

1
𝜃𝐷/𝐴 = JG ΣTL

θD/A =
mm 2
[800 Nm(2m)−300 Nm(3m)+600 Nm(2m)](1000 )
m
π(50 mm)4
(28,000 MPa)
32
360ᵒ
θD/A = 0.1106 rad (2πrad)
𝛉𝐃/𝐀 = 𝟔. 𝟑𝟒ᵒ
Keys
 Mechanics of Deformable Bodies 26

When power is transmitted from one shaft to another, machine member such as
pulley, sprockets and gears are needed. These machine members are normally fastened to
the shafts by means of keys. As the power is transmitted, the key is subjected to shearing as
well as compression while the shaft and hub are subjected to compression.

A key is machine member employed at the interface of pair of a mating male and
female circular cross-sectioned members to prevent relative angular motion between these
mating.
A keyway is a female mating member which forms a groove in the shaft to which the
key fits.
Design Calculation

For the shearing stress on the Key

 Mechanics of Deformable Bodies 27

𝐅 𝟐𝐓
𝛕= =
𝐀 𝐰𝐋𝐃

For the compressive stress on the Key

𝐅 𝟒𝐓
𝛔= =
𝐀 𝐡𝐋𝐃

where: τ = shearing stress (MPa, psi)

σ = compressive stress (MPa, psi)
T = Torque (N-mm, lbf-in)
P = Force (N, lbf)
D = shaft diameter (mm, in)
w = width (mm, in)
h = height (mm, in)
L = length (mm, in)

*Note that if it is a square key, the width (w) and height (h) will be equal

Example:
 Mechanics of Deformable Bodies 28

1. A 200-mm-diameter pulley is prevented from rotating relative to 60-mm-diameter shaft

by a 70-mm-long key, as shown in the figure below. If a torque T = 2.2 kN·m is applied to the
shaft, determine the width w of the key if the allowable shearing stress in the key is 60 MPa.
Solution:

2T
τ = wLD
2T
w = τLD
1000 N 1000 mm
2(2.2 kN−m)( 1 kN )( 1 m )
w= (60 MPa)(70 mm)(60 mm)

𝐰 = 𝟏𝟕. 𝟒𝟔 𝐦𝐦

2. If the key in the problem above has a 10 mm height, calculate the maximum compressive
stress it will receive.
Solution:
4T
σ = hLD
1000 N 1000 mm
4(2.2 kN−m)( )( )
1 kN 1m
σ= (10 mm)(70 mm)(60 mm)

𝛔 = 𝟐𝟎𝟗. 𝟓𝟐 𝐌𝐏𝐚
 Mechanics of Deformable Bodies 29

3. A line shaft is 2 15/16” in diameter and will transmit 50 hp when turning at 200 rpm at
constant rate. This shaft furnishes the power to 10 machines each requiring 5 hp to operate.
Each of the 10 pulleys is keyed to the shaft by standard flat key. If the width and thickness of
the key are ¾” and ½” respectively, find the length of the key based on shear considering that
the allowable shear stress for commercial shafting is 6000 psi.
Solution:
First calculate for the torque in each pulley.
Tn
P = 63,000
63,000 P
T= n
63,000 (5 hp)
T= 200 rpm

T = 1575 lbf − in

Solving for the required length of the key.

2T
τ = wLD
2T
L = τwD
2(1575 lbf−in )
L = (6000 psi)(3/4")(2.9375")

𝐋 = 𝟎. 𝟐𝟑𝟖 𝐢𝐧
 Mechanics of Deformable Bodies 30

Flanged Bolt Couplings

Commonly used connection between two shafts is a flanged bolt coupling. It consists
of flanges rigidly attached to the ends of the shafts and bolted together as shown in the figure
below. The torque is transmitted by the shearing force P created in the bolts.

𝛑𝐝𝟐
𝐓 = 𝐏𝐑𝐧 = 𝛕 ( ) 𝐑𝐧
𝟒
where: T = Torque (N-mm, lbf-in)
P = Force (N, lbf)
τ = shearing stress (MPa, psi)
d = bolt diameter (mm, in)
R = bolt circle radius (mm, in)
n = no. of bolts (pcs.)
 Mechanics of Deformable Bodies 31

Sometimes, there are flanged bolt couplings that has two concentric rows of bolts as
shown in the figure below. The torque capacity of the coupling is

𝐓 = 𝐏𝟏 𝐑 𝟏 𝐧𝟏 + 𝐏𝟐 𝐑 𝟐 𝐧𝟐

The relation between P1 and P2 can be

determined from the fact that the comparatively
rigid flanges cause shear deformations in the bolts
that are proportional to their radial distances from
the shaft axis.
γ1 γ2
=
R1 R 2
Using Hooke’s law for shear
τ
γ=
G
Therefore, we will have

τ1 τ2
=
G1 R1 G2 R 2
Or

P1 P2
=
A1 G1 R1 A2 G2 R 2
But if the bolts have the same size A1 = A2 and same material G1 = G2, therefore the
relation between the force P and bolt circle radius R will be

P1 P2
=
R1 R 2
 Mechanics of Deformable Bodies 32

Example:
1. A flanged bolt coupling consists of ten 20-mm-diameter bolts spaced evenly around a bolt
circle 400 mm in diameter. Determine the torque capacity of the coupling if the allowable
shearing stress in the bolts is 40 MPa.
Solution:
Solving for the torque capacity

πd2
T = PRn = τ ( ) Rn
4
π(20 mm)2
T = 40 MPa [ ] (200 mm)(10)
4

T = 25,132,741.23 N − mm
T = 25,132.74 N − m
𝐓 = 𝟐𝟓. 𝟏𝟑 𝐤𝐍 − 𝐦

2. A flanged bolt coupling consists of eight 10-mm-diameter steel bolts on a bolt circle 400
mm in diameter, and six 10-mm-diameter steel bolts on a concentric bolt circle 300 mm in
diameter, as shown in the figure below. What torque can be applied without exceeding a
shearing stress of 60 MPa in the bolts?
Solution:
Solving for the force in the bolt on the outer
circle.

πd2
P1 = τ1 ( )
4 1

π(10 mm)2
P1 = 60 MPa [ 4
]
P1 = 4712.39 N

Solving for the force in the bolt on the inner circle.

P1 P
= R2
R1 2
4712.39 N P2
=
200 mm 150 mm

P2 = 3534.29 N
 Mechanics of Deformable Bodies 33

Solving for the maximum torque that can be applied in the flanged bolt coupling.

T = P1 R1 n1 + P2 R 2 n2
T = (4712.39 N )(200 mm)(8) + (3534.29 N)(150 mm)(6)
T = 10,720,685 N − mm
T = 10,720.68 N − m
𝐓 = 𝟏𝟎. 𝟕𝟐 𝐤𝐍 − 𝐦

3. A flanged bolt coupling is used to connect a solid shaft 90 mm in diameter to a hollow shaft
100 mm in outside diameter and 90 mm in inside diameter. If the allowable shearing stress
in the shafts and the bolts is 60 MPa, how many 10-mm steel bolts must be used on a 200-
mm bolt circle diameter so that the coupling will be as strong as the weaker shaft?
Solution:
Solving for the torque in each shaft.
16T
τ = πd3
16T
60 MPa = π(90 mm)3

T = 8,588,328.92 N − mm

16TD
τ = π(D4 −d4)
16T(100 mm)
60 MPa = π[(100 mm)4−(90 mm)4]

T = 4,051,476.43 N − mm
Solving for the number of 10-mm steel bolts. Use the weaker shaft (lower torque).

πd2
T= τ ( 4 ) Rn
π(10 mm)2 200 mm
4,051,476.43 N − mm = (60 MPa) [ ]( )n
4 2

n = 8.6 bolts 𝑠𝑎𝑦 𝟗 𝐛𝐨𝐥𝐭𝐬

 Mechanics of Deformable Bodies 34

Helical Spring
The close-coiled helical spring in the figure below is elongated by an axial load P. The
spring is composed of a wire or round rod of diameter d wound into a helix of mean radius R.
The helix angle is small, so that any coil of the spring may be considered as lying
approximately in a plane perpendicular to the axis of the spring.
To determine the stresses produced by P, we follow the general procedure of passing
an exploratory cutting plane m-n through any typical section as shown and then determining
the resisting forces required for equilibrium.

To balance the applied axial load P, the exposed shaded cross section of the spring
must provide the resistance Pr, equal to P. The free body diagram is now in equilibrium as far
as a vertical and horizontal summation of forces is concerned.

The magnified view of the cross section in the figure above shows the stress
distribution that created the resisting forces. The maximum shearing stress occurs at the
inside element is given by the sum of the direct shearing stress τ1 and the maximum value of
the torsional shearing stress τ2.
τ = τ1 + τ2
P 16T
τ= +
A πd3
4P 16PR
τ= +
πd2 πd3
 Mechanics of Deformable Bodies 35

This may also be written as

16PR d
τ= (1 + )
πd3 4R

It should be noted that the preceding discussion contains an error because the torsion
formula derived for the use with straight bars was applied to a curved bar. In the figure below,
the torsion in the straight bar produces the same shearing deformation om fibers AB and CD.
The shearing strain at B and D is the same, since the elements AB and CD have the same
original length.
On the other hand, although fibers AB and CD undergo the same shearing
deformation, the shearing strain at B on the inside element is greater that at D on the outside
element because of the shorter initial length of AB.

Evidently, this difference depends on how sharply curved the spring wire is, that is,
on the ratio of d to R.A.M. Wahl has developed the following formula that takes account of the
initial curvature of the spring wire.
𝟏𝟔𝐏𝐑 𝟒𝐦 − 𝟏 𝟎. 𝟔𝟏𝟓
𝛕𝐦𝐚𝐱 = ( + )
𝛑𝐝𝟑 𝟒𝐦 − 𝟒 𝐦

𝟏𝟔𝐊𝐏𝐑
𝛕𝐦𝐚𝐱 =
𝛑𝐝𝟑
The ratio [(4m-1/4m-4) +0.615/m] is called the Wahl Factor K, and m is the spring
index or ratio of the mean diameter of the spring to the wire diameter of the spring (m =
2R/d = D/d ).

where: τmax = shearing stress (MPa, psi)

T = Torque (N-mm, lbf-in)
P = Force (N, lbf)
d = wire diameter (mm, in)
D = mean diameter (mm, in)
R = mean radius (mm, in)

In light springs, where the ratio of m is large, the first term in the parentheses
approaches unity, therefore it can be written as,
 Mechanics of Deformable Bodies 36

𝟏𝟔𝐏𝐑 𝟎. 𝟓
𝛕𝐦𝐚𝐱 = (𝟏 + )
𝛑𝐝𝟑 𝐦
Spring Deflection

Practically all the spring elongation, measured along its axis, is caused by torsional
deformation of the spring wire. If we temporarily assume all the spring in the figure below to
be rigid except the small length dL, the end A will rotate to D through the small angle dθ.
Because dθ is small, the arc AD = AB dθ may be considered as a straight line perpendicular
to AB. Therefore, from the similar triangle ADE and BAC, we can obtain

AE BC
= AB
AD

dδ R
=
AB dθ AB
Therefore

dδ = R dθ
Applying the formula for the angle of twist dθ

T dL
dδ = R ( )
JG
Since T = PR
(PR)dL
dδ = R [ ]
JG

which is then integrated to get the total elongation contributed by all elements of the spring:
PR2 L
δ=
JG
Since the length of the spring L = 2πRn, where n is the number of coils and J = πd4/32.

PR2 (2πRn)
δ=
πd4
32 G

𝟔𝟒𝐏𝐑𝟑 𝐧
𝛅=
𝐆𝐝𝟒
 Mechanics of Deformable Bodies 37

where: δ = spring deflection (mm, in)

P = Force (N, lbf)
R = mean radius (mm, in)
d = wire diameter (mm, in)
n = number of coils (active)
G = modulus of rigidity (MPa, psi)

*Note: This equation is used to compute the deflection in compression springs provided the
coils are not spaced so closely that they touch when the load is applied.

Spring Constant

The deflection δ of spring is directly proportional to the applied force P. The ratio of P to δ is
called the spring constant k and is equal to
𝐏 𝐆𝐝𝟒
𝐤= =
𝛅 𝟔𝟒𝐑𝟑 𝐧
where: k = spring constant (N/mm, lbf/in)
δ = spring deflection (mm, in)
P = Force (N, lbf)
R = mean radius (mm, in)
d = wire diameter (mm, in)
n = number of coils (active)
G = modulus of rigidity (MPa, psi)

Spring in Series

Some problem may involve two or more springs connected by either series or
parallel. For springs connected in series the following properties will be considered.

For the spring constant

𝟏
𝐤𝐓 = 𝟏 𝟏
+ …..
𝐤𝟏 𝐤𝟐

For the applied force

𝐏𝐓 = 𝐏𝟏 = 𝐏𝟐
For the spring deflection
𝛅𝐓 = 𝛅𝟏 + 𝛅𝟐
 Mechanics of Deformable Bodies 38

Spring in Parallel

For springs connected in parallel the following properties will be considered.

For the spring constant

𝐤𝐓 = 𝐤𝟏 + 𝐤𝟐
For the applied force
𝐏𝐓 = 𝐏𝟏 + 𝐏𝟐
For the spring deflection
𝛅𝐓 = 𝛅𝟏 = 𝛅𝟐

Example:
1. Determine the maximum shearing stress and elongation in a helical steel spring composed
of 20 turns of 20-mm-diameter wire on a mean radius of 90 mm when the spring is
supporting a load of 1.5 kN. Use G = 83 GPa.

Solution:

Solve for the spring index

2R 2(90 mm)
m= = =9
d (20 mm)

Solve for the Wahl Factor

4m−1 0.615
K = 4m−4 + m

4(9)−1 0.615
K = 4(9)−4 + 9

K = 1.162
Solving for the maximum shearing stress.

16KPR
τmax = πd3

16(1.162)(1500 N)(90 mm)

τmax = π(20 mm)3
 Mechanics of Deformable Bodies 39

𝛕𝐦𝐚𝐱 = 𝟗𝟗. 𝟖𝟕 𝐌𝐏𝐚

Solving for the spring deflection

64PR3n
δ=
Gd4
64(1500 N)(90 mm)3(20)
δ= (83,000 MPa)(20 mm)4
𝛅 = 𝟏𝟎𝟓. 𝟒 𝐦𝐦

2. Compute the maximum shearing stress developed in a phosphor bronze spring having
mean diameter of 200 mm and consisting of 24 turns of 20-mm diameter wire when the
spring is stretched 100 mm. Use G = 42 GPa.

Solution:

Solving for the applied force

64PR3n
δ=
Gd4
δGd4
P= 64R3n
(100 mm)(42,000 MPa)(20 mm)4
P= 64(100 mm)3 (24)
P = 437.5 N

Solve for the spring index

D (200 mm)
m= = = 10
d (20 mm)

Solve for the Wahl Factor

4m−1 0.615
K = 4m−4 + m

4(10)−1 0.615
K = 4(10)−4 + 10

K = 1.145
 Mechanics of Deformable Bodies 40

Solving for the maximum shearing stress.

16KPR
τmax = πd3

16(1.145 )(437.5 N)(100 mm)

τmax = π(20 mm)3
𝛕𝐦𝐚𝐱 = 𝟑𝟏. 𝟖𝟗 𝐌𝐏𝐚

3. Two steel springs arranged in series as shown in the figure below supports 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 MPa, compute the maximum
value of P and the total elongation of the assembly. Use G = 83 GPa. Compute the equivalent
spring constant by dividing the load by the total elongation.

Solution:

Solve for the maximum applied force in each spring

For the spring 1(larger spring)

16PR 4m−1 0.615

τmax = ( + )
πd3 4m−4 m

16P(100 mm) 4(8)−1 0.615

200 MPa = ( + )
π(25 mm)3 4(8)−4 8

P = 5182.29 N

For the spring 2(smaller spring)

16PR 4m−1 0.615

τmax = (4m−4 + )
πd3 m

16P(75 mm) 4(7.5)−1 0.615

200 MPa = ( + )
π(20 mm)3 4(7.5)−4 7.5

𝐏 = 𝟑𝟒𝟗𝟖. 𝟐𝟖 𝐍 use this as the safe load

 Mechanics of Deformable Bodies 41

Solving for the total elongation of spring.

δT = δ1 + δ2

64PR3n 64PR3n
δT = ( ) +( )
Gd4 1 Gd4 2

64(3498.28 N)(100 mm)3(12) 64(3498.28 N)(75 mm)3(10)

δT = [ (83,000 MPa)(25 mm)4 ] + [ (83,000 MPa)(20 mm)4 ]
1 2

𝛅𝐓 = 𝟏𝟓𝟑. 𝟗𝟗 𝐦𝐦

Solving for the total spring constant.

P
k=δ
3498.28 N
k = 153.99 mm

𝐍
𝐤 = 𝟐𝟐. 𝟕𝟐 𝐦𝐦
 Mechanics of Deformable Bodies 42

References

Strength of Materials 4th Edition by Andrew Pytel and Ferdinand L. Singer

https://mathalino.com/reviewer/mechanics-and-strength-of-materials/mechanics-and-
strength-of-materials

Solved Problems in Machine Design by J.A. Mandawe and R.S. Capote

 Mechanics of Deformable Bodies 43

Assessing Learning

Activity 3

Name: ____________________________________________ Score: ___________________

Course/Year/Section: __________________________ Date: _____________________

Directions: Research, solve, and compile at least 10 problems in unit 3