Machine Design 1: BASAEN, RV

Substituting 𝛿𝛿 into equation (a)

4. DESIGNING FOR IMPACT LOADS 𝑃𝑃 1 𝑃𝑃
𝑊𝑊 [ℎ + ] = 𝑃𝑃 � �
𝑘𝑘 2 𝑘𝑘
2𝑊𝑊ℎ𝑘𝑘 + 2𝑊𝑊𝑊𝑊 = 𝑃𝑃2
IMPACT LOADING
𝑃𝑃2 − 2𝑊𝑊𝑊𝑊 = 2𝑊𝑊ℎ𝑘𝑘
Case A: Impact loading from Potential Energy Completing the squares,
𝑃𝑃2 − 2𝑊𝑊𝑊𝑊 + 𝑊𝑊 2 = 2𝑊𝑊ℎ𝑘𝑘 + 𝑊𝑊 2
2ℎ𝑘𝑘
elastic body (1) (𝑃𝑃 − 𝑊𝑊)2 = 𝑊𝑊 2 �1 + �
𝑊𝑊
𝟐𝟐𝟐𝟐𝟐𝟐
𝑷𝑷 = 𝑾𝑾 �𝟏𝟏 + �𝟏𝟏 + � → equation (A)
𝑾𝑾
where
rigid body (2)
𝟐𝟐𝟐𝟐𝟐𝟐
𝐏𝐏 stopper of (1) Shock Factor or Impact Factor = �𝟏𝟏 + �𝟏𝟏 + �
ℎ 𝑾𝑾
W = dead load
P = impact load
𝛿𝛿
𝑥𝑥 𝑥𝑥
Note:
- P depends on W, h and k
- If ℎ = 0 then 𝑃𝑃 = 2𝑊𝑊 this is a case of suddenly
where applied load.
W = weight of rigid body (2) - Above expression of 𝑃𝑃 is conservative since body
h = height of drop of (2) from the stopper of (2) is treated to be rigid. Actual collisions involve
body (1) both deformable bodies for (1) and (2) —
P = impact load equivalent of (2) caused on resulting in a smaller magnitude of impact load P.
body (1)
𝛿𝛿 = deformation of the elastic body (1) due to P Case B: Impact loading from Kinetic Energy
x-x = lowest level reached at impact
𝑣𝑣
Analysis: (Energy balance based on level x-x) 𝐏𝐏
𝑚𝑚
[Energy given-off by body (2)]
= [Energy absorbed by body (1)] rigid body (2) elastic body (1)
𝛿𝛿
𝟏𝟏
𝑾𝑾 [𝒉𝒉 + 𝜹𝜹] = 𝑷𝑷𝑷𝑷 → (𝒂𝒂)
𝟐𝟐
𝑊𝑊 𝑃𝑃 [Energy given-off by body (2)]
= [Energy absorbed by body (1)]

[Kinetic Energy] = [Strain Energy]

𝟏𝟏 𝟏𝟏
𝒎𝒎𝒗𝒗𝟐𝟐 = 𝑷𝑷𝑷𝑷
𝟐𝟐 𝟐𝟐

0 0 𝛿𝛿 Substituting 𝑘𝑘,
ℎ + 𝛿𝛿 𝟏𝟏 𝑾𝑾 𝟐𝟐 𝟏𝟏 𝑷𝑷
Potential Energy Strain Energy � � 𝒗𝒗 = 𝑷𝑷 � �
𝟐𝟐 𝒈𝒈 𝟐𝟐 𝒌𝒌

𝑾𝑾𝑾𝑾𝟐𝟐 𝒌𝒌
Introducing the spring constant or stiffness of the elastic 𝑷𝑷 = � → equation (B)
material, 𝑘𝑘 𝒈𝒈
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑃𝑃 𝑃𝑃
𝑘𝑘 = = ; 𝑜𝑜𝑜𝑜 𝛿𝛿 = Note:
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝛿𝛿 𝑘𝑘
The above equations in (A) and (B) neglect the effect
of the elastic body’s mass 𝑚𝑚𝑒𝑒 . If its mass or weight is
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 Machine Design 1: BASAEN, RV
considered, the resulting impact load tends to be smaller. b. Bending Loads (Special Cases Only):
From the law of conservation of momentum 1. Cantilever w/ load at its free end

𝑚𝑚1 𝑣𝑣1 + 𝑚𝑚2 𝑣𝑣2 = (𝑚𝑚1 + 𝑚𝑚2 )𝑣𝑣𝑐𝑐 𝐏𝐏

𝑚𝑚2
𝑣𝑣𝑐𝑐 = 𝑣𝑣 � � : 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒: 𝑚𝑚1 = 𝑚𝑚𝑒𝑒 ; 𝑚𝑚2 = 𝑚𝑚
𝑚𝑚1 + 𝑚𝑚2 𝑦𝑦
Or, expressed in terms of weights

𝐿𝐿
𝟏𝟏
𝒗𝒗𝒄𝒄 = 𝒗𝒗 � �
𝑾𝑾
𝟏𝟏 + 𝑾𝑾𝒆𝒆
𝑃𝑃 𝑃𝑃𝑃𝑃3
𝑘𝑘 = ; 𝑏𝑏𝑏𝑏𝑏𝑏 𝑦𝑦 =
𝑦𝑦 3𝐸𝐸𝐸𝐸
where
𝑣𝑣𝑐𝑐 = corrected contact velocity 𝟑𝟑𝟑𝟑𝟑𝟑
𝑊𝑊𝑒𝑒 = equivalent weight of the beam or of the ∴ 𝒌𝒌 =
𝑳𝑳𝟑𝟑
elastic body

Thus, if the beam’s mass or the elastic body’s mass is 2. Simple beam loaded at midpoint
considered, the contact velocity becomes smaller. Also, in 𝐏𝐏
equation (A), the height of drop is modified with
𝑦𝑦
𝟏𝟏
𝒉𝒉′ = 𝒉𝒉 � �
𝑾𝑾
𝟏𝟏 + 𝑾𝑾𝒆𝒆 L/2
L
The above conditions result in a lower impact load P. Thus,
aside from being simpler in the resulting impact load 𝑃𝑃 𝑃𝑃𝑃𝑃3 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒
𝑘𝑘 = ; 𝑏𝑏𝑏𝑏𝑏𝑏 𝑦𝑦 = : ∴ 𝒌𝒌 =
equations, the analysis is more conservative if the elastic 𝑦𝑦 48𝐸𝐸𝐸𝐸 𝑳𝑳𝟑𝟑
body’s mass is neglected.

In designing members for impact loading, it is generally on 3. Fixed Beam with Load at Midpoint
the safer side if the mass of the elastic body is neglected.
𝑃𝑃

• Expressions of the Stiffness or Spring Constant, 𝒌𝒌 𝑦𝑦

a. Axial Loading
𝐏𝐏 𝐿𝐿/2
𝐿𝐿
𝛿𝛿 𝑃𝑃 𝑃𝑃𝑃𝑃3 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝑘𝑘 = ∶ 𝑏𝑏𝑏𝑏𝑏𝑏 𝑦𝑦 = ; ∴ 𝒌𝒌 =
𝑦𝑦 192𝐸𝐸𝐸𝐸 𝑳𝑳𝟑𝟑
𝐿𝐿
Note:
Only three special bending cases are listed above.
Other bending cases are not included since they result in a
more complicated expression of beam deflection y, and
𝑃𝑃 𝑃𝑃𝑃𝑃 stiffness k.
𝑘𝑘 = : 𝑏𝑏𝑏𝑏𝑏𝑏 𝛿𝛿 =
𝛿𝛿 𝐴𝐴𝐴𝐴
𝑨𝑨𝑨𝑨
∴ 𝒌𝒌 =
𝑳𝑳

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 Machine Design 1: BASAEN, RV
Example 1: e. Cantilever beam, struck by a horizontally
Determine the maximum stress caused by the 20- moving body at a velocity equal to that of a)
lb rigid body in each of the following cases. The absorbing
body is a steel rod with 1” diameter and a length of 12".
The height of drop is ℎ = 4”. v
W
a. Straight rod, axially loaded

W
L
h = 4”

L = 12" steel rod 1" dia.

Solution:
a. 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 =?: Axial loading

b. Cantilever Beam, loaded at free end

c. Simply-supported Beam, loaded at midpoint

h b. 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 =?: Bending; cantilever beam loaded at free

end

𝐿𝐿/2
𝐿𝐿

d. Beam with fixed ends, loaded at midpoint

𝐿𝐿/2
𝐿𝐿

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 Machine Design 1: BASAEN, RV
c. 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 =? Bending; simply-supported beam loaded at
midpoint

Example 2:
A platform is to be designed to absorb the impact
from a 20-lb rigid body that may accidentally drop from a
height of 6” as shown. The platform has a width of 8” and
an unknown thickness “H”. If the platform material has E
= 20 x 106 psi and Sy =30 ksi, what thickness “H” should
be recommended? Use N=3, based on the yield point.
d. 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 =? Bending; beam with fixed ends, loaded at
midpoint W
h = 6 in.
H=?
B = 8"
L=5ft=60in.

Solution

e. 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 =? Bending; cantilever beam loaded at free end

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 Machine Design 1: BASAEN, RV

Example 3:
A beam of hollow circular section with D = 3” and
d = 1.5”, is made of an AISI C 1020, as rolled steel. It is
simply-supported at a length of 4 ft. A rigid body weighing
30 lb is to strike the beam horizontally at its midpoint, STRESS RAISER AND STRESS
resulting in a stress equal to the material’s yield point CONCENTRATION
value. Calculate this striking velocity v in kph.
Stress Raiser
Solution: Any discontinuity or change of section such as
holes, grooves, bends, scratches, notches, etc. is a stress
Simply-supported, raiser. Its presence in a loaded member results in a
hollow section beam concentration of stress or “localized stress” 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚, which is
greater than the average or nominal stress value, 𝑆𝑆𝑜𝑜.
If a loaded member has a stress raiser, the
maximum magnitude of stress becomes …
𝑣𝑣 =?

𝑺𝑺𝒎𝒎𝒎𝒎𝒎𝒎 = 𝑲𝑲𝒕𝒕 𝑺𝑺𝒐𝒐

Nominal or average
stress value
Theoretical stress
concentration factor
(SCF)
Maximum magnitude of
stress (at the location of
the stress raiser)

Where, Kt >1, for members affected by stress concentration

Note:
• The nominal stress 𝑆𝑆𝑜𝑜 can be any of the following
expressions
𝐹𝐹 𝑇𝑇𝑇𝑇 𝑀𝑀𝑀𝑀
� , , 𝑜𝑜𝑜𝑜 �
𝐴𝐴 𝐽𝐽 𝐼𝐼
• The value of the stress concentration factor depends on
both the stress raiser involved and the nature of
loading (i.e. 𝐾𝐾𝑡𝑡 for torsion is different from 𝐾𝐾𝑡𝑡 for
bending). Values of 𝐾𝐾𝑡𝑡 can be seen from Charts AF8
up to AF14 of DME by Faires. Other similar charts
from other sources give almost equal factors.

• The effect of a stress raiser is more pronounced if the

member is loaded with several repetitions (repeated

loading). Failure in such case is generally known as

“fatigue failure”.

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 Machine Design 1: BASAEN, RV
Comparison between loaded members with and
without a Stress Raiser
T
• Tension
d = 1” D = 1.5”
𝑟𝑟
𝐅𝐅𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑆𝑆𝑜𝑜 𝐅𝐅 𝐅𝐅 𝐅𝐅 r= 0.25"
ℎ ℎ 𝑆𝑆𝑜𝑜 𝐻𝐻 T

𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑆𝑆𝑜𝑜 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚

𝑏𝑏 Solution:
(Uniform bar)
𝑏𝑏
𝑭𝑭 𝑭𝑭
𝑺𝑺𝒎𝒎𝒎𝒎𝒎𝒎 = 𝑺𝑺𝒐𝒐 = = (Bar with a notch)
𝑨𝑨 𝒃𝒃𝒃𝒃
𝑭𝑭 𝑭𝑭
𝑺𝑺𝒎𝒎𝒎𝒎𝒎𝒎 = 𝑲𝑲𝒕𝒕 𝑺𝑺𝒐𝒐 = 𝑲𝑲𝒕𝒕 = 𝑲𝑲𝒕𝒕
𝑨𝑨 𝒃𝒃𝒃𝒃

• Bending

𝐌𝐌 𝐌𝐌 𝐌𝐌
𝐌𝐌
ℎ ℎ 𝑆𝑆𝑜𝑜 𝐻𝐻 a) Tension

𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑆𝑆𝑜𝑜 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚

(Uniform bar) 𝑏𝑏 𝑟𝑟 𝑏𝑏
𝑴𝑴𝑴𝑴 𝟔𝟔𝟔𝟔
𝑺𝑺𝒎𝒎𝒎𝒎𝒎𝒎 = 𝑺𝑺𝒐𝒐 = = (Bar with a notch)
𝑰𝑰 𝒃𝒃𝒉𝒉𝟐𝟐 𝑴𝑴𝑴𝑴 𝟔𝟔𝟔𝟔
𝑺𝑺𝒎𝒎𝒎𝒎𝒎𝒎 = 𝑲𝑲𝒕𝒕 𝑺𝑺𝒐𝒐 = 𝑲𝑲𝒕𝒕 = 𝑲𝑲𝒕𝒕 𝟐𝟐
𝑰𝑰 𝒃𝒃𝒉𝒉
b) Bending

• Torsion
𝐓𝐓 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑆𝑆𝑜𝑜

𝑑𝑑

𝐓𝐓
(Uniform shaft) c) Torsion
𝑻𝑻𝑻𝑻 𝟏𝟏𝟏𝟏𝟏𝟏
𝑺𝑺𝒎𝒎𝒎𝒎𝒎𝒎 = 𝑺𝑺𝒐𝒐 = =
𝑱𝑱 𝝅𝝅𝒅𝒅𝟑𝟑
𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚
𝑆𝑆𝑜𝑜
𝐓𝐓

𝐓𝐓 d) Compression
(Shaft with a notch or groove) )
𝑻𝑻𝑻𝑻 𝟏𝟏𝟏𝟏𝟏𝟏
𝑺𝑺𝒎𝒎𝒎𝒎𝒎𝒎 = 𝑲𝑲𝒕𝒕 𝑺𝑺𝒐𝒐 = 𝑲𝑲𝒕𝒕 = 𝑲𝑲𝒕𝒕
𝑱𝑱 𝝅𝝅𝒅𝒅𝟑𝟑

Example 1
The stepped shaft with given dimensions is
loaded with the following loads as specified. Determine
the maximum stress in each case.
a. 𝐹𝐹 = 1000 𝑙𝑙𝑙𝑙 (Tension)
b. 𝑀𝑀 = 2500 𝑖𝑖𝑖𝑖 − 𝑙𝑙𝑙𝑙 (Bending)
c. 𝑇𝑇 = 3500 𝑖𝑖𝑖𝑖 − 𝑙𝑙𝑙𝑙 (Torsion)
d. 𝐹𝐹 = 1000 𝑙𝑙𝑙𝑙 (Compression)

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 Machine Design 1: BASAEN, RV
Example 2:
Determine the stress concentration factor 𝐾𝐾𝑡𝑡 and the
maximum stress on each of the following members loaded
as shown.
a. Plate with a central hole

𝑏𝑏
c) 𝐾𝐾𝑡𝑡 , 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 =?
𝑭𝑭 𝑭𝑭=1500 lb
ℎ
ℎ = 2”
𝑡𝑡 = 0.25”
𝑑𝑑 𝑑𝑑 = 0.5”
𝑡𝑡 𝑏𝑏 = 2”

b. Shaft with a radial hole

𝑻𝑻 = 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 𝒍𝒍𝒍𝒍 − 𝒊𝒊𝒊𝒊

𝑑𝑑 = 0.25"

𝐷𝐷 = 1"

c. Plate with an eccentric hole

𝑴𝑴 𝑴𝑴 = 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒 𝒊𝒊𝒊𝒊 − 𝒍𝒍𝒍𝒍

ℎ = 2”
𝑏𝑏 𝑡𝑡 = 0.25”
𝑐𝑐 𝑑𝑑 = 0.5”
𝑒𝑒
ℎ 𝑏𝑏 = 1”
𝑒𝑒 = 0.5”

𝑑𝑑 𝑡𝑡
Solution:
a) 𝐾𝐾𝑡𝑡 , 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 =?

b) 𝐾𝐾𝑡𝑡 , 𝑆𝑆𝑚𝑚𝑚𝑚𝑚𝑚 =?

43