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10. ENERGY METHODS
h
m
10.3
IMPACT LOADING
(tvid- 10.3)
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10.3.1 Our Purpose in This Chapter is to calculate the stresses and deformations that occur on a solid object as a result of a weight falling
or being applied suddenly on it, and to compare it with the values in the case of static loading. The subject will be explained through examples.
Example 10.3.1
10.3.2 General Solution Logic: The dropped mass hits the other object with its initial energy and an impact force occurs. The work done by this force on the struck object turns into strain energy in the struck object. As a result, the initial energy of the dropped object is stored as deformation energy in the object it hits. Some energy may be lost during impact, but this is neglected.
An object of mass M, whose dimensions are negligible, is dropped from a height h onto a rod of mass m, elastic modulus E and cross-section A. Calculate the amount of deflection in the rod.
Energy Methods / Impact Loading
M
h
A
m, E
The potential energy of mass M in its initial position is:
Strain Energy in last position:
m, E
M
h
Deflection in axial loading..>>
(Same as equation 3.3.a)
If mass M is placed slowly on the rod, static loading occurs and the axial compressive force becomes equal to the weight force:
In this case..>>
If the loading was static:
(Dynamic deflection value):
(Static deflection value):
Figure 10.3.1
Figure 10.3.2
Figure 10.3.3
rod
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Energy Methods / Impact Loading
The positive one (the one with a + sign) should be taken from the roots found. Acoording to this:>>
Dynamic Multiplier :
If we consider the energy balance in Dynamic Loading (i.e. impact):
Amount of deflection as a result of impact:
If equations 10.3.21 and 10.3.22 are substituted in equation 10.3.26 ..>>
From equ.10.3.23:
If we calculate the roots of this last equation of the second degree:
If we rearrange the last equation :
(From equ.10.3.24) static deflection value:
Strength of Materials - Lecture Notes / Mehmet Zor
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Energy Methods / Impact Loading
If we expand the equation 10.3.29:
If 𝑀 is suddenly dropped from a height of zero
For forces in dynamic (i.e impact) and static loading:
For normal stresses in Dynamic and Static loading:
A Special Case :
M
A
m, E
Dynamic multiplier from equation 10.3.28:
from equ. 10.3.31 :
(The stress is twice as high as in the static situation)
from equ. 10.3. 29:
(The amount of deflection (shortening) is twice as much as in the static situation)
If the energy lost during the impact (𝑈lost) is not neglected:
The 𝜉 term comes into the dynamic multiplier equation. Namely:
𝜉 is calculated by the side equation:
If the mass of the struck object (i.e. the fixed rod) is negligible (𝑚≈0), 𝜉 = 1 is found from equation 10.3.33. Equations 10.3.32 and 10.3.28 become the same. This means that the lost energy 𝑈lost can be neglected. (𝑈lost ≈0)
If we divide both sides of equation 10.3.30 by area A:
Figure 10.3.4
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Example 10.3.2
h
L
M
A
B
a
b
A mass M, whose dimensions are negligible, is dropped from a height h onto the right free end of the beam of mass 2kg, whose left end is fixed. Calculate the maximum stress occurring in the beam a-) neglecting the beam mass, b-) without neglecting the beam mass.
(M=20kg, m=2kg, L=2m, h=20cm, a=4cm, b=8cm, E=200GPa)
Energy Methods / Impact Loading
m
E
If there was a static loading State:
Deflection of end B :
(Found in example 3.2.1.)
x
y
z
(Since the bending moment is in the z direction)
Solution:
(If M is slowly placed at point B):
M
B
x
y
L
A
Since the dimensions of the mass M can be neglected, we can think of the force Ps as acting on the end point B.
Dynamic Loading State
(If mass M is dropped from height h):
Dynamic multiplier: From Eq. 10.3.28
a-) If m is neglected, 𝜉=1 and energy is conserved.
b-) If m is not neglected, energy loss is taken into account.
From equ.10.3.33 :
From equ. 10.3.32:
From equ.10.1.1:
The maximum stress occurs at the points (along the upper line) with y coordinate b/2 in the cross section at end A:
x
Figure 10.3.5
(a)
Figure 10.3.7
(b)
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Example10.3.3
Objects of mass M are dropped from the same height h onto the non-stepped poles I and the stepped poles number 2. The poles are made of the same material and their mass can be neglected. Accordingly, calculate the maximum stresses occurring in the poles.
L=2m, h=20cm, L1=1m, L2=1m, M=50kg, A1=10cm2, A2=50cm2,
For poles: E=80GPa
Energy Methods / Impact Loading
M
h
A1
L
I
M
h
A2
A1
L1
L2
II
Solution:
For pole no. I
If there was static loading
Dynamic Loading:
If we calculate the amount of static deflection:
Static loading stress:
Dynamic multiplier :
Dynamic loading stress:
Since the masses of the poles are neglected, energy is conserved (𝜉=1).
If mass M was placed slowly on pole I, static loading would occur.
Mass M is dropped from height h.
M
M
h
(from equ. 10.3.31)
Dynamic loading deflection:
Figure 10.3.8
Figure 10.3.9
Figure 10.3.10
Figure 10.3.11
Strength of Materials - Lecture Notes / Mehmet Zor
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L=2m, h=20cm, L1=1m, L2=1m, M=50kg, A1=10cm2, A2=50cm2, direkler için E=80GPa
Energy Methods / Impact Loading
M
h
A2
A1
L1
L2
II
For pole no. II (We divide the pole into two parts.)
The total static deflection is the sum of the deflections of each region.
For static loading max stress:
(Since the internal force is the same in the regions, the maximum stress occurs in region 1, which has a small cross-section..)
Dynamic multiplier :
Dynamic loading stress:
M
1
M
1
2
Static loading:
1
2
Dynamic loading:
Dynamic loading deflection::
Figure 10.3.12
Figure 10.3.13
Figure 10.3.14
Strength of Materials - Lecture Notes / Mehmet Zor
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If an 80 kg mass is dropped from a height of 10 mm into the middle of a steel beam, calculate the location and intensity of the maximum normal stress that will occur in the beam. (For steel: E=200GPa).
Answer: in the middle cross-section, 143.5MPa
Question 10.3.4
A load of 600kN is suddenly dropped from a height h = 0 on an Aluminum pipe whose bottom part is fixed to the ground. Find the amount of deflection occurring in the aluminum pipe. Ignore the energy loss during impact. (EAl= 70GPa)
Answer: 1.19 mm
Question 10.3.5
Questions with Answers
Energy Methods / Impact Loading
Figure 10.3.15
Figure 10.3.16