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Design 05

The document discusses dynamic loading and impact stresses. It defines impact loading as loading with a time less than half the natural period of vibration. Impact loading results in higher stresses than static loading. The maximum stress under impact σ' is related to the static stress σ by a factor that depends on the height h of impact and the static deflection δ. Similar equations are derived for stresses in bending and torsion. An example problem calculates the diameter and maximum deflection of a beam under impact loading.

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Neya Sebastian
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52 views8 pages

Design 05

The document discusses dynamic loading and impact stresses. It defines impact loading as loading with a time less than half the natural period of vibration. Impact loading results in higher stresses than static loading. The maximum stress under impact σ' is related to the static stress σ by a factor that depends on the height h of impact and the static deflection δ. Similar equations are derived for stresses in bending and torsion. An example problem calculates the diameter and maximum deflection of a beam under impact loading.

Uploaded by

Neya Sebastian
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Dynamic Loading

 Impact Loading
 Also called shock, sudden or impulsive load
 Ex: Driving a nail with a hammer, Automobile collisions, etc.
 Time of loading is less than half the natural period of vibration
 If the time of loading is greater than three times the natural period of
vibration, the load is classified as a static load
 A stress impact factor is normally used in design
( = 4.0 while designing automotive suspension parts)
 Parts subjected to impact are designed to absorb energy
 Statically loaded parts are designed to carry loads
Impact Stresses
 Consider a freely falling mass of weight W, impacting a structure
 The structure may be idealized as a spring since all structures
have some elasticity
 Assume –
 Mass of the structure is negligible
 Deflections within the falling mass are negligible
 Damping is negligible
 Change in potential energy = Strain energy stored

ẟ′
Impact Stresses (Contd...)
1 1
Energy equation W (h + δ ′) = Feq δ ′ = Aσ ′δ ′
2 2
δ′ = Maximum deflection; Feq = Equivalent force; σ′ = Maximum stress
W Feq
Define, σ = = σ st σ′=
A A

W Feq Feq σ′
Feq A
Then, = =k and δ′ =δ =δ =δ
δ δ′ W W A σ

ẟ′
Impact Stresses (Contd...)
1 1
From the energy equation, ie. from W (h + δ ′) = Feq δ ′ = Aσ ′δ ′
2 2
 σ ′
2σ  h + δ 
2W ( h + δ ′)  σ 
σ′= i.e. σ ′ =
Aδ ′ σ′
δ
σ
W
Where, δ = δ st = ; k = spring stiffness
k

 σ′  ẟ′
2h + δ 
σ′ σ
=  .......(1 )
σ σ′
δ
σ
Impact Stresses (Contd...)
 2h 
 Solving (1), σ ′ = σ 1 + 1 + 

 δ 
 2h 
Therefore, Feq = W 1 + 1 + 
 ẟ′
 δ 

 2h 
Also, δ ′ = δ 1 + 1 + 

 δ 

Wl W 2hEA 
For direct load , δ = σ ′ = 1 + 1 + 
Wl 
; So,
AE A
Impact Stresses in Bending

 Change in potential energy


= Strain energy stored
 This leads to similar kinds of
equations

 2h 
y ′ = y 1 + 1 + 

 y 

 2h 
σ ′ = σ 1 + 1 + 
 y 

Wl 3
y is the static deflection at the load y= for a cantilever beam
3EI
Impact Stresses in Torsion

 Change in potential energy = Strain energy stored

 2h 
θ ′ = θ 1 + 1 +  ; r is the moment arm of load W
 rθ 

 2h 
τ ′ = τ 1 + 1 + 
 rθ 
Example
A 10 N load drops from a height of 250 mm to impact the free
end of a horizontal cantilever beam 300 mm long. The beam has a
circular cross section. If the maximum allowable bending stress is
180 MPa, find the diameter of the beam. Also determine the
maximum deflection if the Young’s modulus of elasticity is 200 GPa.
 2h  Wl 3
σ ′ = σ 1 + 1 +  y= = Static_deflection
 y  3EI
 2h  d
y ′ = y 1 + 1 +  Mc
Wl
 y  σ= =
π 4
2 = Static_stress
I d
64

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