CE 809 - Structural Dynamics
Lecture 5: Response of SDF Systems to Impulse Loading
Semester β Fall 2020
Dr. Fawad A. Najam
Department of Structural Engineering
NUST Institute of Civil Engineering (NICE)
National University of Sciences and Technology (NUST)
H-12 Islamabad, Pakistan
Cell: 92-334-5192533, Email: [email protected]
Prof. Dr. Pennung Warnitchai
Head, Department of Civil and Infrastructure Engineering
School of Engineering and Technology (SET)
Asian Institute of Technology (AIT)
Bangkok, Thailand
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Impulse Loading
β’ Impulsive forces, shock loads, short duration loads.
β’ Impact, blast wave, explosion
β’ Sudden stop/break of trucks & automobiles & travelling cranes
The study of impulse response is also important for the analysis of response to arbitrary
loadings.
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Response to Impulse Loading
Response
π : Step Impulse
ππ
π
π‘No Load
Decayed
Oscillation
βπ‘
Phase I: Loading Phase
Phase II: Free Vibration Phase
(π‘)
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Response to Impulse Loading
Impulse Force Magnitude = ππ, start at π‘ = 0, Duration = Ξπ‘, where Ξπ‘/π << 1
Initial at-rest: π’ = 0, π’ = 0Structure (0) (0)
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Response to Impulse Loading
Phase 1:
The particular solution to a step
loading is simply a static deflection:
ππ
π‘βπ‘
π π‘
π‘ β¦β¦β¦.. (1)π’π =πππ
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Response to Impulse Loading
This solution satisfies the equation of motion:
π’π = 0
π’π = 0
Substitute the above relations into the equation of
motion:
Phase 1:
0 + 0 + ππ0π
= π0
π’π =πππ
π‘
π‘
π‘
π π’π + π π’π + π π’π = π0π‘ π‘ π‘
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Response to Impulse Loading
A general solution:
π’ = π’π + π’β
π’ =πππ+ πβπ π π‘ π΄ sin ππ·π‘ + π΅ cos ππ·π‘
Where
π΄ and π΅ are determined such that the at-rest initial conditions are satisfied.
Phase 1:
β¦β¦β¦.. (2)
(π‘) (π‘) (π‘)
π‘
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Response to Impulse Loading
π’ =πππ+ π΅ = 0
π’ = ππ·π΄ β πππ΅ = 0
So we obtain:
Phase 1:
π΅ = βπππ
π΄ =πππ΅
ππ·
β¦β¦β¦.. (3)
0
0
π’ =πππ+ πβπ π π‘ β
π π
ππ·
πππsin ππ·π‘ β
πππcos ππ·π‘
β¦β¦β¦.. (4)π‘
(valid in the range 0 < π‘ < Ξπ‘)
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Response to Impulse Loading
Due to fact that 0 < π‘ < Ξπ‘ and Ξπ‘/π << 1 and π << 1
π’ βπππβπππcos ππ‘ β
πππ
1 β cos ππ‘
Phase 1:
β¦β¦β¦.. (5)
Equation (5) shows the approximate response during the loading phase.
π‘
π’ =πππ+ πβπ π π‘ β
π π
ππ·
πππsin ππ·π‘ β
πππcos ππ·π‘
β¦β¦β¦.. (4)π‘
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Response to Impulse Loading
Phase 2:
π‘
π‘ = βπ‘
π‘ = 0Phase II
No Load: Free vibration response
depends on the initial conditions at
π‘ = βπ‘
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Response to Impulse Loading
Phase 2:
π’ =πππ
1 β cos ππ‘π‘
π’ =πππ
π sin ππ‘π‘
π’ =πππ
π sin π Ξπ‘ β¦β¦β¦.. (6)βπ‘
π’ =πππ
1 β cos π Ξπ‘βπ‘
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Response to Impulse Loading
Phase 2:
Employing Tylorβs expansion:
sin π = π βπ3
3!+π5
5!β¦β¦
cos π = 1 βπ2
2!+π4
4!β¦β¦
For small π, by neglecting the second order term and higher terms, we get
sin π β π, cos π β 1 β¦β¦β¦.. (7)
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Response to Impulse Loading
Phase 2:
Introducing this approximation in Equation (7) into Equation (6), we obtain
π’ =πππ
1 β cos π Ξπ‘ β 0
π’ =πππ
π sin π Ξπ‘ β ππ π
2 βπ‘
π=ππ βπ‘
π
Note that ππ βπ‘ is an impulse.
Equation (8) says that impulse β the change in momentum (of the mass)
The impulse introduces βmomentumβ into a structure but the duration of impulse is so
short that the displacement has not been developed yet.
β¦β¦β¦.. (8)
Ξπ‘
Ξπ‘
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Response to Impulse Loading
Using the Equation (8) as the initial conditions for free vibration in Phase 2,
π’ = πβπ π (π‘ββπ‘) π’
ππ·sin ππ· π‘ β βπ‘
Since βπ‘/π‘ << 1 , it is justified to let π‘ β βπ‘ β π‘, that is
π’ β πβπ π π‘ππ Ξπ‘
π ππ·sin(ππ·π‘)
Equation (10) will be used when we analyze the response to arbitrary loading in the next
section.
Phase 2 (t > βπ ):
β¦β¦β¦.. (9)
β¦β¦β¦.. (10)
π‘βπ‘
π‘
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Response to Impulse Loading
πβπ π π‘ππ Ξπ‘
π ππ·sin(ππ·π‘)
ππ
π‘βπ‘
Total Response
Thank you