PMCE Structural Dynamics - Home - StructuresPro
Transcript of PMCE Structural Dynamics - Home - StructuresPro
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CE 809 - Structural Dynamics
Lecture 5: Response of SDF Systems to Impulse Loading
Semester – 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
• Impulsive forces, shock loads, short duration loads.
• Impact, blast wave, explosion
• 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
𝑝 : Step Impulse
𝑝𝑜
𝑇
𝑡No Load
Decayed
Oscillation
∆𝑡
Phase I: Loading Phase
Phase II: Free Vibration Phase
(𝑡)
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Response to Impulse Loading
Impulse Force Magnitude = 𝑝𝑜, start at 𝑡 = 0, Duration = Δ𝑡, where Δ𝑡/𝑇 << 1
Initial at-rest: 𝑢 = 0, 𝑢 = 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:
𝑝𝑜
𝑡∆𝑡
𝑝 𝑡
𝑡 ……….. (1)𝑢𝑝 =𝑝𝑜𝑘
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Response to Impulse Loading
This solution satisfies the equation of motion:
𝑢𝑝 = 0
𝑢𝑝 = 0
Substitute the above relations into the equation of
motion:
Phase 1:
0 + 0 + 𝑘𝑝0𝑘
= 𝑝0
𝑢𝑝 =𝑝𝑜𝑘
𝑡
𝑡
𝑡
𝑚 𝑢𝑝 + 𝑐 𝑢𝑝 + 𝑘 𝑢𝑝 = 𝑝0𝑡 𝑡 𝑡
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Response to Impulse Loading
A general solution:
𝑢 = 𝑢𝑝 + 𝑢ℎ
𝑢 =𝑝𝑜𝑘+ 𝑒−𝜉 𝜔 𝑡 𝐴 sin 𝜔𝐷𝑡 + 𝐵 cos 𝜔𝐷𝑡
Where
𝐴 and 𝐵 are determined such that the at-rest initial conditions are satisfied.
Phase 1:
……….. (2)
(𝑡) (𝑡) (𝑡)
𝑡
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Response to Impulse Loading
𝑢 =𝑝𝑜𝑘+ 𝐵 = 0
𝑢 = 𝜔𝐷𝐴 − 𝜉𝜔𝐵 = 0
So we obtain:
Phase 1:
𝐵 = −𝑝𝑜𝑘
𝐴 =𝜉𝜔𝐵
𝜔𝐷
……….. (3)
0
0
𝑢 =𝑝𝑜𝑘+ 𝑒−𝜉 𝜔 𝑡 −
𝜉 𝜔
𝜔𝐷
𝑝𝑜𝑘sin 𝜔𝐷𝑡 −
𝑝𝑜𝑘cos 𝜔𝐷𝑡
……….. (4)𝑡
(valid in the range 0 < 𝑡 < Δ𝑡)
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Response to Impulse Loading
Due to fact that 0 < 𝑡 < Δ𝑡 and Δ𝑡/𝑇 << 1 and 𝜉 << 1
𝑢 ≈𝑝𝑜𝑘−𝑝𝑜𝑘cos 𝜔𝑡 ≈
𝑝𝑜𝑘
1 − cos 𝜔𝑡
Phase 1:
……….. (5)
Equation (5) shows the approximate response during the loading phase.
𝑡
𝑢 =𝑝𝑜𝑘+ 𝑒−𝜉 𝜔 𝑡 −
𝜉 𝜔
𝜔𝐷
𝑝𝑜𝑘sin 𝜔𝐷𝑡 −
𝑝𝑜𝑘cos 𝜔𝐷𝑡
……….. (4)𝑡
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Response to Impulse Loading
Phase 2:
𝑡
𝑡 = ∆𝑡
𝑡 = 0Phase II
No Load: Free vibration response
depends on the initial conditions at
𝑡 = ∆𝑡
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Response to Impulse Loading
Phase 2:
𝑢 =𝑝𝑜𝑘
1 − cos 𝜔𝑡𝑡
𝑢 =𝑝𝑜𝑘
𝜔 sin 𝜔𝑡𝑡
𝑢 =𝑝𝑜𝑘
𝜔 sin 𝜔 Δ𝑡 ……….. (6)∆𝑡
𝑢 =𝑝𝑜𝑘
1 − cos 𝜔 Δ𝑡∆𝑡
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Response to Impulse Loading
Phase 2:
Employing Tylor’s expansion:
sin 𝜃 = 𝜃 −𝜃3
3!+𝜃5
5!……
cos 𝜃 = 1 −𝜃2
2!+𝜃4
4!……
For small 𝜃, by neglecting the second order term and higher terms, we get
sin 𝜃 ≈ 𝜃, cos 𝜃 ≈ 1 ……….. (7)
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Response to Impulse Loading
Phase 2:
Introducing this approximation in Equation (7) into Equation (6), we obtain
𝑢 =𝑝𝑜𝑘
1 − cos 𝜔 Δ𝑡 ≅ 0
𝑢 =𝑝𝑜𝑘
𝜔 sin 𝜔 Δ𝑡 ≅𝑝𝑜 𝜔
2 ∆𝑡
𝑘=𝑝𝑜 ∆𝑡
𝑚
Note that 𝑝𝑜 ∆𝑡 is an impulse.
Equation (8) says that impulse ≈ the change in momentum (of the mass)
The impulse introduces “momentum” into a structure but the duration of impulse is so
short that the displacement has not been developed yet.
……….. (8)
Δ𝑡
Δ𝑡
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Response to Impulse Loading
Using the Equation (8) as the initial conditions for free vibration in Phase 2,
𝑢 = 𝑒−𝜉 𝜔 (𝑡−∆𝑡) 𝑢
𝜔𝐷sin 𝜔𝐷 𝑡 − ∆𝑡
Since ∆𝑡/𝑡 << 1 , it is justified to let 𝑡 − ∆𝑡 ≈ 𝑡, that is
𝑢 ≈ 𝑒−𝜉 𝜔 𝑡𝑝𝑜 Δ𝑡
𝑚 𝜔𝐷sin(𝜔𝐷𝑡)
Equation (10) will be used when we analyze the response to arbitrary loading in the next
section.
Phase 2 (t > ∆𝒕 ):
……….. (9)
……….. (10)
𝑡∆𝑡
𝑡
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Response to Impulse Loading
𝑒−𝜉 𝜔 𝑡𝑝𝑜 Δ𝑡
𝑚 𝜔𝐷sin(𝜔𝐷𝑡)
𝑝𝑜
𝑡∆𝑡
Total Response
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