class12 2005SImpact Print · Linear & Bending Impact Assumptions: •No mass of structure •No...
Transcript of class12 2005SImpact Print · Linear & Bending Impact Assumptions: •No mass of structure •No...
also called shock, sudden or impulsive loadingImpact
Compare Time of Load Application t with Natural Period of Vibration τ
a) Rapidly moving
dashpot
■driving a nail with a hammer, automobile collisions ….
b) Suddenly applied c) Direct Impact■Combustion in
engine cylinder■ vehicles crossing
a bridge■ pile drive,■ drop forge,■ car crash
To distinguish:
km
π= 2
Carrying loads
Absorbing energy
t > 3 τStatic load “Gray area” Impact load
3 τ > t > 1/2 τ t < 1/2 τ
Design for
Impact Factor
material properties vary with loading speed Impact
Su
Sy
Sy/Su
elongation
ksi %•Works favorably
•Promotes brittle fracture
•Loading rate only approximated
In praxis:•Static propertiesknown
■ x4 for automobile suspension parts
Linear & Bending Impact
Assumptions:
•No massof structure
•No deflectionwithin impacting mass
•No damping
dynamic = static deflection curveonly multiplied by impact factor
all energy goes into structure
most severe impact
W
Wk k
hδ
Fe
W
Guide rod
δst δ h
Elastic-strain energy stored in structure= 0.5Feδ
Work of falling weight = W(h+δ)
W (h + δ) = 0.5 Fe δFe = k δ
k = W / δstFe = W δ/δst
⎟⎟⎠
⎞⎜⎜⎝
⎛++=
stst
h211δ
δδ
⎟⎟⎠
⎞⎜⎜⎝
⎛++=
ste
h211WFδ
Impact Factor
Impact Deflection …
Impact Load …
⎟⎟⎠
⎞⎜⎜⎝
⎛++=
stste
h211δ
σσImpact Stress …
Linear & Bending ImpactW
Wk k
hδ
Fe
W
Guide rod
δst δ h
Elastic-strain energy stored in structure= 0.5Feδ
Work of falling weight = W(h+δ)
W (h + δ) = 0.5 Fe δFe = k δ
k = W / δstFe = W δ/δst
v2 = 2gh → h= v2/2g
h = 0 or v=0suddenly applied load
Impact Factor = 2 (Double Safety Factor)
Impact Factor
For h >> δst
Linear & Bending ImpactW
Wk k
hδ
Fe
W
Guide rod
δst δ h
Elastic-strain energy stored in structure= 0.5Feδ
Work of falling weight = W(h+δ)
W (h + δ) = 0.5 Fe δFe = k δ
k = W / δstFe = W δ/δst
Impact Factor
For h >> δst
Vertical movement
Horizontal movement
W = k δst = g m
Impact kin. Energy … U = 0.5 m v2
As greater Stiffness and kin. Energy as greater Equivalent static force
Linear Impact of Straight Bar
σ = Fe / A k = A E / L
Impact Energy Capacity of a straight rod
• Basic relationship → Gives optimistic results
• Calculates stress lower than actual stressbecause of- Non-uniform loading & stress distribution- Mass of the impacted rod
σ
Linear Impact of Straight Bar
How compare Energy-Absorbing Capacaties ?
Set δ = Sy
But Vb = 2½ Va
…mass specific energy capacity
Shows Importance of Section Uniformity
Impact Absorption Capacity
A
bumper
U = 0.5 Fe δFe = Se A
δ = Fe L / AE
…mass specific energy capacity
Se…elastic limit
Bending Impact
Scan p285 u equationsScan p286 o equations
Effect of Stress Raisers on Impact
Ratio 2.25 : 1 : 1/64
225% : 100% : 1.56%
Effect of Stress Raisers on Impact
Due to combination of Impact and Stress Raiser Brittle fracture is promoted
with stress concentration factor almost equal Kt
Effect of Stress Raisers on Impact
ImprovedDesign
pilot
Reduce Stress Concentration and design for Uniform Stress Distribution
< Head/Pilot
Shank
E f f e c t o f S t r e s s R a i s e r s o n I m p a c t
ImprovedDesign
pilot
Reduce Stress Concentration and design for Uniform Stress Distribution Raiser
Axial holepilot
Torsional Impact
Te…eq. static Torque [Nm]Fe…eq. static force [N]
U…kin. energy [Nm]U…kin. energy [Nm]ω…velocity [rad/s]v…velocity [m/s]K…spring rate [Nm/rad]k…spring rate [N/m]I…moment of inertia [kgm2]m…mass [kg]
θ…deflection [rad]δ…deflection [m]TorsionalLinear
Torsional Impact
sheave
2400rpm
G=79GPa
ρ=2000kg/m3
suddenly stoppedU=0.5 I ω2 …kin. energy
I=0.5 m r2 …moment of inertia of wheelm= π r2 b ρ …mass of wheel
U=0.25 π r4 b ρ ω2
U=0.25 π (0.060)4 (0.002) (2000) (2400x2π/60)2 NmU=25.72 Nm
T=τJ/r
Exam
Monday April 18
Chapter 5: Elastic Strain Deflection Stability
Chapter 6: Failure TheoriesSafety Factors
Chapter 7: Impact
Chapter 8: Fatigue
5.1-2, 5.5-7, 5.10-12
6.1-2, 6.5-8, 6.10-12
7.1-2, 7.4
8.1-11