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### Transcript of class12 2005SImpact Print · PDF file Linear & Bending Impact Assumptions: •No mass...

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:

k m

π= 2

Absorbing energy

3 τ > t > 1/2 τ t < 1/2 τ

Design for

• Impact Factor

Su

Sy

Sy/Su

elongation

ksi % •Works favorably

•Promotes brittle fracture

In praxis: •Static properties known

■ x4 for automobile suspension parts

• Linear & Bending Impact

Assumptions:

•No mass of structure

•No deflection within impacting mass

•No damping

dynamic = static deflection curve only multiplied by impact factor

all energy goes into structure

most severe impact

W

W k k

h δ

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 δ Fe = k δ

k = W / δst Fe = W δ/δst

⎟ ⎟ ⎠

⎞ ⎜ ⎜ ⎝

⎛ ++=

st st

h211 δ

δδ

⎟ ⎟ ⎠

⎞ ⎜ ⎜ ⎝

⎛ ++=

st e

h211WF δ

Impact Factor

Impact Deflection …

⎟ ⎟ ⎠

⎞ ⎜ ⎜ ⎝

⎛ ++=

st ste

h211 δ

σσImpact Stress …

• Linear & Bending Impact W

W k k

h δ

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 δ Fe = k δ

k = W / δst Fe = W δ/δst

v2 = 2gh → h= v2/2g

h = 0 or v=0 suddenly applied load

Impact Factor = 2 (Double Safety Factor)

Impact Factor

For h >> δst

• Linear & Bending Impact W

W k k

h δ

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 δ Fe = k δ

k = W / δst Fe = 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 stress because 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 δ Fe = Se A

δ = Fe L / AE

…mass specific energy capacity

Se…elastic limit

• Bending Impact

Scan p285 u equations Scan 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

Improved Design

pilot

Reduce Stress Concentration and design for Uniform Stress Distribution

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

Improved Design

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]

• Torsional Impact

sheave

2400rpm

G=79GPa

ρ=2000kg/m3

suddenly stopped U=0.5 I ω2 …kin. energy

I=0.5 m r2 …moment of inertia of wheel m= 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 Nm U=25.72 Nm

T=τJ/r

• Exam

Monday April 18

Chapter 5: Elastic Strain Deflection Stability

Chapter 6: Failure Theories Safety 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