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