Impact

also called shock, sudden or impulsive loading

■driving a nail with a hammer, automobile collisions ….

dashpot

a) Rapidly moving b) Suddenly applied c) Direct Impact

■ vehicles crossing ■ Combustion in ■ pile drive,
a bridge engine cylinder ■ drop forge,
■ car crash
To distinguish:
Compare Time of Load Application t with
m
Natural Period of Vibration τ = 2π
k
Static load “Gray area” Impact load
t >3τ 3 τ > t > 1/2 τ t < 1/2 τ
Carrying Absorbing
Design for
loads energy
 Impact
material properties vary with loading speed

ksi Sy/Su %
•Works favorably

•Promotes
brittle fracture
Su

elongation

Sy

In praxis:
•Static properties •Loading rate
Impact Factor
known only approximated
■ x4 for automobile suspension parts
 Linear & Bending Impact
Elastic-strain energy
W stored in structure
Guide rod
= 0.5Feδ
h Fe
δ Work of falling weight
W = W(h+δ)
W
k k
δst δ h
Assumptions:
•No mass dynamic = static deflection curve
of structure only multiplied by impact factor
•No deflection all energy goes into structure
within impacting mass
•No damping most severe impact

W (h + δ) = 0.5 Fe δ
Fe = k δ
k = W / δst
Fe = W δ/δst
⎛ 2h ⎞
Impact Deflection … δ = δ st ⎜ 1 + 1+ ⎟
⎜ δ st ⎟
⎝ ⎠
⎛ 2h ⎞ Impact Factor
Impact Load F
… e = W ⎜1 + 1+ ⎟
⎜ δ st ⎟
⎝ ⎠
⎛ 2h ⎞
Impact Stress … σ e = σ st ⎜ 1 + 1 + ⎟
⎜ δ st ⎟
⎝ ⎠
Linear & Bending Impact
Elastic-strain energy
W stored in structure
Guide rod
= 0.5Feδ
h Fe
δ Work of falling weight
W = W(h+δ)
W
k k
δst δ h
W (h + δ) = 0.5 Fe δ
Fe = k δ
k = W / δst
Fe = W δ/δst For h >> δst

v2 = 2gh → h= v2/2g

h = 0 or v=0
suddenly applied load
Impact Factor = 2 Impact Factor
(Double Safety Factor)
Linear & Bending Impact
Elastic-strain energy
W stored in structure
Guide rod
= 0.5Feδ
h Fe
δ Work of falling weight
W = W(h+δ)
W
k k
δst δ h
W (h + δ) = 0.5 Fe δ
Fe = k δ For h >> δst
k = W / δst
Fe = W δ/δst

Vertical movement

Impact Factor

W = k δst = g m
Impact kin. Energy … U = 0.5 m v2

Horizontal movement

As greater Stiffness and kin. Energy

as greater Equivalent static force
Linear Impact of Straight Bar

σ = Fe / A k=AE/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
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
δ = Fe L / AE
U = 0.5 Fe δ
Fe = Se A

A
…mass specific
bumper 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
pilot

Improved
Design

Reduce Stress Concentration and

design for Uniform Stress Distribution

< Head/Pilot

Shank
E f f e ct o f St r e ss R a i se r s o n Im p a ct

pilot

Improved
Design

Reduce Stress Concentration and

design for Uniform Stress Distribution Raiser

pilot Axial hole

 Torsional Impact
Linear Torsional
δ…deflection [m] θ…deflection [rad]
Fe…eq. static force [N] Te…eq. static Torque [Nm]
m…mass [kg] I…moment of inertia [kgm2]
k…spring rate [N/m] K…spring rate [Nm/rad]
v…velocity [m/s] ω…velocity [rad/s]
U…kin. energy [Nm] U…kin. energy [Nm]
 Torsional Impact
ρ=2000kg/m3
sheave

G=79GPa

2400rpm

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

5.1-2, 5.5-7, Deflection
5.10-12
Stability
Chapter 6: Failure Theories
6.1-2, 6.5-8, Safety Factors
6.10-12
Chapter 7: Impact
7.1-2, 7.4

Chapter 8: Fatigue
8.1-11