mechanical physics lab / project

Impact Loading

Ref: Roymechx.co.uk at http://www.roymech.co.uk/index3.htm#Tables

Generally when the strength of machine elements are considered it is assumed that the loading is

static or applied gradually. This loading condition is often not the case, the loading may be

cyclic requiring assessment for fatigue. Fatigue or it may involve impact or suddenly applied

loads. When loads are applied suddenly and when the loads are applied as impact loads the

resulting transient stresses (and deformations) induced in the machine elements are much higher

than if the loads are applied gradually. This effect is shown in the diagram below

It is normal practice to design machines such that impact loads are eliminated or reduced by

inclusion of shock absorbers. Inclusion of low cost, mass produced, shock absorbers can

virtually eliminate the increased stresses and deformations resulting from impact loads.

Most ductile materials have strength properties which are a function of the loading speed. The

more rapid the loading the higher the tensile and ultimate strengths of the materials. . Two

standard tests, the Charpy and Izod, measure the impact energy (the energy required to fracture a

test piece under an impact load), also called the notch toughness.

The detailed assessment of the strength of machine elements under impact loading regimes

involves use of advanced techniques including Finite Element Analysis. Impact loads result in

shock waves propogating through the elements with possible serious consequences. It is possible

to complete a relatively simply stress evaluation for suddenly applied and impact loads by using

the principle of conservation of energy and conditional that the materials considered are

operating within their elastic regions.

The equations are most accurate for relatively heavy impacting masses moving at low velocities

on impact.

Notation

 A = Area (m 2 )

 E = Modulus of Elasticity

(N/m 2 )

 h = Drop distance (m)

 V = Velocity (m/s)

 w = Specific Weight (kg/m 3 )

 σ = stress (N/m 2 )

 σ stat = stress resulting from static

 k =Stiffness (N/m)

 M = Mass-moving body (kg)

 M1 = Mass of static bar or beam

(kg)

 K = Factor to allow for energy

loss at impact

 l = length of bar (l)

 tp = time period (s)

 v = velocity (m/s)

 W = Weight - moving body(N)

load(N/m 2 )

 δ = deflection (m)

 δ stat = deflection resulting from

static load(m)

 μ = Ratio Moving

Mass/Stationary bar

 β = Constant = A Sqrt(wEg/W) -

see text

Linear Impact deflections and stresses (gravity loads)

Important note: The notes below represent a very simple view of the loading condition and do

not consider more real case involving shock waves being propagated through the loaded member

or the moving mass

Consider a loading regime as shown below with a ring of Mass M(kg) with weight W= M.g(N)

being dropped through a distance h onto a collar supported by a vertical bar which behaves as a

spring with a stiffness of k (N/m).

The support bar has a length l (m), an Area A (m 2 ) With a modulus of elasticity E (N/m

2 )

In practice the weight would impact onto the support which would elastically deform until all of

the potential energy has been absorbed. The support would then contract initiating damped

oscillations until the system assumes a stable static position. The equations below determine the

initial maximum deformation which provides the most highly stressed condition.

In accordance with conservation of energy the potential energy of the weight is converted to

elastic strain energy.

This may be expressed as a quadratic equation

This is solved for the maximum deflection δmax as follows

The weight applied gradually would result in a deflection δst thus

Note:

The stiffness k = Force /Deflection = F /δ: E = stress /strain = σ /e = (F /A) /(δ/l) Therefore k

= EA/l

Substituting this into the equation for δmax results in

This can be expressed as

The resultant maximum force is simply

Pmax = k δmax

and the resultant maximum stress =

σmax = Pmax/ A

This may be expressed as

For the calculation of the stress due to a suddenly applied load with h = 0

σmax = 2 σst

Linear impact deflections and stresses (kinetic impacts )

Important note: The notes below represent a very simple view of the loading condition and do

not consider more real case involving shock waves being propagated through the loaded member

or the moving mass

Impact loads based pimarily on kinetic energy e.g horizontal impacts are treated slightly

differently. For these applications the kinetic energy is converted into stored energy due to

elasticity of the resisting element.

Consider a Mass M(kg) with a velocity of v impacting on a collar which is supported by a bar

with a stiffness of K (N/m) - Ignoring gravitational forces.

The kinetic energy of the mass Mv 2 /2 is transformed into stored energy in the support.

The resulting equation is

The resultant maximum deflection equals..

Noting that the static deflection = Wl/AE this equation can be written

The equivalent maximum stress =

Noting that the static deflection = Wl/AE this equation can be written

Beams

Using similar principles as expressed above it can be easily proved using principles of

conservation of energy that.

and

If the impact is horizontal instead of vertical . The resulting deformation and stress resulting

from the impact are

Note: The above approximate relationships can been applied generally to most structural systems

subject to distortion with the elastic range when subject to impact loading

Energy losses on impact

The above equations are very approximate and include many assumptions. A very important

assumption is that all of the energy ( based on h or v 2 ) is used up in producing the same

distortion as would result from static loading. In reality, some kinetic energy is lost in internal

friction. Account can be taken for these losses by multiplying h or v 2 by an appropriate factor

K. This factor is derived from the Mass of the moving body (M) and the mass of the beam or bar

(M1). The factor is different for different loading systems ass follows.

In the equations above h or v 2 would be replaced by K.h or K v

2

The K factor is nearest unity when M is large compared to M1 . As an example for the axial

impact

if M1/M is say 0,1 then K = 0,95..

if M1/M is say 10 then K = 0,15..

This is illustrated in the figure below

1) A moving mass M striking axially one end of a bar of mass M1. The other end of the bar being

fixed...

2) A moving mass M striking traversely the end of a beam of mass M1...

3) A moving mass M striking traversely the center of a beam with simple supported ends of mass

M1....

4) A moving mass M striking traversely the center of a beam with fixed ends of mass M1. ..

Impact stresses considering propagation of shock waves- Unsupported bar

When a impact force is suddenly applied to and elastic body , a wave of stress is propogated

traveling through the body with a velocity..

w = weight /unit volume (kg/m 2 ), v = velocity (m/s)

The unsupported bar subject to the longitudonal impact from a rigid body with velocity v

experiences a wave of compressive stress of intensity σ.

If the mass of the moving body is very large compared to the mass of the bar the wave of

compression bounces back from the far end of the bar as a wave of tension and returns to the

struck end after a time period .

t p = 2L / V

If the mass of the moving body is very large compared to the bar so that it can be considered

infinite then after breaking contact the moving bar will move away from the impacting mass with

a velocity of vb= 2v. The moving bar will be stress free.

If the mass of the impinging body is μ time the mass of the bar then the bar will move away with

an average velocity of

The moving bar is left vibrating with a stress intensity of

Impacts considering propagation of shock waves- bar with one end supported

For the case of a bar with one end fixed , the wave of compressive stress resulting from the

impact on the unsupported end is reflected back unchanged from the supported (fixed end) and

combines with the advancing waves to produce a maximum stress approximately equal to..

μ = Mass of Moving Body/Mass of Bar

Finite Life for Aluminum-1.pdf