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Impact Load Factors

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Description

Rice University

Impact Load Factors for Static Analysis Often a designer has a mass,

hitting an object and thereby causing a suddenly applied impact load

Rather than conduct a dynamic analysis an amplified static analysis is used,

at least for the preliminary design

In a static stress analysis the static force (or weight of the mass) must be increased by an “Impact Factor” so as to obtain a good approximation of the maximum dynamic deflection and stress

For hard elastic bodies the Impact Factor,

is always greater than or equal to two

It is not unusual for the vertical Impact Factor to be more than 10,

The following figure shows an amplified static load,

used to estimate the maximum deflection and stress in a beam due to the dynamic response of a cantilever beam having a weight,

A static force P approximates the dynamic effects of a falling weight W

In the study of the mechanics of solids,

an energy balance approximation is used to estimate the required static load

That approximation assumes that all of the kinetic energy of the moving mass is converted,

If you assume that no noise,

and neglect the mass of the struck member then

? = 1 and the collision is 100% efficient

There are several handbook equations that provide corrections to the elementary theory based on the ratio of the striking mass to the member mass

Those corrections seldom give an efficiency of less than

In the case of a weight dropped vertically from a height,

? is the deflection of the member due to a static force (W) applied at the impact point in the impact direction

even if h = 0 this factor is equal to two

Similarly,

a mass moving horizontally with a velocity of v has an horizontal Impact Factor of

where g is the acceleration of gravity

There are handbook equations for the static deflections of bars and beams that can be used to define Page 1 of 13

Rice University the corresponding member stiffness,

In general the static deflection is

where k is the stiffness of the member,

in the direction of the impact

For any object,

a finite element static stress analysis can be probed to obtain the static deflection,

in the direction of the force,

at the point of loading to obtain the value of

the mechanics of materials solutions,

and a TK Solver worksheet will all be used to estimate the Impact Factor for example bars and beams

The first example (solved by all three methods) is for a weight dropped on the middle of a simply supported beam of rectangular cross‐ section

The mass of 80 kg is dropped vertically 10 mm from rest onto the center of a horizontal simply supported beam of length 1

The beam is steel (E = 200 GPa) with a rectangular cross‐section area having a width of 60 mm and a height of 30 mm

The first example causes an Impact Factor of about 6

The second example will yield an Impact Factor above 1,200

SolidWorks Finite Element Simulation: The SolidWorks model begins the first example with the cross‐section,

extrudes it to the beam length,

and adds vertical split lines to the solid to allow loading at the middle and/or quarter points,

and to recover the deflection and stresses there

Generating the symmetric beam member,

The beam member is pinned at the right and supported with rollers on the left

The static weight (W = 784

Static force applied vertically at beam center

After assigning the material properties,

the static deflection at the point of loading is obtained:

Page 2 of 13

Rice University

The static deflection at the load point

Its value,

001049 mm,

is substituted into the above vertical factor equation,

010)1 = 5

so the actual static force needs to be

48 (784

Applying that load gives a maximum dynamic horizontal fiber stress estimate of about SX = 143 MPa,

Maximum horizontal stress (143 MPa) is at the center plane top and bottom

Of course,

since the static model is linear there is no need to re-run the model (unless you want pretty plots)

The static maximum fiber stress (SX) of 26

Page 3 of 13

Rice University TK Solver Model The TK Solver worksheet begins by writing a set of rules (from mechanics of solids and particle dynamics)

The basic TK Solver rules for impact forces

TK rules for five common member structural stiffnesses

The initial input variables (below) at the first solution (the first click on the TK lightbulb icon) were incomplete since neither the static deflection nor member stiffness was known

several rules were not satisfied as indicated by the lack of values in the Output column

Page 4 of 13

Rice University

Initial velocity results,

without the beam stiffness However,

the stiffness of a simply supported beam at the center (load) point was calculated in the second solve (Lightbulb icon pick) from additional rules by inputting the beam properties,

Additional output of the beam stiffness for input for the member stiffness Page 5 of 13

Rice University There were enough inputs provided to calculate the stiffness of three of five typical members including the simply supported beam specified in this example (as seen above)

Since the weight is dropped on the center of a simple beam,

the value of that beam stiffness (k = 7

and pasted into the input column of k

the next solve gives the desired Impact Factor values for both a horizontal or vertical impact

Inputting the member stiffness at the load point gives the impact factors

Now the static deflection (

?) matches the finite element study well,

as does the vertical impact factor of about 5

(Had the weight been held at the beam surface without toughing it (h = 0) and then released that calculation gives the vertical Impact Factor equal to two,

the vertical value was input as the impact factor and the next solve gave the equivalent force value as

and the dynamic deflection was estimated as about

The final solve gives the equivalent static force (about 4,300 N)

The following figure is an image of a hand solution example taken from the “Mechanics of Materials” text by A

Pytel and J

Kiusalaas,

Thomson,

Those results match the finite element solution,

Page 6 of 13

Rice University

Page 7 of 13

Rice University

The second example models a 1 kg mass moving horizontally along a steel (E = 207 GPa) bar 100 mm long with a 10 mm diameter and striking the end of the bar with a velocity of 1 m/sec

Norton,

Machine Design,

Prentice Hall,

) It is assumed that the correction for the mass of the struck bar reduces the efficiency to η = 0

These values were input in the first solve,

but the impact factor cannot be obtained without a member stiffness

So the first solution was to actually find the stiffness of the axial bar,

so it could be copied to the member stiffness input

First solve mainly finds the bar member stiffness to be copied

Page 8 of 13

Rice University Having the member stiffness,

the impact factors are obtained along with a value of the “static deflection”

However,

the latter (vertical load) value is only for reference since the impact is horizontal

The member stiffness yields the horizontal factor to copy

An impact factor defines the equivalent static force and the dynamic displacement This is a very large impact factor

However,

the dynamic deflection still leads to a small elastic strain:

energy loss due to plastic action does not have to be estimated

However,

there are corrections published to account for the mass of the bar

Typical additional rules to correct the efficiency for the mass of the member are listed below

It is well known that for bars of a constant diameter the Impact Factor decreases rapidly with increased bar length

The equations for the Impact Factors neglect the mass of the member being struck

There are handbook equations that reduce the efficiency to less than unity by including the mass of the struck member

Some of those corrects are also provided as rules,

but they rarely reduce the efficiency below 0

In this case,

the bar mass correction only reduces efficiency to η = 0

Rice University as shown in the figures below

Correction rules for the member mass

Efficiency correction due to member mass

The member mass correction has little effect on the maximum dynamic displacement For a third example the effects of dropping an 80 kg mass from different heights onto the tip of a cantilever beam of 1,200 mm length will be illustrated

The stiffness of the cantilever is 46,875 N/m

Using that member stiffness and drop heights of 0

and 1 meter yields vertical impact factors of 2

respectively as seen in the following outputs

Page 10 of 13

Rice University

Cantilever beam Impact Factor values for drop heights of 0

Hibbeler,

Mechanics of Materials,

Prentice Hall,

An 80 Mg railway car moves horizontally with a speed of 0

The steel cantilever post is 200 mm by 200 mm in cross-section and the length of the post,

Determine the maximum dynamic displacement of the post tip and the equivalent static force for determining the post stresses

The first solve is just used to find the bending inertia of the post for two common shapes

Cross-section inertia values

Page 11 of 13

Rice University The rectangular inertia value is copied into the beam member inertia so that typical beam and bar stiffnesses are evaluated,

The cantilever stiffness value is copied into the member stiffness input so the third solve yields the horizontal impact factor of about 0

Input a member inertia and obtain stiffness for a cantilever beam

The member stiffness yields the horizontal impact factor to be copied

The final solve gives the dynamic displacement as about δmax = 11

There are enough data to obtain the correction factor for the post mass but it gives (below) almost complete efficiency,

so there is no need to re-run this set of calculations

Page 12 of 13

Rice University

The final horizontal equivalent static force

A correction for the member mass is not required

Page 13 of 13