blast course work

8/9/2019 Blast Course Work

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Blast Loading Capacity of a OneWay Spanning Carbon Fibre

Reinforced Plastic Beam

Introduction

This report will investigate the blast loading capability of a simple supported one way spanning carbon

fibre reinforced plastic panel. Units ofKPa (kN / m2), kN, kg, meters, and secondswill be used thru the

report.

The properties for this material are:

Yield Strength Shear Strength Density Elastic Modulus

205000 KPa 118000 KPa 1612 kg/m3 111 x 10

6 KPa

Carbon Fibre Reinforced Plastic is a high yield stress, relatively high stiff material, which exhibits little or

no ductility. Therefore, for the analysis, the material will be assumed to behave fully elastically, and will

assumed to fail in a brittle way.

The investigation will be made on beam ofa section ofdepth of 0.42 meters, and width of 0.27 meters,

spanning 8.97 meters which is simply supported at both ends.

Fundamentals of Blast Waves

An explosion can be defined as a large

scale, rapid and sudden release of energy. The detonation of ahigh explosive generates a large amount of energy. The energy expands forcing out the volume it

occupies, and as a result of this a layer of compressed air (blast waves) forms.

When these blast waves reach to a surface, they apply pressure for a period of time. As these blast

waves expand, they lose their density, i.e. their pressure value; however, they tend to act on the

structure for a longer time.

Because the blast waves expand

in a circle way in 3 dimensions,

they act on surfaces in different

ways depending on distance

between the explosion point and

the structures surface. When the

surface is too close to the explosion point, the radius is very low, so the blast waves in the middle reach

the surface first, and in time the surrounding waves hit the surface . When the distance is too long, the

radius gets bigger, so the perimeter, so it is assumed that the waves act at the same time on the surface.

This is calledfarfield blastloading andwill be considered for this reportas well for simplicity.


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Blast Waves Pressure on Structures

Blast waves increase to a value of pressure above

the ambient pressure. This is referred to as the

side on overpressure that decays as the shock

wave expands outward from the explosion

source. After a short time, the pressure behind

the front may drop below the ambient pressure.

During such a negative phase, a partial vacuum is

created and air is sucked in.

This can be shown like the figure at the left hand

side. The exponential decay is called the

freidlander decay. When dealing with far field

blast loading, the exponential part can be

idealized as a reverse ramp load, with the same

overall impulse, which gives the total load acting on the structure = total area under the pressure time graph, and the negative part is neglected.

Static Failure Loading and the Failure Deflection

Now that the magnitude of the loading and the shape of the loading are assumed, the failure mode will

be derived for static loading case, and then willbe used for help for analysis of dynamical loading case,

which is the blast type of loading.

The assumed loading, since far field blast loading analysis is being made, is uniformly distributed on the

structure. If this loading was to act on the structure as a statically load; the maximum moment acting onthe beam would be given by well known formula = q l2 / 8. The ultimate uniformly distributed that this

beam can resist can be found by equating the ultimate bending moment to the resistance of the

material and the section.

For the given material and the section, by applying simple knowledge from mechanics of structures, for

static loading conditions, the ultimate moment resistance is found to be 1627.3 kN.m. And assuming

perfectly elastic behavior, for the given geometric and loading conditions the deflection that this beam

makes at failure state is found to be 0.072 meters.

The shear resistance of the given material is 118000 KPa. The loading that causes the maximum moment

value is generated by a uniformly distributed 161.8 kN / m. This value causes a shear value of 725.7 kNat the supports, and a shear stress of 6399.23 KPa in the cross section, which means the failure is

happens by flexure before shear in static load case.

However, as can be seen in the following pages, the shear check will be considered for dynamic analysis

as well.


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Dynamic Loading Analysis

When static analysis is made, it is assumed that inertial force of the structure is not important. However,

in dynamic analysis, when a great load is applied to a structure in a very short duration, the inertial

force also becomes important.

In reality, structures have distributed mass, loading and resistance. But when doing blast analysis, a

simplified method is used. The properties that have been mentioned are transformed, or assumed to be

lumpedat a single point. The factors for the given situation can be seen in the table below:

Method of Solution

The simplest way for blast wave solutions is free vibration solutions. Basically, the beam has been

reduced to a single degree of freedom structure withequivalent factors, and the (assumed) uniformly

distributed blast loading will be now assumed to be concatenated on this assumed lumped SDOF

freedom, that is assumed to deform (and fail) in the first mode of (which is given by the statically

deformed shape) deformed shape / vibration.

After the simplifications are made, dynamic analysiscan be done depending on the properties of the

SDOF model, and the loading properties, namely the maximum loading value, and the duration of the

loading.

Depending on the maximum deflection and comparing with the static analysis deflection, a judgment

can be made if this beam will fail or not; in flexural failure. (Shear failure has not been discussed yet.)


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The solution will be done by an iterative method. Dynamic SDOF analysis will be run for various

equivalent loads, on the equivalent system the assuming reverse ramp load situations, to find out for

how long that loading must be applied to make the beam fail. One example is shown below:

Using the factors given in the table in the previous page;

Equivalent stiffness = 12601.35 kN / m

Equivalent mass = 819.9 kg

Equivalent systems natural period:0.05 seconds

By trying, it is found that when 3242.67 kPa of pressure is loaded on the structure for 0.00293

seconds, the first half cycle of the response of the structure is found to be: (maximum deflection is

equal to the maximum deflection that is calculated for static analysis which stands for flexural fail.)

There are few comments that can be done at this moment.

This beam can resist a pressure of 3242.67 KPa, if the pressure is applied only for 0.00293 seconds,which is only 5% of natural period of the structure. This loading also corresponds to an impulse value of

= 3242.67 0.00293 / 2 = 4.75 kPA.sec. (reminder: divided by 2 reverse ramp loading assumption)

As can be seen, when the natural period of the beam is greater than the duration of the load, the

beam is able to resist a very high pressure. This type of loading is called: impulsive loading. ( td / tstructure

< 0.4). In this instance the load is applied so rapidly that the structure has not enough time to respond,

which means, the load has been released before the structure has time to do peak deflections. In this

instance, the load that can be resisted can be higher than if it was statically loaded.

Opposite to this, when the loading is applied for a long duration, ( td / tstructure > 4), this is called a quasi

static loading. In this instance, when structure makes the peak deflection due to instant loading,

there is still loading acting on the structure. When this is the case, the structure can resist less loading

compared to static loading case.

In between two cases lies the dynamic realm.


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PressureImpulse Diagram

As exampled in the previous page, by

trial and error method, for different

values of pressure, the loading

duration, meaning that the impulse

can be found that will cause the failure

deflection.

This diagram can be seen above, called

the pressureimpulse diagram.

This line in this pressure impulse

diagram shows failure deflection of the

beam. The vertical asymptote gives the

impulse failure limit and the horizontal line gives the pressure. The shaded area with gray shows the

failure states of the beamwhen the deflections, therefore the flexural failure are considered.

Dynamic Reactions - Failure Condition when the Shear Force Reaction is Considered

In addition to calculating structural deflections, the determination of reaction forces is required when

designing structures to resist dynamic loading.

The simplified formula for checking

the shear failure is 0.39 R + 0.11 F

where Rmaks = 8 Mm / l = 1451.32 kN

F is given by the applied pressure

times the surface area of the beam.

The shear failure value of the beam is

given by = 118000 cross area of the

beam = 13381.2 kN.

The loading conditions that are going

to give us shear failure values can be

found by taking the Rmin by zero and

thus finding the shear failure zones.

The shear failure values will be found to be between: 13381.2 = 2 0.11 F, thus F = 60822.73 which

gives a pressure value of: 23393.60 KPa. Where R is 1451.32 kN, the shear failure pressure will be

9450.40 KPa. These values are in a zone much more higher than the moment curvature zone. This only

concludes us that, the limiting impulse value by flexural failure mode is not usable anymore, meaning

that below the value of the impulse of 5, the pressure cannot be up to infinity as it is found by the

flexural fail mode, but will be limited to 23393.60 KPa by shear failure, but even convservatively

thinking, 9450.40 Kpa.


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Minimum Safe StandOff Distance

In this part of the report, the minimum safe standoff distance for this beam will be investigated under

a blast incident of 1000 kg of TNT.

To find out the pressure and the impulse value from a detonation these two formulas are used,

depending on many experimental data:

Where z is given by = w / s, where s is the standoff and w is 10001/3 for this instance. For different values

it is found that the Pressure and Impulse values change as in this figure: (Only flexural failure mode is

investigated because the limiting shear failure is much higher)

The value that is in the circle is for a standoff value of 11 meters. The node on the Stand Off line at top

is for 10 meters. All the nodes go in 1 meters each. As it can be seen, when the bomb is detonated at 11

meters, the beam is ok. However, as it gets closer, its damage ability gets much higher and the beams

fails in flexure.

5432 )(214.0)(036.0)(15.1)(335.0)(89.269.3)log( zLogzLogzLogzLogzLogprSCALED

5432 )(00015.0)(0003.0)(064.0)(222.0)(31.175.2)log( zLogzLogzLogzLogzLogirSCALED


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Conclusions

In this report, the blast resistance of a particular beam has been tried to investigated, and tried to be

made clear for a person who knows nothing about blast loading, and the safe stand off distance has

been derived for a detonation of 1000 kgs of TNT bomb.

Thru the report, all the assumptions have been written in italics, and for a very quick summary only theparts in bold can be read.

Blast loading is a very complex area of loading.

There are many simplifications and assumptions.

Perhaps the biggest assumption that has been made is the far field loading assumption. This actually

varies with the distance of the detonation point and the surface of the object.

Also, when assuming reverse ramp loading, due to trying to get the same impulse value, the duration of

the loading is being modified.

Also, there is a very big simplification when the properties of the structure are lumped to a SDOFsystem. In more complex scenarios, Finite Element Modeling of structures can be done, and the

behavior of the structure can be investigated in very small time steps, either in 2D or 3D.

Also in the last part, it is assumed that the bomb is detonated in free air, and there is no other surface

close to our structure. However in real life, waves can reflect from different surfaces causing much more

complex loading variations on the structure.

And also it is assumed that, when getting the flexural fail mode, the beam was deforming same where it

is loaded statically. However the beam clearly has more than one natural deforming shapes, and can be

triggered by different blast loading conditions, which makes this way of solving very questionable.

For a more detailed analysis, finite elements models should be investigated.

For blast loading, there is only one thing that is clear: as can be seen from the last diagram, as the

bomb gets away, the loading decreases.

So for blast safety: get away from the bomb. As the structure is unable to move clearly, do not let the

bomb explode close to you: design your structure with a perimeter where the bomb cannot pass thru.

This will also help making the far field loading assumption correct, and so that being in the safe side

will be more likely.

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