Part 9 - Estimation of magnitude of the unbalanced centrifugal forces driving tectonic movement pptx

In the analysis of this walled cylinders subjected to internal pressures it is assumed that the radial plans remains radial and the wall thickness dose not change due to internal pressur

Part 9 - Estimation of magnitude of the unbalanced

centrifugal forces driving tectonic movement

Based on the above observations, let us assume as a working hypothesis that the Earth can be modelled as a rotating body where the centre of mass is offset from the principal axis of rotation For the purposes of this paper the author will consider the two principal approaches to determine the circumferential forces associated with an unbalanced rotating body

Model 1 Rigid body dynamics

As discussed in Section 5, the Pacific plate has all the appearances of being in compression while the almost diametrically opposed African plate appears to be subject to tensile forces

The simplest model is to consider the Earth as an eccentrically rotating solid body such as an unbalanced flywheel Although this model (shown in Figs 8 & 9 and enumerated in Appendix 1) accounts for the compressive and tensile stresses developed in the outer rim, it does not describe the unbalanced centripetal forces which the author believes to be linked to tectonic forces

resulting in plate movement

Fig 8 below Force/Vector diagram showing the centripetal force P and the circumferential or

tensile force F in the outer rim of a solid rotating body Rotating Machines are designed to ensure that the developed circumferential stress F, is less than the design hoop stress

Fig 9 below Force/Vector diagram showing the differential circumferential stress induced in the

outer rim of a solid rotating body, whose centre of mass is not co-incident with the principal axis

of rotation Under these conditions the tensile forces are increased on the 'heavier' side and are decreased on the 'lighter' side Thus the hoop stress in the rim will be in tension on the 'heavier side' and in compression on the 'lighter side'

Model 2 Outer rim is able to slide relative to main body

In order to determine the magnitude and direction of the forces postulated as being responsible for tectonic movement the model used in one in which the thin crust is able to slide relative to the solid body at the crust /mantle interface By way of illustration Fig 10 shows that if an unbalanced disc with an outer annular ring containing fluid is rotated about its principal axis, the liquid will move to the ‘lighter’ side Fig 11 shows an analogous situation with the sliding continental plates

If we consider the crust as being able to move relative to the mantle, albeit it over a long

geological time scale, then a simple force diagram (Fig 12) can be constructed by making the following assumptions: the crust is a thin shell that is able to slide relative to the mantle, the

forces due to eccentricity are superimposed on the stress caused by the general rotation and gravity, and the stress that is of interest for the purposes of tectonic movement is the differential stress due to this eccentricity

By approaching the problem in terms of a thin shell moving relative to the mantle, it is possible

to consider what increments of the tensile force are responsible for putting the Pacific basin under compression (note crumpled profile) and the African Plate under tension (note the Rift Valley) The calculations to derive the expression of the circumferential stress at the surface of the earth, which are based on the consideration of the eccentrically induced loads on the thin crust are detailed in Appendix 11 The term radius of eccentricity’ was introduced to denote the distance between the centre of mass and the major axis of rotation

From Appendix 2 the following relationship was derived:

Eq.2 shows that the magnitude of the circumferential forces or in this case the derived

circumferential stress is dependent on the distance between the geometric centre and the centre

of mass, i.e the ‘radius of eccentricity’ In a limiting case, if the ‘radius of eccentricity’ is zero, the rotating body will be balanced and the net force will be zero Fig 13 shows this relationship

in between F (circumferential stress) and the E (radius of eccentricity)

Fig 13 below Relationship between the radius of eccentricity and the circumferential stress

Having derived an equation which relates the circumferential stress with the rotational velocity and the centre of eccentricity it would seem appropriate to consider the possibility of tectonic activity on the planet Venus Due to the low peripheral velocity of Venus (1 Revolution in 243 days = 6.5 kmhr -1 ) as compared with 531.5 kmhr -1 on Earth, the centrifugal forces available as compared to the similar sized planet Earth will be in the ratio of (42.5) 2 / (531.5)2 =

1806.25/282492.25 = 0.006: 1 This would give a stress value of 3.9x10-3 Nmm-2 (0.059 psig) The unbalanced centripetal forces thus needed for tectonic activity are negligibly small

In order to better understand the magnitude of the calculated circumferential stress in the

continental crust, it is helpful to relate the model to more familiar applications (These are shown pictorially in Cartoons 2, 3 &4) The stress value of 7.29x10-2 Nmm-2 if applied to a 1 tonne braked motor vehicle with a rear surface area of 1000 mm x 1300 mm=1.3x106 mm2 will yield a push force 94770 N In imperial units this equates to a push of 21305 lbf (pound force) or 9.5 tonf (ton force) Rounded up and put more simply, this equates to the vehicle being pushed by

118 people each of whom weighs 180 pounds (81.8 kg) (see Cartoon 1) As the incline between the height of the Andes (taken as 5 km) and distance between the Peru- Chile trench and the Cordillera –Real (taken as c.1000 km) is c 1:250, the vehicle can be considered to be on a level surface for purposes of scaling However a 3 tonne hoist will easily pull the vehicle up a 1:3 incline onto a pick-up truck It is also worth noting that an upward acting net force of 2.37 x 10-2N/mm2 (3.5 psig) on a 60 metre long wing span of an aircraft is sufficient to keep a large 350 tonne aircraft flying

A puff of wind with dynamic pressure as low as 0.135 x 10-2 N/mm2 (0.2 psig) acting on the large surface area of a ship’s sail will cause a boat to move across water Thus the unbalanced centrifugal forces created by placing the centre of mass of the Earth 1 km off centre are large and cannot be ignored The calculated circumferential forces if applied to the cross sectional area of the South American plate are more than sufficient to push it over the Nazca plate

What is important to remember is longitudinal stresses are half as much

as the circumferential stresses Therefore, we can say that longitudinal strength is twice as strong as circumferential strength This is only true for illustration purposes

LECTURE 15

Members Subjected to Axisymmetric Loads

Pressurized thin walled cylinder:

Preamble : Pressure vessels are exceedingly important in industry Normally two types of

pressure vessel are used in common practice such as cylindrical pressure vessel and spherical pressure vessel

In the analysis of this walled cylinders subjected to internal pressures it is assumed that the radial plans remains radial and the wall thickness dose not change due to internal pressure Although the internal pressure acting on the wall causes a local compressive stresses (equal to pressure) but its value is neglibly small as compared to other stresses & hence the sate of stress of an element of a thin walled pressure is considered a biaxial one

Further in the analysis of them walled cylinders, the weight of the fluid is considered neglible Let us consider a long cylinder of circular cross - section with an internal radius of R 2 and a constant wall thickness‘t' as showing fig

This cylinder is subjected to a difference of hydrostatic pressure of ‘p' between its inner and outer surfaces In many cases, ‘p' between gage pressure within the cylinder, taking outside pressure to be ambient

By thin walled cylinder we mean that the thickness‘t' is very much smaller than the radius Ri and

we may quantify this by stating than the ratio t / Ri of thickness of radius should be less than 0.1

An appropriate co-ordinate system to be used to describe such a system is the cylindrical polar one r, q , z shown, where z axis lies along the axis of the cylinder, r is radial to it and q is the angular co-ordinate about the axis

The small piece of the cylinder wall is shown in isolation, and stresses in respective direction

have also been shown

Type of failure:

Such a component fails in since when subjected to an excessively high internal pressure While it might fail by bursting along a path following the circumference of the cylinder Under normal circumstance it fails by circumstances it fails by bursting along a path parallel to the axis This suggests that the hoop stress is significantly higher than the axial stress

In order to derive the expressions for various stresses we make following

• There are no shear stresses acting in the wall

• The longitudinal and hoop stresses do not vary through the wall

• Radial stresses sr which acts normal to the curved plane of the isolated element

are neglibly small as compared to other two stresses especially when

The state of tress for an element of a thin walled pressure vessel is considered to be biaxial, although the internal pressure acting normal to the wall causes a local compressive stress equal

to the internal pressure, Actually a state of tri-axial stress exists on the inside of the vessel However, for then walled pressure vessel the third stress is much smaller than the other two stresses and for this reason in can be neglected

Thin Cylinders Subjected to Internal Pressure:

When a thin – walled cylinder is subjected to internal pressure, three mutually perpendicular principal stresses will be set up in the cylinder materials, namely

• Circumferential or hoop stress

• The radial stress

• Longitudinal stress

now let us define these stresses and determine the expressions for them

Hoop or circumferential stress:

This is the stress which is set up in resisting the bursting effect of the applied pressure and can be most conveniently treated by considering the equilibrium of the cylinder

In the figure we have shown a one half of the cylinder This cylinder is subjected to an internal pressure p

i.e p = internal pressure

d = inside diametre

L = Length of the cylinder

t = thickness of the wall Total force on one half of the cylinder owing to the internal pressure 'p'

Because s H.L.t is the force in the one wall of the half cylinder

the equations (1) & (2) we get

Consider now again the same figure and the vessel could be considered to have closed ends and contains a fluid under a gage pressure p.Then the walls of the cylinder will have a longitudinal stress as well as a ciccumferential stress

Total force on the end of the cylinder owing to internal pressure

= pressure x area

= p x p d2 /4

Area of metal resisting this force = pd.t (approximately)

because pd is the circumference and this is multiplied by the wall thickness

Assessing the Biomechanical Impact of Medical Devices Using MSC.Software at Texas A&M University

Contest

Winner

By Lucas H Timmins, PhD Student, Texas A&M University, College Station, TX

The Vascular Biomechanics Group in the Department of Biomedical Engineering at Texas A&M University investigates the role of biomechanics in the treatment of vascular disease In

particular, we use MSC.Software to investigate the solid mechanical implications of implanting vascular stents and aortic stent grafts (Figure 1)

Figure 1: A Vascular stents on balloon catheters B Aortic stent grafts

Cardiovascular diseases are the principal cause of death in the United States, with total cost associated with the disease in upwards of $450 billion USD As a result, an extensive effort has gone into developing biomedical devices to treat these diseases As the placement of these

devices alters the biomechanical environment inside an artery, proper computational mechanical analysis must be carried out to ensure their treatment efficacy, but also to examine the

mechanical loads that they place on the arterial wall MSC.Software allows for proper analysis of these inherently difficult (e.g non-homogeneous, non-linear) contact mechanics problem

MSC.Software allows for construction and simulation of vascular stent implantation to examine the mechanical implications of varying specific design parameters on the arterial wall mechanics (Figure 2) Through the modeling of biological tissues and the application of physiologic

boundary conditions, appropriate finite element models can be developed in MSC.Patran and solved with MSC.Marc

Figure 2: Stent Geometry: MSC.Patran models of vascular stent designs created by varying specific design parameters (f,ρ,h) The designs on the right are approximately 5 and 20 mm

in outer diameter and length, respectively Computational analyses of these models were then carried out to assess their biomechanical impact on the arterial wall.

Figure 3: Stented Artery Model: An MSC.Software simulation illustrating the

circumferential stress distribution on the arterial wall after modeling stent implantation.

Patient specific computational models of abdominal aortic aneurysms (AAA) are also

investigated in our laboratory to assess treatment methods and/or device design MSC.Patran provides an excellent means of constructing finite element models from various medical imaging techniques to examine the biomechanical environment after aortic stent graft implantation

Figure 4: Aortic Stent Graft Model: A Computational mesh of an AAA with different biological tissue components B Max principal strain field on outer vessel wall.

Through proper computational analysis, which MSC.Software provides, our laboratory is able to examine the mechanical impact of implanted vascular medical devices Such analysis, along with clinical evidence of device success/failure, can provide information in optimizing biomedical device designs and treatments

For further information, please refer to the following publications:

Bedoya, J., Meyer, C.A., Timmins, L.H., Moreno, M.R., Moore, J.E Effects of Stent Design

Parameters on Normal Artery Wall Mechanics J Biomech Eng 2006; 128:757-675.

Timmins, L.H., Moreno, M.R., Meyer, C.A., Criscione, J.C., Rachev, A., Moore, J.E Stented

Artery Biomechanics and Device Design Optimization Med Biol Eng Comput 2007;

45:505-513

Thin Walled Cylinders Under Pressure.

The three principal Stresses in the Shell are the Circumferential or Hoop Stress; the Longitudinal Stress; and the Radial Stress

If the Cylinder walls are thin and the ratio of the thickness to the Internal diameter is less than about 1/20 then it can be assumed that the hoop and longitudinal stresses are constant across the thickness It may also

be assumed that the radial stress is small and can be neglected In point

of fact it must have a value equal to the pressure on the inside surface and zero at the outside surface These assumptions are within the bounds of reasonable accuracy

The above cylinder has an internal diameter d and a wall thickness of t If the applied internal pressure is p then the Hoop stress is

and the Longitudinal stress is

Sectioning the above cylinder through a diametral plane and consider the equilibrium of the resulting half cylinder

acts upon at area of The resultant vertical pressure force is found from the projected horizontal area

Clearly the Pressure force upwards is equal to the hoop Stress times the area over which it acts

Hence

(1)

(2)

Now consider the equilibrium of a section cut by a transverse plane The

diameter taken should really be the mean diameter) and the pressure p

this is true no matter the actual shape of the end of the cylinder) Equating longitudinal Forces:-

(3)

(4) Where long cylinders or tubes are braced or carried in brackets, the longitudinal Stress may be much less than given by equation (4) and is sometimes neglected.

Thin Spherical Shells Under Internal Pressure.

As in the previous section the radial Stress will be neglected and the circumferential or hoop Stress is assumed to be constant

By symmetry the two principal Stresses are equal and the stress in any tangential direction is

(5)

d is the internal diameter The equation can be re-written as:-

(6)

Cylindrical Shells With Hemispherical Ends.

Let the thickness of the cylinder walls is

and the thickness of the hemisphere It is assumed that the internal diameter of both are the same

If the internal pressure of the shell is then:-

The Stresses in the cylinder walls are:-

(7)

(16) Taking a value for Poisson's ratio of 0.3

(17) Note that the maximum Stress will then occur in the hemispherical ends i.e

(18)

Which is greater than the hoop stress

in the cylinder For equal maximum stress

should equal 0.5

Volumetric Strain On The Capacity Of A Cylinder.

the dimensions increase by

there will be an increase

in volume and:-

(20) Neglecting the products of small quantities

(21)

(22) Since the length of the Circumference = the diameter X a constant

(23)

It should be noticed that this is the sum of the linear Strains in three Principal directions It is worth comparing this with "Engineering Materials: Compound Stress and strain Equation (97)"

Similar reasoning for a Spherical Shell shows that:-

(24) i.e To increase the capacity it is only necessary to multiply the Volumetric Strain by the original Volume.

Example 1:

A Boiler drum consists of a cylindrical portion 8 ft long ; 4 ft in diameter and 1 in thick.It is closed at both ends by hemispherical shells In a hydraulic test to 1000

lb./sq.in, how much additional water will be pumped in after an initial filling at atmospheric pressure? Assume that the circumferential Strain at the junction of cylinder and hemisphere is the same for both For the drum material

For the hemispherical ends:-

The question states that the hoop Strain in the ends is the same as in the cylinder thus:-

Example 2:

A Cylindrical tank is 6 ft in diameter and 8 ft long It is made of 1/2 inch thick steel

The ends are flat and are joined by nine, equally spaced, tie bars each 1 1/2 in

diameter If the tie bars are stressed to to 3 tons/sq.in and the tank filled with

water, find the increase in capacity when the pressure is raised to 200 lb./sq.in and

find the final stress in the tie bars

(39)

Initially As the tie bars are in tension they are compressing the walls of the Tank It is therefore possible to write an equation which equates the compressive force on the cylinder walls to the tensile force in the tie bars In this initial condition there is no Hoop Stress

Stress in the cylinder walls The equilibrium equation is therefore:-

(40)

(41) After th tank has been filled with water so that the pressure raised to

200 lb./sq.in

Let be the final tensile stress in the tie

By substituting from equation (43) into equation (48)

(49)

(50) And using equation (43)

(51) The increase in the capacity of the tank is (2 X increase in hoop strain + increase in longitudinal strain) X volume

Assuming that the cylinder remains full then the increase in the volume of water must equal the increase in the capacity of the cylinder Since the Volumetric Strain

is determined by the change of stresses only the stresses due to the drop in pressure need to be considered

(63)

(64)

(65)

Note This assumes that the Stress is uniform

The Maximum and Minimum Principal Stresses (See the pages on "Compound Stress and Strain")

Thin Walled Cylinders Under Pressure.

The three principal Stresses in the Shell are the Circumferential or Hoop Stress; the Longitudinal Stress; and the Radial Stress

If the Cylinder walls are thin and the ratio of the thickness to the Internal diameter is less than about 1/20 then it can be assumed that the hoop and longitudinal stresses are constant across the thickness

It may also be assumed that the radial stress is small and can be neglected In point of fact it must have a value equal to the pressure

on the inside surface and zero at the outside surface These assumptions are within the bounds of reasonable accuracy

The above cylinder has an internal diameter d and a wall thickness of t If the applied internal

Sectioning the above cylinder through a diametral plane and consider the equilibrium of the

The resultant vertical pressure force is found from the

Clearly the Pressure force upwards is equal to the hoop Stress times the area over which it acts Hence

(1)

(2)

Now consider the equilibrium of a section cut by a transverse plane The longitudinal Stress

acts on an area of approximately

( Note the diameter taken should really be the mean

diameter) and the pressure p acts upon a projected area of

( Note this is true no matter the actual shape of the end of the cylinder)

Equating longitudinal Forces:-

(3)

Where long cylinders or tubes are braced or carried in brackets, the longitudinal Stress may be much less than given by equation (4) and is sometimes neglected

Thin Spherical Shells Under Internal Pressure.

As in the previous section the radial Stress will be neglected and the circumferential or hoop Stress is assumed to be constant

By symmetry the two principal Stresses are equal and the stress in any tangential direction is

(5)

d is the internal diameter The equation can be re-written as:-

Cylindrical Shells With Hemispherical Ends.

Let the thickness of the cylinder walls is and the thickness of the

the same

If the internal pressure of the shell is then:-

The Stresses in the cylinder walls are:-

Which is greater than the hoop stress in the cylinder For

Strain On The Capacity Of A Cylinder.

there will be an increase in volume and:-

(24)

i.e To increase the capacity it is only necessary to multiply the Volumetric Strain by

the original Volume

Example 1:

A Boiler drum consists of a cylindrical portion 8 ft long ; 4 ft in diameter and 1 in

thick.It is closed at both ends by hemispherical shells In a hydraulic test to 1000

lb./sq.in, how much additional water will be pumped in after an initial filling at

atmospheric pressure? Assume that the circumferential Strain at the junction of

cylinder and hemisphere is the same for both For the drum material

For the hemispherical ends:-

The question states that the hoop Strain in the ends is the same as in the cylinder thus:-

(38)

Example 2:

A Cylindrical tank is 6 ft in diameter and 8 ft long It is made of 1/2 inch thick steel

The ends are flat and are joined by nine, equally spaced, tie bars each 1 1/2 in

diameter If the tie bars are stressed to to 3 tons/sq.in and the tank filled with

water, find the increase in capacity when the pressure is raised to 200 lb./sq.in and

find the final stress in the tie bars

Initially As the tie bars are in tension they are compressing the walls of the Tank It is therefore possible to write an equation which equates the compressive force on the cylinder walls to the tensile force in the tie bars In this initial condition there is no Hoop Stress

Stress in the cylinder walls The equilibrium equation is therefore:-

(40)

(41) After th tank has been filled with water so that the pressure raised to

200 lb./sq.in

Cylinder

The equilibrium equation is now:-

(50) And using equation (43)

(51) The increase in the capacity of the tank is (2 X increase in hoop strain + increase in longitudinal strain) X volume

(63)

(64)

(65)

Note This assumes that the Stress is uniform

The Maximum and Minimum Principal Stresses (See the pages on "Compound Stress and Strain")

Wire Winding Of Thin Walled Cylinders.

One way to strengthen a thin walled tube against an Internal Pressure, is to wind the outside with wire under tension This puts the tube into compression and consequently reduces the Hoop Stress In many applications the maximum Stress will be in the wire which must be made of high-tensile material.