RESEARCH POSTER PRESENTATION DESIGN © 2015

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Currently, the City of Baltimore, Department of Public Works (DPW) oversees the operation of an aging water distribution network that has increasingly presented the City with costly water main ruptures with the consequences of urgent repairs associated with disruptions of traffic, commerce and potentially electricity, telephone, cable and internet services. To help mitigate this, our team will produce a predictive failure model that can be used to assess the state of “health” of the water distribution network for the city of Baltimore while providing the DPW critical diagnostic tool. Failure criteria due to shear stress, shear strain, fracture toughness, and fatigue will be incorporated into the model. The model will incorporate evolution laws needed to account for changes in the water pipe material constitution, geometry as well as operating conditions, including changes in loading and environment. The model will be used to conduct parametric studies needed to establish failure maps capable of exploring the effects of several agents contributing to the degradation and failure of water mains. Through close collaboration with the DPW, we will utilize the wealth of data they have collected allowing the model to transition from purely theoretical to a practical working model. This transition will require the modification of the parameters for the parametric studies and refinement of the parametric studies themselves. The model will produce contour maps to easily and quickly understand the results of the parametric study in addition to a color map of the Baltimore water main system to display the final failure times calculated by the model. The addition of a Graphical User Interface (GUI) will afford the DPW an intuitive program tailored to meet their specific needs. Displaying the final failure times from this wealth of data will be handled by a color map of the Baltimore water main system.

ABSTRACT

We are choosing to model a water pipe as a thin wall pressure vessel. This vessel has an axial loading (N), a torsional loading (T), an inner and outer pressureand respectively), and some thickness. These pipes are typically given a year range of failure from the manufacturer of when replacement is needed. However to analyze why it is breaking we look at the pipe as a compilation of stress elements. Above denotes an element of the outer surface. Eq. (1a) is the hoop stress (), due to internal and external pressure; Eq. (1b) shows the axial stress (), due to the internal and external pressures with an axial force; and the shear stress () due to the torsion.

THEORETICAL DEVELOPMENT

METHODS AND IMPLEMENTATION

RESULTS

CONCLUSIONS AND FURTHER RESEARCH

Being able to make a predictive model for the City of Baltimore will help the city and taxpayers a lot of money, as well as prevent water breaks from happening as often as the age of the water pipes increases. To continue our research, we will be able to incorporate a Graphic User Interface, that makes this code easy to use for all of those involved as well as testing the model against actual Department of Public Works data, and customize our model to be indicative to the City of Baltimore’s pipes and water mains. Relating to improvements of the code we currently have only failure due to stress and plan to add back in fracture mechanics failure and strain failure, which we have in our older model but the newest version does not yet include these criteria. This will require multiple fsolve functions. We noticed that the prevailing cause of failure was fracture mechanics so adding in this criteria will give us more insight into what characteristics cause a pipe to break and focus our attention on to make a more precise model. We currently have internal pressure, external pressure, thickness 1 and E 2. Continuing work will be done to parametrize axial force and torsion to add into the stress equations and consequently all dependent on them.

REFERENCES

Anderson, T. L. Fracture Mechanics: Fundamentals and Applications. 3rd ed. Boca Raton: Taylor & Francis Group, 2005. Print.P. Charalambides and K. Kalayeh, “Water Pipe Response and Failure Model – ENME220h,” UMBC, Baltimore City, MD, Oct. 2015.

ACKNOWLEDGEMENTSMadeline Driscoll1 , Art Shapiro1 , Dr. Rudolph Chow1

Honors Mechanics of Materials – FA 2015 Class1 Department of Public WorksUMBC Breaking Ground

University of Maryland, Baltimore County, Baltimore, MD 21250

Aakash Bajpai, Justin Taylor, Matthew Bleakney, Nicholas Rabuck Kourosh Kayaleh, Dr. Panos Charalambides

Assisting the City of Baltimore in Maintaining and Upgrading its Aging Water Distribution Infrastructure

(1𝑎 )𝜎h=𝑝𝑖𝑟𝑡 −

𝑝𝑜𝑟𝑡

(1𝑏)𝜎𝑎=𝑝𝑖𝑟2 𝑡 −𝐾

𝑝𝑜𝑟𝑡 − 𝑁

2𝜋 𝑟𝑡 (1𝑐 )𝜏 h𝑎 =𝑇𝑟𝐽

(2𝑎 )𝜀𝑎=𝜎 𝑎

𝐸 − 𝑣𝐸 (𝜎h+𝜎 𝑧)+𝛼∆𝑇

(2𝑏)𝜀h=𝜎h

𝐸 − 𝑣𝐸 (𝜎𝑎+𝜎 𝑧 )+𝛼 ∆𝑇

(2𝑐 )𝜀𝑧=𝜎𝑧

𝐸 − 𝑣𝐸 (𝜎 𝑎+𝜎 h )+𝛼 ∆𝑇

Knowing that the stresses can be compared with allowable stresses (which we call failure stresses) to know if the material can withstand the stress. If it exceeds the allowable limit we designate that as a failure. However it is important to also include other failure criteria such as failure by strain. Below are generalized hooks law equations which assume a homogeneous, linear, and isotropic material. Eq.s (2a), (2b), and (2c) calculate the strain in the axial, hoop, and out of plane z direction.

(3𝑎 )𝜎h1=

(𝑝¿¿ 𝑖−𝑝𝑖𝑛𝑡 )𝑟 1

𝑡 1¿

(3𝑏 )𝜎h1=

(𝑝¿¿ 𝑖−𝑝𝑖𝑛𝑡)𝑟1

2 𝑡1− 𝑁

2𝜋𝑟1 𝑡1¿

( 4𝑎 )𝜎h2=

(𝑝𝑖𝑛𝑡−𝐾𝑝𝑜)𝑟 2

𝑡2

( 4𝑏 )𝜎 𝑎2=

(𝑝𝑖𝑛𝑡−𝐾𝑝𝑜)𝑟 2

2 𝑡2− 𝑁

2𝜋 𝑟2 𝑡2

The failure may not occur within one of these planes so we also add equations that convert these stresses and strains to find the principles of them. These are the maximums and are at a transposition with respect to the current coordinate axis by some specific degree angle. Water pipes will change over time. There will be a buildup on the inside of the pipe that is a mixture of several mineral deposits as well as eroded pipe matter. We assume it to be homogeneous, isotropic, and having an inner pipe geometry. This develops a biomaterial pipe model. There is a pressure at the area of contact between the two pipes called the interface pressure (. Eq.s (3a) and (3b) denote the stress equations from this conclusion..

𝐺=1−𝑣𝐸 [𝐾 1

2 ] 𝐴(𝑉 )≤ 2𝛾𝐺=1.118𝑃 (𝑡)√𝑟 (𝑡)

𝑡 (𝑡) √𝜋 𝑎(𝑡)≤2𝛾

Fig. (1) General Pipe Diagram

Fig. (2) Schematic of composite pipe

(5a) ) =)

(5 b )𝑝𝑖𝑛𝑡=1

1+𝜁 −𝜁 𝑣1′ 𝑡1

𝑟1+𝑣2

′ 𝑡1

𝑟1

(𝜁 𝑝𝑖+𝐾 𝑝𝑜+𝑁

2𝜋 (𝜁 𝑣1′ 1𝑟1

2 +𝑣2′ 1𝑟 2

2 )−2 𝑡 2𝐸2

𝑟 2 ( 2−𝑣2 )(𝛼1 −𝛼2)∆𝑇 )

𝜁=2 −𝑣1

2 −𝑣2

𝑟1

𝑟2

𝑡 2

𝑡1

𝐸2

𝐸1

occurs at the interface which occurs at some where the hoop strain for both are equal leading to the compatibility equation (Eq. (5)) below. In Eq. (5b) is derived from 5

𝑣1′ =

2𝑣1

2−𝑣1𝑣2

′ =2𝑣2

2 −𝑣2

A key component of our code was to make it non-dimensional. This changed the inputs to have some units with dimensions to dividing these variables with characteristic values outputting a dimensionless value. The characteristic length is defined as and characteristic pressure is defined as. All length values are divided by the characteristic length and all pressure values (including Elastic modulus as it has units in terms of pressure) are divided by the characteristic pressure. Some examples are shown below in Eq. (7a) through (7d).

(7𝑎 ) 𝑡1=𝑡1

𝑙𝑐(7𝑏 )𝑟1=

𝑟 2

𝑙𝑐(7𝑐 )𝑝𝑖𝑛𝑡=

𝑝𝑖𝑛𝑡

𝑙𝑐(7𝑑)𝑝𝑖=

𝑝𝑖

𝑝𝑐=1

The third failure criteria added to the model is fracture mechanics. Equation 6a describes fracture in a non-quasistatic condition defined using the pressure at time t (P(t)), the small scale yielding at time t (r(t)), half the crack length at time t (a(t)), and the thickness at time t (t(t)). This simplifies to Equation 6b under a quasistatic state defined by the stress intensity factor (K1), the Poisson’s ratio (v), the elastic modulus (E), and the instantaneous crack length speed at time t (A(V)). All of this is related to 2Ɣ, which is the surface energy of the material. We choose 2Ɣ because that is the stored elastic strain energy that is released as the crack grows, and is the driving thermodynamic force for the material to crack. This has to be greater than the value of fracture (G) because of the relation G=2Ɣ + GP, which denotes that since both 2Ɣ and Gp have to be negative as it is the energy dissipated, G must inherently be a greater negative, thus smaller than the 2Ɣ value. The equation we used in our research is Equation 6b as it allows us to remain this in a quasistatic condition, which is ideal for our research over a number of years. For our model the pipe is a smooth surface but in reality degrades non-uniformly there are peaks and trophs. The exterior radius of the pipe diminishes with time through rusting. The unequal deterioration of the surface leads to the formation of crevices in the surface, in this model it is assumed that the deepest of these crevices is one-third of the overall change in the exterior radius. This assumed ratio is provided by the user when the query is presented. This allows the user to assign a different, possibly more accurate, ratio. The remainder of the time-dependent variables are linear interpolations given by the loading, material properties and geometry scripts.

(6a) (6b)

Define all of the variables that stay constant during the time

interpolation of the parameter.

Redefine selected parameter into a range of values for t=0 (initial) and for t=final (final)

Plot the time of failures in a contour and mesh graphs,

showing which values of the initial and final values they correspond to for their time.

Begin matching each initial values for the parameter into the

functions that define time of failure using a for loop.

Solve for the time of failure.

Store the time of failure in a matrix.

Begin matching each of the final values for the parameter into the

functions that define time of failure using a for loop.

FLOWCHART OF CODE

𝜎 𝑓 (𝑡 ) −𝜎𝑝1(𝑡 )=0

𝝈𝒇 { 𝝈 𝒇𝒊𝒏𝒊=𝟓𝟎→𝟏𝟎𝟎

𝝈 𝒇 (𝑻=𝟏𝟎𝟎 𝒚 )=𝟒𝟎→𝟔𝟎

𝑬𝟐{ 𝑬𝟐𝒊𝒏𝒊=𝟏

𝑬𝟐 (𝑻=𝟏𝟎𝟎 )=𝟎 .𝟓→𝟐𝑷 𝒊{ 𝑷 𝒊𝒊𝒏𝒊=𝟏

𝑷 𝒊 (𝑻=𝟏𝟎𝟎)=𝟏→𝟓

𝐸2𝑖𝑛𝑖=

𝐸2𝑖𝑛𝑖

𝐸2𝑖𝑛𝑖=1

𝐸1 (𝑇=100 𝑦 )=𝐸1 (𝑇=100 𝑦 )

𝐸2𝑖𝑛𝑖 =0.13

𝐸1𝑖𝑛𝑖=0

𝐸2 (𝑇=100 𝑦 )=𝐸2 (𝑇=100 𝑦 )

𝐸2𝑖𝑛𝑖 =0.98

𝜎 𝑓 (𝑇=100 𝑦 )=𝜎 𝑓 (𝑇=100 𝑦 )

𝑃𝑖𝑖𝑛𝑖 =50

𝜎 𝑓𝑖𝑛𝑖=

𝜎 𝑓𝑖𝑛𝑖

𝑃 𝑖𝑖𝑛𝑖=100

𝑃𝑜𝑖𝑛𝑖=

𝑃𝑜𝑖𝑛𝑖

𝑃 𝑖𝑖𝑛𝑖=0.5

𝑃 𝑖 (𝑇=100 𝑦 )=𝑃 𝑖 (𝑇=100 𝑦 )

𝑃 𝑖𝑖𝑛𝑖 =2

𝑃 𝑖𝑖𝑛𝑖=

𝑃 𝑖𝑖𝑛𝑖

𝑃 𝑖𝑖𝑛𝑖=1

𝑃𝑜 (𝑇=100 𝑦 )=𝑃𝑜 (𝑇=100 𝑦 )

𝑃𝑜𝑖𝑛𝑖 =0.5

Fig. (3) MATLAB function fsolve finds tf Fig. (4) Parametric study of

Fig. (5) Parametric study of Fig. (6) Parametric study of

Relevant Constants:

𝑟2𝑖𝑛𝑖=

𝑟2𝑖𝑛𝑖

𝑟2𝑖𝑛𝑖=1

𝑟2 (𝑇=100 𝑦 )=𝑟2 (𝑇=100 𝑦 )

𝑟2𝑖𝑛𝑖 =1.01

𝑡 2𝑖𝑛𝑖=

𝑡 2𝑖𝑛𝑖

𝑟 2𝑖𝑛𝑖=0.05

𝑡 2 (𝑇=100 𝑦 )=𝑡2 (𝑇=100 𝑦 )

𝑟2𝑖𝑛𝑖 =0.03

𝑡1𝑖𝑛𝑖=

𝑡1𝑖𝑛𝑖

𝑟 2𝑖𝑛𝑖=0

𝑡1 (𝑇=100 𝑦 )=𝑡1 (𝑇=100 𝑦 )

𝑟2𝑖𝑛𝑖 =0.01