Railway Transportation Engineering

Al-Mustansiriya University

College of Engineering - Highway & Transportation Dept.

Railway Transportation Engineering (RTE)

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Railway Transportation Engineering Syllabus

1- Introduction ( Classification of transportation System, Historical Development, Railway Network in the World and in Iraq , Fright (mainly railway) and Passenger Data, Motive Power).

2- Railway Track Elements ( The permanent way ,Track gauge and clearance gauge, Ballast and subballast , Rails ,Tie plates, Cross – ties (sleepers), Ballast – free track, Continuous control (slab track), b- Frame slab track).

3- Beams on Elastic Foundation. ( Differential equation for the deflection curve of a beam on Winker foundation , Deflection curve and bending moment for infinite beam on Winker springs due to a concentrated;

1. Replacing separate elastic supports by an imaging continuous foundation. 2. Modulus of the foundation after Timoshenko and Hetenyi. 3. Maximum tie pressure.

4- Bending of a Rail due to the Wheel Pressures of a Train ( Method of superposition for varbus concentrated load , Various cases for two and three concentrated loads , Application).

5- Stresses in Rails ( Bending stresses , Thermal stresses , Vertical and lateral buckling , Factor of safety , Allowable stresses for jointed rails and continuously welded rails (CWR) , Contact stresses ( The rail wheel contact., Contact stress computation after Hertzain theory, Contact stresses defects, Corrective procedure.)

6- Train rolling dynamics ( Purpose of train – track dynamics., Horse power output of a locomotive, Resistances, The rolling gradient fright traffic, Tonnage rating., Train size.)

7- Geometric design. ( Design speed, Horizontal alignment ( Circular curve, Transition curve, good appearance, more comfort, Supper elevation ,Length of transition curves) , Vertical alignment ,Turnouts, sidetrack, and yard.)

References:

1- Hey,W., W. : Railway Engineering. 2- Fastemraths, F.: Railroad Track, Theory and Practice. 3- Commb, D. H.: British Railway Track, Design Construction and Maintenances. 4- Magee, G. H.: Rail Transportation Engineering. 5- Eisenmanm, J.: Railroad Track Structure for High Speed Lines. 6- Fredrick, C. O.: the Effect of Lateral Loads on Track Movement. 7- Kern, A. D.: Lateral Buckling of Railroad Tracks Due to Constrained. 8- Timonshenko, S.: Strength of Materials. 9- Hitenyi, M. Beams on Elastic Foundations.

10- Ministry of Transport and Communication, New Railways Design Manual: NRIA (New Railway Implementation Anthony)

11- Vaziram, V. W. and Chandula, S. P.: Railways, Bridge and Tunnels. 12- Saxena, S.C. and Arora, S.: A Text Book of Railway Engineering. 13- Hennes, R.G. and Eksoe, M.: Fundamental of Transportation Engineering . 14- Hag, W.W.: An Introduction of Transportation Engineering. 15- Paguette, R.J., Ashford, N.J. and Wright, P.: Transportation Engineering :Planning and Design.




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Introduction

: أن الغرض الأساسي من أنشا خط السكة الحديدية الأسباب التالية (عملية نقل البضائع تكون أكثر اقتصادية باستخدام خط السكة الحديدية )اقتصادية (1 (نقل المعدات العسكرية الثقيلة )عسكرية (2

خواص النقل باستخدام خط السكة الحديدية1) Fright traffic ( من خدمات النقل توفرها خط السكة % 50 تشكل (عملية نقل البضائع 2) Passenger traffic ( من الخدمات % 10عملية نقل الأشخاص وتشكل)

الوحدات المستخدمة لغرض وصف عملية النقل هي 1) Fright traffic

a) Car load )وهي عبارة عن وحدة مترية تستخدم لوصف حمولة العربة او وزنها ولكنها لا تعكس كمية المواد المنقولة خلال السنة) b) Tons ( عبارة عن وحدة تعكس الوزن الكلي المنقول ولكنها لا تأخذ بنظر الاعتبار مسافة النقل ) c) Tons / mile ( وهي الوحدة الأكثر استخداما

2) Passenger traffic a) Passenger. b) Passenger / mile.

Classification of Transport Systems Hesto Development

Classification of transport systems on the basis of : 1- Surface of Transport

a. Land Transport b. Water Transport c. Air Transport

2- Degree of Freedom a. System of one degree of freedom b. System of two degree of freedom c. System with three degree of freedom

Water Transport: provide facilities for transport of heavy and bulk where time may not be of movement importance.

Air Transport : obtains maximum utility where safely of time is almost importance rather than the other.

Land Transport: Door to door service Railways have the greatest utilization in the transport of large volumes of heavy of bulk commodities over long distance. UIC = Union International des Chemin de ( France). 1759 J, Watt the ideal during wheels by means of steam engine 1814 Stephenson, Locomothe, solp أشاء اه خػ عنل حذذ ف انيرشا 1821-1825

1838 – Germany. 1837 – France. 1830 - U.S.A. 1960 Europe , 300-830 km

Asia 288, 694 km Africa 73,475 km N. M., U.S.A. 492,560 km S. America 107,776 km

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Australia 49,770 km total= 1,321,106 km Railway Network in Iraq

Baghdad – Mosul 412 km Mosul – Rabi 112 km Baghdad – Basrah 542 km Karkuk – 272 حذثح تج km

km 5,000,000 ton – 10,000,000 ton 376 شعثح– تغذاد km 3.4 million ton 150 عناشاخ– قائ

Capacity Baghdad line (Istanbul – Baghdad) 2400 km 1903-1940 Movement of fright constitute the much source of revenues of the railway 1960 W-Germany

Passenger per traffic is 1.53*109 p/year. - 1.9 * 109 $ Fright traffic 284 * 109 ton/year. - 4.2 * 109 $

. ف عشض خ1898 دضه عشض امح صعا 1900 ف الأىا

خػ اعرخذ اىقج اىنشتائح ف أهاىقطاساخ ذخض اىقذسج اىنشتائح ف . اعطثه – تغذاد 1903تذاح اىغنل ف اىعشاق

America 1895 2009 Iraq

Passenger per traffic is 187772 p/year. Fright traffic 708678 ton/year.

Disadvantage a. High initial cost of installation of a power plant. b. High initial cost of installation of transition lines. c. Generally more costly locomotives.

Advantage a. Clean operation: free for smoke and gas

This mode of transportation more suitable in tunnels and area where some be or forms make legally protected

مو اىذ اىنثشج ذغرخذ اىقاغشاخ اىنشتائح مزىل مافح خطغ ف اىعاى

While the motor are derived by wheel of locomotive, the function as generators and much of the energy of down grade lane is regenerated into transition lines for operating of upside lanes. Motive Power:-

1. Steam motive power. 2. Diesel motive power. 3. Electric motive power.

Parts of Diesel Electric Engine. 1- Diesel Engine. 2- The generator directly connected to the diesel engine design to produce simultaneously sufficient

direct current (D.C) for the traction motors and alternating current (A.C) for lighting, electrical controls, air compressor, blowers and other auxiliaries.

3- The traction motors which operate the driving wheels through a chain of gears




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Hemis ; Estimated in land interiaty traffic in USA (%)

Carrier Fright % Passenger %

1916 1939 1950 1916 1939 1950

Railways 77 62 58.7 98 8.6 8.3

In land water ways 18.4 17.9 16.2 2 0.5 0.3

Petroleum pipelines 4.4 9.5 0.7 --- ---- ----

Highways --- 10 12.4 --- 90.7 89.4

Airports --- --- 0.03 --- 0.2 2




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Railways Track Elements

The permanent way of rail road track consist generally of two lines of parallel steel rails which bear and guide the flange wheel of cars and locomotive and which are support on tie plates, ties and ballast. Track gauge = distance between the inside heads of rail 5/8 in below the top of rail Standard Gauge Track on Tangent Gauge can be divided into ; 1)Track gauge and. 2) clearance gauge. a. structure gage. b. loading gauge. Landing gauge: a limit above and to both sides of the track which no part of rolling stock or their loads are allowable to encroach upon. Structure Gauge:

1. Structure Gauge ( I ). 2. Structure Gauge ( II ). 3. Structure Gauge (III ).

Structure gauge is the limit above and to both side of track within which permanent structure or installation is permitted to be built. Structure Gauge I: no structure / installation of any type is permitted within this gauge. This gauge is the exceptional minimum clearance. First isolated structure like signal in between track where structure gauge I cannot be followed Structure Gauge II: recommended minimum clearance for isolated structures such as signals, columns and posts. It is the exceptional minimum clearance for other structure. Structure Gauge III: it is recommended within clearance for all other structures. Structure Gauge in Tunnels: this shall be determined individually for each case taking into account all relevant factors such as speeds, No. of tracks, (both present and future), length of tunnels, geotechnical features, …. etc.

Gauge (mm) % Length of World Network

1000 9

1067 7.8

1435 61.6

1524 8.8

1600 1.3

1667 6.1

21 other 5.4

∑ = 100%

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1- BALLAST

1. General. Ballast is a select material placed on the subgrade to: a. Restrain the track laterally, longitudinally, and vertically under the dynamic loads imposed by trains and the thermal stresses induced in the rails by changing temperature. b. Provide adequate drainage of the track. c. Distribute the load of the track and trains to prevent overstressing the subgrade.

1- Distribute loads uniformly over the subgrade. 2- To hold the track structure to line and grade, (prevent or reduce possibility of buckling. 3- To reduce the excess pore water pressure developed in clay subgrade. 4- To provide good drainage of track structure. 5- To prevent dust. 6- To prevent growth of brush and weeds. 7- Frost heavy is reduced. 8- Maintenance operation are facilitated صعتح سفع اىطأمثشعىح سفع اىحجش

Irregularities in surface can be corrected by tamping ballast under the ties and the track can be shifted to restore line Repeated load – fatigue – replacement of tie and sleepers.

9- Resilience to absorb shock from dynamic loading is formed in ballast. The rough particle interlocked together will move a slight amount giving to a limited extent a spring – like action to the ballast as it support the imposed loads.

2. Selection criteria.




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a. Considerations for selecting materials to be used as ballast include: (1) Size and gradation. (2) Shape (angularity). (3) Weight. (4) Strength. (5) Durability (6) Cleanliness. (7) Economics.

b. New ballast materials used in the maintenance shall meet the requirements specified in the AREA Manual For Railway Engineering, chapter 1, part 2, for the gradation requirements given in table 4-1.( Ballast Specifications):

Fine grades of crushed stone ballast (5 degree)

Size No.

Nominal Size Square

opening

Percentage Passing by Weight

3 in 2 ½ in

2 in 1 ½ in

1 in ¾ in ½ in

3/8 in

No. 4

No. 200

24 2 ½ - ¾ in 100 90.1 --- 25.6 --- 0.1 0.5 --- --- ---

3 2 – 1 in

4 ½ - ¾ in --- --- 100 90.1 20.55 0.15 --- 0.5 --- ---

5 1 – 3/8 in

High quality is recommended for heavy loads, continuously welded rails and concrete ties. Railway : Grading Specification for Subballast

Gravel Size 2 in 1 in 3/8 in No. 10 No. 40 No. 200

% Passing 100 90-100 50-84 26-50 12-30 0-10

3. Types of Ballast 1. Crushed stone. 2. Prepared gravel. 3. Patron gravel. 4. Sand. 5. Other material.

Crushed Stone Ballast:- best of available ballast. It is probably the most expensive in first cost. Result from:

Granite and limestone may be used when crushed to size varying from (3/4) in to 3(1/2) in




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Crushed stone ballast is expensive especially when much transportation is involved.

Tire life is prolonged by the use of ballast Prepared gravel : pit nam gravel displays varying qualities By washing and screening excessively fine and silt materials, a ballast of uniform quality and characteristic is obtained. Large pieces are crushed to proper size and deleterious are removed. Gravel lacks the strength and durability of stone and can not ordinary be cleaned. Easy to work with frequently relatively inexpensive. Pit rum Gravel :- pits with reasonably uniform quality material and free from deleterious substance. (minimum cost) Sand : sand can be used in the same location as cinders. Cinders are suitable for yards and sidings, light traffic, branch lines and may also be used as subballast. It lacks stability, particularly if the particles are rounded or posses excess moisture and will blow and drift in windy areas. It use is largely limited to secondary lines. Even soil can be used in emergencies when other materials are not available or when the track under construction is temporary and subject to little use and light loading 4. Maintenance. The ballast section should be clean, free draining, and free of vegetation, soil (mud), and other foreign materials. b. During major maintenance or track rehabilitations, dirty or fouled crushed stone or slag ballast meeting the requirements may be cleaned or reconditioned and reused. c. Ballast materials shall not be allowed to cover or be at a level above the top of the ties. Ballast subgrade system:- according to Telbots formula

𝑕𝑚𝑖𝑛 . = [16.8 𝑃𝑡

𝑃𝑠]4/5

Where hmin.= Minimum depth of ballast (and sub ballast if any below the bottom of the tie and over the subgrade (in).

Pt = Allowable bearing pressure over the beneath area of the tie (psi). Ps = Bearing capacity of subgrade . Assume

Ps = 20 psi (average allowable bearing capacity of a normally fine subgrade soil) Pt = 65 psi (maximum desirable unit tie pressure especially for ballast at less that top quality) Pt = 85 psi for concrete ties is allowable

For Pt = 65 psi, Ps = 20 psi then 𝑕𝑚𝑖𝑛 . = [16.8∗65

20]

4

5 = 24.5 𝑖𝑛

For Pt = 85 psi, Ps = 20 psi then 𝑕𝑚𝑖𝑛 . = [16.8∗85

20]

4

5 = 30.41 𝑖𝑛

𝑃𝑡 = 𝑃 (𝑤𝑕𝑒𝑒𝑙 𝑙𝑜𝑎𝑑 )

𝑇𝑖𝑒 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑎𝑟𝑒𝑎

Where Ps,bc = Pressure in pound per square inch under the tie center line. Pt = Uniform distributed pressure over the tie face.

subballast حغابأسدخ إرا -1

a. دساعح اىرقح ىيرشتحBearing Capacity b. حغاب اىغطاء فقSubgrade

.خرثش Bearing capacity for subballast أعطاء تالإنا -2 . ىيرشتحbearing capacity اىخرثش اجا دsubballast غرطع تعذا حغاب غطاء فق -3 . ىيرشتح subgradeغرطع تعذا حغاب غطاء اىطيب فق -4




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Ballast depth equal to tie spacing is thus a desirable minimum. AASHTO 1986

SN = ∑ ai mi Di mi = Drainage factor Dj = Quality of drainage granular material (permeable).

2-TIES

1. General. a. The functions of a tie are to:

(1) Maintain gage. (2) Maintain surface. (3) Maintain alignment. (4) Distribute the load from the rail to the ballast and subgrade.

The inability of a tie to adequately perform any of the above functions constitutes a defective tie. b. Tie selection and treatment. (1) Tie selection. New ties selected for use in the maintenance of Army track shall meet the requirements specified in the AREA Manual For Railway Engineering, chapter 8, part 1 for 6-inch for ties are the following hardwoods: Red Oak, White Oak, Hickory, Ash, Beech, and Gum. Where softwoods are used, the Pine and Fir species are (2) Treatment. Ties shall be pressure treated in accordance with the most current version of American Wood-Preservers Association (AWPA) Standard C6. As a general recommendation, the preferred preservative for ties is a creosote-coal tar solution (60/40) as specified in AWPA Standard P2, “Standard for Creosote and Creosote Solutions”. (3) Switch ties. It is recommended that switch ties be hardwood. (c). installation. Ties shall be installed perpendicular to the rails and properly tamped and spiked. Ties shall be installed with the top of the tie (or the tie plate) in full contact with the base of the rail and the bottom of the tie near the rail seat in full contact with the ballast. 2. Identification of defective ties. a. Defective ties. A tie is defective if it is:

(1) Broken through. (2) Split or otherwise impaired to the extent that it will not hold spikes or other rail fasteners. (3) So deteriorated that the tie plate can preferred. move laterally more than ½ inch relative to the

crosstie. (4) Cut by the tie plate more than 2 inches. (5) Cut by wheel flanges, dragging equipment, fire, etc., to a depth of more than 2 inches within 12

inches of the base of the rail, frog, or load-bearing area. (6) Rotted, hollow, or generally deteriorated to a point where a substantial amount of the material is

decayed or missing. These defects are shown in figure (1).

Classification by use (sleeper) 1- Plain tie (intermediate tie) 2- Joint ties in track (ties occurring under rail joints are referred as joint ties). The best

ties immediately available should be placed under joints 3- Bridge ties are longer and heavier usually 12 in wide, 14 in deep and 10-12 in long.




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4- Switch ties are placed under turn cast, crossover and crossing Timber sleepers: shall be of hard wood , Basic dimensions:

Length 260 cm Width 26 cm intermediate sleeper Height 15 cm

Steel Sleeper (advantage): Steel ties: - it gives good an coverage to the track because ballast is packed inside the hollow shell as well as around it. Steel ties are free from decay and insect attack.

Disadvantage : they are subjected to rust and wear at the rail seat (mechanical wear) b. Improper tie support (down or hanging ties). Ties that do not support both rails are considered defective. If these down ties are not materially defective , they shall be tamped up and re-spiked to fully support the rails. 3. Tie requirements. a. Consecutive defective ties. The occurrence of consecutive defective ties in Categories A and B track requires operating restrictions as specified below:




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b. Joint ties. All joints shall be supported by at least one non-defective tie whose centerline is within 18 inches of the rail ends as shown in figure 5-2. At any location where a rail joint is not supported by at least one non-defective tie, operations shall not exceed 10 mph. c. Tie spacing. rail length, the desired spacing should be established during the next major maintenance cycle. For track constructed with an average tie spacing greater than 22 inches, the desired spacing should be established during the next track rehabilitation. d. Missing or skewed ties. Missing or skewed (crooked) ties are undesirable in track. At any location where the center-to-center tie spacing measured along either rail exceeds 48 inches, operations shall not exceed 10 mph until additional is provided, or skewed ties are straightened to reduce the spacing. Slightly skewed ties should be straightened during the next track rehabilitation.

3-TIE PLATES, RAIL FASTENINGS, AND OTHER TRACK MATERIALS

1. General. a. Tie plates, rail fastenings, and other track materials shall be the proper size and type as specified in

TM 5-850-2/AFM 88-7, Chap. 2 and TM 5- 627/AFM 91—33 and shall conform to the requirements of chapters 4 and 5 of the AREA Manual For Railway Engineering.

b. Tie plates, rail fastenings, and other track materials shall not be flame cut or otherwise al tered. c. Tie plates, rail fastenings, and other track materials which are of improper type, broken, or

otherwise defective shall be replaced with the proper size and type material. d. Function of Tie Plate:-

1) Provide a layer bearing surface on the tie, thus reducing the intensity of bearing pressure. 2) Protects the tie a pressure longitudinal and lateral moments of the rail – base. 3) Makes spikes more effective in the holding the rail to gauge.

2. Tie plates. a. Use.

Tie plates distribute the applied loads from the rail to the tie as well as assist in keeping the rail in position. Their use is especially important on curves where they provide additional lateral restraint.

b. Type. Tie plates may be of either the single shoulder type (fig 6-1) or the double shoulder type (fig 6-2).

Classification by material (sleeper):- 1) Wood tie or timber sleeper, usedly treated by a

creosote- oil and zine clloride presenmtive to retord decay and insect attack

2) Steel ties 3) Prestress

Classification by use (sleeper) 1- Plain tie (intermediate tie) 2- Joint ties in track (ties occurring under rail joints are referred

as joint ties). The best ties immediately available should be placed under joints




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e. Installation. For track without tie plates, plates shall be installed during the next track rehabilitation or rail replacement.

3. Spikes a. The rail shall have a sufficient number of fasteners (spikes) to effectively maintain gage and provide sufficient rail restraint. b. Spikes shall be: (1) Of proper size for the tie plates used. (2) Driven vertical and square with the rail. (3) Either of the cut or screw type. (4) Driven with approximately 1.875 inches of space remaining between the head of the spike and the base of the rail. c. Spiking pattern. (1) On tangent track and curves less than 4 degrees, spikes shall be installed as shown in figure 6-3. (2) On curves 4 degrees and greater and on the curved side of turnouts, the use of an additional spike as shown in figure 6-4 is recommended.




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(3) Spikes in angle bars. Spikes shall not be installed through the slots in skirted-type, slotted joint bars (angle bars) as shown in figure 6-5. (4) Correction on incorrect spiking patterns. An incorrect spiking pattern by itself is not a cause for removing and re-driving spikes. However, an incorrect spiking pattern shall be corrected when other maintenance requires the spikes to be removed. Old spike holes shall be plugged to prevent accelerated tie deterioration.

f. Missing and loose spikes. Missing spikes shall be replaced. Old spike holes shall be plugged with a treated tie plug prior to replacing the spikes. Loose spikes shall be removed, the holes plugged, and the spikes re-driven.

g. Spikes beneath rail base. Where the head of a spike has become lodged beneath the base of a rail, the spike shall be removed and properly re-driven. A spike lodged beneath the rail base is an undesirable condition that can lead to a broken rail base. 4. Joints. a. Joint bars. Rails shall be joined with proper factory designed and constructed joint bars. Joint bars may not be altered with a flame in any manner, including the bolt holes. Joint bars not meeting these requirements shall be replaced. b. Compromise joints. Rails of different size or section shall be joined with proper factory designed and constructed compromise bars, taper rails, or offset welds. c. Cracked or broken joint bars. Cracked or broken joint bars shall be replaced.

(1) If one joint bar at a rail joint is cracked or broken between the center holes, operations over that location shall not exceed 10 mph.

(2) If both joint bars at a rail joint are cracked between the center holes, operations over that location shall not exceed 5 mph. (3) Operations shall not be permitted over any location where both joint bars are broken, or where worn or loose joint bars allow vertical movement of either rail with respect to the other.

d. Bolts. Each joint shall be bolted with at least two bolts in each rail (1) All bolts shall be of proper size and tightly in place. A bolt is not considered tight unless the spring

washer is compressed closed and

is flat against the joint bar, as

shown in figure 6-6.




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(2) Bolts which cannot be tightened shall be replaced. Missing bolts shall be replaced. (3) Bolts shall be installed with spring lock washers. Nuts shall be installed against the spring

lock washer as shown in figure 6-6. At least one full bolt thread shall extend past the outside of the nut. (4) Bolts shall be installed so that the nuts will be alternately on the inside and outside of the rail as shown in figure 6-6. (5) Whenever bolts and joint bars are removed, the rail in the joint area and the contact surfaces of the joint bar shall be cleaned (wire brushed) and lubrication applied to the joint bar and the bolt threads. New joint bars and bolts shall be lubricated before installation. e. Loose and missing bolts. (1) If all bolts at a joint are loose or if there is only one bolt through each rail, operations shall not exceed 10 mph. (2) Operations shall not be permitted over locations where all bolts in one rail are missing. f. Rail end mismatch.

(1) At any location where rail end mismatch exceeds 3/16 inch on the tread portion or gage side of the rail, operations shall not exceed 10 mph.

(2) Operations shall not be permitted over locations where the rail end mismatch exceeds ¼ inch. g. Joint gap. (1) At any location where the gap between rail ends exceeds 1 inch, operations shall not exceed 10 mph. (2) Operations shall not be permitted over locations where the gap exceeds 2 inches. h. Joints in restricted areas. Whenever a joint is installed within 20 feet of a road crossing, the outer perimeter of any structure, or any location which restricts access to the joint, it is recommended that the joint be welded. 5. Rail anchors. a. Rail anchors help prevent the longitudinal movement of rails commonly known as “running” or “creeping”. Rail anchors should be used at locations where the track is subject to serious movement from rail expansion or traffic conditions. b. Rules for anchor application. General rules on the use of rail anchors are: (1) Anchors shall be applied to the gage side of the rail against the same tie face on opposite rails. (2) Anchors shall grip the base of the rail firmly and have full bearing against the face of the tie. (3) When the bearing of the rail anchor against the tie has been disturbed by removal of the tie, the anchor shall be removed and reset. (4) Anchors shall not be moved by driving them along the rail. (5) Skewed ties shall be straightened before applying rail anchors. c. Anchor locations.




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(1) Where used, a minimum of eight anchors per 39 foot of rail is recommended as shown in figure 6-7. Additional anchors should be used as needed, and they should be installed four per tie with approximately uniform spacing along the rail.

(2) At open deck bridges. Where anchors are used on track approaching open deck bridges, every third tie should be box anchored (four anchors per tie) for at least two rail lengths off each end of the bridge. No anchors shall be applied on the bridge itself. (3) At rail crossings. Where anchors are used on track approaching rail crossings, every third tie should be box anchored (four anchors per tie) for at least two rail lengths in all directions from the crossing. 6. Gage rods. a. Use. Gage rods are used to help maintain proper track gage but are not a substitute for good track maintenance and good tie conditions. Gage rods are sometimes used at the following locations: (1) On sharp curves where there is difficulty holding the gage. (2) In turnouts just ahead of the switch points and on the curved closure rails. b. Spacing. Where gage rods are used in sharp curves, two to four rods should be installed for each rail length. Rods should be installed at evenly spaced intervals along the rail length. c. Application. Gage rods should be installed at right angles to the rail with the jaws firmly gripping the base of the rail. d. Maintenance. (1) Gage rods shall be kept tight while maintaining the proper track gage. (2) Bent or broken gage rods shall be replaced where the track conditions warrant their continued use.

4- RAIL

1. Defective rail and remedial actions. a. Remedial actions for rail defects are presented in table 7-1. Where rail defects have been identified but

remedial action has not been completed, the operating restrictions presented in table 7-1 shall apply. Appendix B provides brief descriptions of the common rail defects that may be observed in track.

b. Multiple defects. Any individual rail having two or more of the fissure or fracture type defects listed in table 7-1, whether they are the same or different, shall be removed and replaced. c. Worn rails.




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On rail suspected of being worn more than the allowances provided for in table 7- 1, wear measurements shall be taken at the center and at each end of the rail not more than 1 foot from the end of the joint bar. Rail wear measurements shall consist of a vertical head wear measurement and a side wear measurement as shown in figure 7-1. Table C-1 presents a table of details and properties for various rail sections. This table may be used to assist in identifying rail sections and estimating the amount of rail wear. d. Base corrosion. Rail shall be removed from track if the base is corroded more than ¼ inch as shown in figure 7-2. e. The rail is referred to by the weight (Ib) per yard and the design of the cross section. For example:

132 RE

132 Ib/yard Designated for the rail is designed by AREA




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Properties of Typical Rail Sections Section Weight per

yard Ib Moment of inertia in4

Base to natural axis Yr

section modulus base in3

Area in2 Height in Base width in

132 RE 133.4 86 3.2 27 13.08 7 1/16 6 3

100 RE 101.5 49 2.75 17.8 9.95 6 5 3/8 2 11/10

155 Ps 155 129 3.38 36.7 15.2 8 6 ¾

Ps: significance Pennsylvania rail road design




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UCI 60 ف اىطشق اىاصفاخ ف اىقد اىحاظش ثو خػ عناشاخ غرعي

UIC 60 60.4 kg/m

Area 76.6 cm2

Height 172mm

Base width 150 mm

Bed width 72 mm

Moment of inertia 3055 cm4

6. Length of Rails 1) Jointed rails shall not be less than 15 m long (the preferred length being 36 m different in

transportation). 2) Welded rails no joint.

e. End batter. Rail end batter is measured ½ inch from the rail end with an 18-inch straightedge laid only on the rail being measured as shown in figure 7-3. Table 7-1 presents limits and remedial actions for end batter. f. Running surface damage. Rail running surface damage, such as deep engine burns, dents, etc., is measured at the midpoint of an 18-inch straightedge laid on the railhead over the defect. Table 7-1 presents limits and remedial actions for rail surface damage. 2. Rail maintenance. a. Rail. New rail used in the maintenance of Force track shall meet the requirements specified in the AREA Manual For Railway Engineering, chapter 4. b. Internal defect inspection. A “continuous search” internal rail defect inspection shall be performed at least once every 3 to 6 years on Categories A and B track. This internal rail defect inspection should be conducted using ultrasonic inspection techniques. Table 7-2 presents the recommended frequency for rail defect inspections.




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c. Torch cut. Rail shall not be flame cut in any manner. This includes cropping of the rail end, Rail shall be cut using a rail saw or other appropriate cutting tool. d. Short rail. Rail less than 13 feet in length shall not be installed in track. 3. Lightweight rail. is defined as rail weighing less than 90 pounds/yard. Research has shown that lightweight rail may not be suitable for use in track which is subjected to heavy wheel loads. a. Rail weights of 70 pounds/yard or less should be replaced if that rail is to experience car loads of more than 50 tons (25,000 pounds axle loads). b. Rail weighing 75 to 85 pounds/yard may be adequate depending upon support conditions. A structural evaluation and stress analysis is necessary to determine the adequacy of these rail weights. Rail not adequate to support the desired wheel loads should be replaced. c. The existence of lightweight rail in Categories A and B track should be considered when planning major repair and/or rehabilitation projects. 4. Function of Rails

1- The rail resists high contact pressure hence the rail; should be hard 2- Capable of distributing load 3- Rail should be able to do so as that breaking ( flexural and fatigue strength for repeated loading ,

sufficient strength fatigue) : عما ذرطية ذشمة أعلا اىطاىةجع

5. Chemical Composition of Rails Rails are made of steel and the basic element in steel is iron. With iron are combined small quantities of carbon, magnetic, silicon and less desirable sulfur and phosphors Carbon add hardness (very high contact stress) standard carbon rail steel has a brinall hardness of (250-278). High strength steel may be have brinell hardness (321-388). Manganese give strength and toughness (the manganese content in intermediate alloy steel may be 1.3-1.6% Silicon its high affinity for oxygen aids in remaining gases during the pouring and rolling processes

ذؤخز ىغشض ( x-rays) ظش اىخيو ا اىيحظعف فقاعاخ فيزىل ن اىقطع أشناه اىصة ذحر عي اء غاصاخ عي أثاء اىعاد

قيو احراه جد اىفقاعاخ اىائح ف اىقطعالأمغج رص اىغين جد فقاعاخ

Phosphors : makes steel brittle (sudden failure) and likely to break under impact Sulfur : causes breaks during the rolling process

.س ذشققاخ ف اىعادعفف صاعح اىحذذ غثة اهأثاء

5-TURNOUTS

1. General. Turnouts are designed to divert trains from one track to another. Good turnout maintenance is essential for the safe and efficient operation of trains. Major components of a turnout are the switch, frog, and guard rails. Figure 8-1 shows a typical turnout with the various parts identified. 2. General requirements Materials. All materials used within the limits of a turnout shall: (1) Be factory designed and constructed. (2) Be the proper type and size. (3) Be properly installed.




20

(4) Not be flame cut or otherwise altered. b. Rail. All rail used within the limits of a turnout shall be of the same weight and section. Compromise joints are not permitted within the limits of a turnout. c. Ties. The standards in chapter 5 of this manual shall apply to ties within the limits of a turnout. d. Track geometry. Turnout track geometry shall conform with the standards in chapter 12 as well as those in this chapter. e. Reversing tangent. It is recommended that the tangent between the frog and any reverse curve past the frog be no less than 50 feet in length as shown in figure 8-2.




21

. f. Switch stand placement. (1) The switch stand stall be installed so that when the switch is lined for the normal (main) route, the connecting rod keeps the points closed with a pulling (rather than a pushing) force. In most cases this will mean installing the stand on the diverging side of the turnout as shown in figure 8-2. (2) When a ground-throw stand is used, the handle shall point toward the frog when the switch lined for the normal (main) route. g. Cleaning, lubrication, and adjustment. Switches and switch stands shall be cleaned, lubricated, and adjusted at least annually. 3.Switches and switch stands. a. Switches shall be inspected for the following defects:

(1) Switch difficult to operate. (2) Gap greater than 1.875 inch between the switch point and the stock rail. Checked with points in

both positions. (3) Worn or chipped switch points. (See fig 8-3.) (4) Point of switch higher than stock rail. See fig 8-4.) (5) Point rail beyond taper lower than stock rail. (See fig 8-4.) (6) Damaged or missing switch stand lever cracked or improper heel joint bars or heel filler. (7) Insecure, loose, damaged, or improperly installed switch stand. (See 8-2g) (8) Loose, damaged, or missing jam nut at the switch stand end of the connecting rod. (9) Bent, damaged, loose, binding, or improperly installed connecting rod, switch rods, or switch clips. (10) Loose, damaged, or missing switch clip, switch rod, or connecting rod bolts. (11) Loose, damaged, or missing heel bolts; (12) Loose, damaged, or missing rail braces.

(13) Loose, damaged, or missing slide latches or switch point lock. plates; dirt and debris buildup on slide plates. (14) Missing cotter keys on switch rod and switch clip bolts. (15) Debris in flange ways. (16) Debris obstructing switch rods and connecting rod.




22

b. Switch point gap. (1) If the switch can be thrown and locked in either direction with a 1.875 inch spacer between the switch point and the stock rail, operations through the turnout shall not exceed 10 mph. (2) If the switch can be thrown and locked in either direction with a ¼ inch spacer between the switch point and the stock rail, operations through that side of the turnout shall not be permitted. c. Switch points. A switch point shall be restored or replaced if the point is chipped, broken, or worn more than ½ inch down and 6 inches back from the point. (See fig 8-3). Metal flow shall be removed to ensure proper closure. If a switch point is worn or damaged beyond these limits, operations through the turnout shall not exceed 10 mph. d. Figure 8-4 shows the proper elevation of the switch point and point rail with respect to the stock rail. (1) If the top surface of the switch point is higher than the top of the stock rail, operations through the turnout shall not exceed 5 mph. (2) If the point rail beyond the taper is lower than the stock rail, operations through the turnout shall not be permitted. e. Switch stand lever latches and point locks. Switch stand lever latches or switch point locks shall be installed on all switches. Where latches or locks are missing, damaged, insecure, or otherwise inoperative, operations through the turnout shall not exceed 10 mph. f. Switch stand. The switch stand shall be fully secured to the head block ties to prevent any unintentional movement of the switch points. Where operations through the switch result in visible lateral movement of the switch stand or opening of the switch points (point gap), operations through the turnout shall not be permitted. g. Connecting rod, switch rods, and switch clips. (1) These parts shall be installed and maintained to allow unobstructed motion when the switch is thrown. Rod ends and clips shall not contact adjacent ties. Damaged parts shall be replaced, and improperly




23

installed parts shall be adjusted. Washers or similar spacers shall not be permitted between the switch clip and the switch point. (2) If the connecting rod, switch rod, or switch clip is insecurely fastened or is damaged, operations through the turnout shall not be permitted. (3) The jam nut at the switch stand end of the connecting rod shall be kept tight against the switch stand clevis. (4) Where washers or similar spacers are between the clip and switch point, operations through the turnout shall not exceed 10 mph. h. Connecting rod bolts, switch rod bolts, and clip bolts.

Connecting rod and switch rod bolts shall be installed with the nut on top and cotter keys in place. Clip bolts shall have cotter keys. All bolts shall be kept tight. i. Switch heel (bolts, fillers, and joint bars).

The heel of the switch shall be secure and the supporting switch ties solidly tamped. The inside joint bar (nearest and track center) should be a bent bar per AREA design. If any heel bolts are loose or missing, or the heel is otherwise not fully secure, operations through the turnout shall not exceed 5 mph.

j. Rail braces. (1) Rail braces are essential to provide proper lateral support to the stock rails. Rail braces shall be fully secured to the tie and tight against the outside of the stock rail on both sides of the turnout. (2) It is recommended that rail braces be in stalled on each tie from the point of switch to within two ties of the switch heel. (3) If there are less than four rail braces fully secured to the tie and tight against the outside of each stock rail, operations through the turnout shall not exceed 5 mph. k. Debris in flange ways. Flange ways shall be kept clear of debris. Any obstructions, including ice and packed snow, shall be removed. 4. Frogs. a. Frog point. A frog shall be restored or replaced if the point is chipped, broken, or worn more than ½ inch below the original top surface and 6 inches back from the original point location.(See fig 8-5) b. Frog surface. A frog shall be restored or replaced if the tread surface is worn more than 3/8 inch below its original contour. (See fig 8-5) c. Guarding face of self guard end frogs. The raised guarding face on a self-guarded frog shall not be worn more than 3/8




24

inch. (See fig 8-6) d. Repairing self guarded frogs. If repairs are made to a self-guarded frog without removing it from service, the raised guarding face must be re- restored before rebuilding the point. e. Frog bolts. All frog bolts shall be in place and tight. f. Flange way width. Standard frog flange way width is 1 / 8 inches. Where existing flange way width is less than this amount, the restored striation in table 8-1 shall be applied. (See fig 8-7)




25

g. Flange way depth. Standard frog flangeway depth is 1.875 inches. Where existing flangeway depth is less than this amount, the restrictions in table 8-1 shall be applied. (See fig 8-7) 8-5. Guard rails. a. Guard rails shall be in place, properly positioned, and fully secured. b. Guard rails shall be installed so that the straight guarding face (the portion of the guard rail parallel with the running rail) extends in advance of the frog point a distance at least equal the values given in table 8-2. c. Guard check gage. Standard guard check gage is 54% (54.625) inches. Where guard check gage measures less than this amount, the restrictions in table 8-1 shall be applied. (See fig 8- d. Guard face gage. Standard guard face gage is 53¾ (52.75) inches. Where guard face gage measures more than this amount, the restrictions in table 8—1 shall be applied. (See fig 8—7) e. Flangeway width. Standard guard rail flange way width is 1.875 inches. If the flange way width is less than this amount, the restrictions in table 8-1 shall be applied. (See fig 8-7).

6-RAIL CROSSINGS

9-1. General. Rail crossings are designed to carry one track across another at grade. 2. Requirements. a. Size. Rail crossings shall be the proper size and section for the rails being joined. b. Flange way width. Standard flange way width for rail crossings is 1.875 inches. Where existing flange way width is less than this amount, the following restrictions shall be applied. (1) If the flangeway width is 1% (1.625) inches or less, operations through the crossing shall not exceed 10 mph.




26

(2) No operations shall be permitted through any crossing where the flangeway width is (1.50) inches or less. c. Flangeway depth. New or fully restored flangeway depth for rail crossings is at least (1.875) inches. Where existing flangeway depth is less than this amount, the following restrictions shall be applied. (1) If the flangeway depth is (1.50) inches or less, operations through the crossing shall (2) No operations shall be permitted not exceed 10 mph. through any crossing where the flangeway depth is 1% (1.375) inches or less. d. Debris in flangeways. Flangeways shall be kept clear of debris. Any obstructions, including ice and packed snow, shall be removed. e. Bolts. All crossing bolts shall be in place and tight. Loose bolts shall be tightened, and bolts which cannot be tightened shall be replaced. Missing bolts shall be replaced. 3. Anchors. Where rail anchors are used on track approaching rail crossings, every third tie should be box anchored (four anchors per tie) for at least two rail lengths in all directions from the crossing

7-ROAD CROSSINGS

1. Drainage. a. Adequate drainage is essential for satisfactory long-term performance of the track and road crossing. b. Water shall not be allowed to pond on or near the track at a road crossing. c. Catch-basins, gutters, ditches, subdrains, and culverts should be properly installed and kept free of debris. 2. Flangeways. a. Flangeway width. For normal operations, flangeway width in a road crossing shall not be less than (2.50) inches nor greater than 3 inches. (1) If the flangeway width is less than (1.7 5) inches, operations through the crossing shall not exceed 10 mph. (2) No operations shall be permitted through any crossing where the flangeway width is (1.50) inches or less. b. Flangeway depth. For normal operations, the flangeway depth in a road crossing shall not be less than (2.00) inches. (1) If the flangeway depth is less than (1.50) inches, operations through the crossing shall not exceed 10 mph. (2) No operations shall be permitted through any crossing where the flangeway depth is 1% (1.375) inches or less. c. Debris. Flangeways shall be kept clear of debris. Any obstructions, including ice and packed snow, shall be removed immediately. 3. Track. a. Ties, tie plates, and spikes. When crossings are rebuilt, all ties within the crossing limits and for at least 20 feet beyond each end of the crossing shall be replaced, fully tie plated, and spiked with eight rail-holding spikes on each tie. (See fig




27

10-1) For road crossings carrying large volumes of traffic and/or heavy loads, the use of hardwood ties is recommended. b. Tie pads. When crossings are rebuilt, it is recommended that rubber tie pads be installed between the tie and tie plate on all ties within the crossing area and for at least 20 feet beyond each end of the crossing.

c. Joints. Bolted rail joints are not desirable in road crossings. When crossings are rebuilt, it is recommended that all joints within the crossing and up to 20 feet outside the crossing be welded. d. Rail anchors. Where the track on either side of the crossing is anchored, it is recommended that the anchoring pattern be continued through the crossing. 4. Crossing surfaces and materials. a. It is essential that the crossing surface be maintained to provide a smooth crossing for vehicles and to

prevent vehicle tires from striking the rails. The crossing surface shall be maintained at an elevation level with or slightly (not more than ¼ inch) above the top of the rails. Additionally, there shall be a smooth transition between the crossing surface and the adjoining pavement.

b. During routine track inspections the inspector should take note of the general condition of the crossing materials and report any damage or condition requiring repair or replacement. Any condition observed in a road crossing which would cause a hazard to motor vehicles using the crossing should be corrected immediately.

5. Crossing protection. a. Signs and signals.

During routine track inspections the inspector should observe the condition of all whistle posts, highway

warning signs, and signals. Signs and signals should be easily legible and clearly visible.

(1) Signs and signals shall confirm to the requirements of the Manual On Uniform Traffic

Control Devices (MUTCD), part VIII, “Traffic Control Systems for Railroad-Highway Grade

Crossings”.

b. Electric/electromechanical signals. Inspection and maintenance of electric and/or electromechanical

signals should conform to manufacturers recommendations, state/local requirements, and the minimum

requirements given below.




28

(1) Inspection of electric/electromechanical signals shall be performed at the same frequency as track

inspections ; however, inspections of signals shall be performed not less than once per quarter (at

approximately 3-month intervals).

(2) Minimum inspection procedures for electric/electromechanical signals are as follows:

(a) Open case.

(b) Turn on signal with switch.

(c) Make sure signal is working properly and all bulbs are burning.

(d) Check to make sure lenses are clean. Clean lenses if needed.

(e) Check power source for obvious problems.

(f) Check charge on standby batteries.

(g) Note date, time condition (ok, existing problem, repaired, etc), and inspectors initials in a small note

book that is to be kept in the signal case.

(3) At least once per quarter a detailed inspection of the signal shall be performed. The detailed inspection

shall include all of the requirements of 10-5b(2) plus the following:

(a) Check all rail bonds.

(b) Check all connections.

(c) Measure and record voltages at critical locations (relays).

c. Any deficiency or defect in road crossing

warning signs or signals shall be repaired immediately.

8-CLEARANCES




29

1. Measurement.

Vertical clearance shall be measured vertically from the top surface of the rail. Side clearance shall be

measured horizontally from the centerline of the track.

2. Clearance requirements, tangent track.

Clearances for tangent track shall not be less than those listed in table 13-1 and shown in figure 13-1.

3 Clearance requirements, curved track.

a. For each degree of curvature, side clearances shall be increased 1½ (1.50) inches over

that required in table 13-1 and figure 13-1.

b. When an obstruction is located adjacent to tangent track but the track is curved within 80 feet

of the obstruction, the side clearances shall be increased by the following amounts:




30

4. Track centers.

The minimum spacing between the centerlines of adjacent tracks shall be maintained as follows:

5. Changes to clearances.

Whenever changes in alignment or elevation are made, clearances shall be checked for compliance with

the criteria given in this chapter. This is especially important in the vicinity of buildings, bridges, overhead

structures, platforms, and tunnels.




31




32




33




34




35




36




37




38




39




40

Buckling of Railway Tracks

Cross – Ties

Early analysis of timber ties under load assumed a uniform distribution of ballast pressure across the length and width of the tie. Ballast pressure = w/b ( b; tie width). W = 2p/L Ballast pressure = 2p/Lb Mr = Bending moment at rail seat. Mc = Bending moment at center of tie sleeper.

𝑀𝑟 = 2𝑝

𝐿

𝐿22

2=

𝑝𝐿22

𝐿

𝑀𝑐 = 2𝑝

𝐿

𝐿

4

𝐿

2− 𝑝 ∗

𝐿1

2

𝑀𝑐 = 𝑃

4 𝐿 − 2𝐿1 =

𝑝

4 𝐿1 + 2𝐿2 − 2𝐿1 =

𝑝

4 2𝐿2 − 𝐿1 = −

𝑝

4 [𝐿1 − 2𝐿2]

Example :- Calculate the permissible tie axis load for the case of wood tie:

1- 7in*9in*8.5ft in standard gauge line knowing that the maximum allowable bending stress in wood tie is 1200 psi.

2- 7in*9in * 8ft 1) L1= 60 in ( standard gauge tie ), L = 8.5 ft, L = 2L2 + L1

𝐿2 =𝐿−𝐿1

2=

8.5∗12−60

2= 21 𝑖𝑛

Uniformly distributed Pressures

Uniformly Distributed Pressures

L2 L2 L1

L=L1+ 2L2

P( Wheel load) P( Wheel load)

W=2P/L ( line load ) 2P axle load

L ; length of tie.

L1 ; Distance between beamy points of the two wheels on the rail head (60 in) for standard gauge railway line.

c ( Center of seat) r ( Rail seat) r ( Rail seat)

P P

L

3 Lecture No.




41

𝑀𝑟 =𝑝𝐿2

𝐿=

𝑝 (12)2

8.5 12 = 4.32 𝑝

𝑀𝑐 = −𝑝

4 𝐿1 − 2𝐿2 =

−𝑝

4 60 − 2 21 = −4.5𝑝

Mc govern the design

𝜎𝑚𝑎𝑥 . = 𝑀

𝐼𝑐 =

𝑀

𝑏𝑑3 12

∗ 𝑑

2=

6𝑀

𝑏𝑑2

1200 =6 4.5𝑝

9 ∗ 72

𝑝 = 19600 𝐼𝑏 ∴ 2𝑝 = 2 ∗ 19600 = 39.2 𝑘𝑖𝑝𝑠 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑡𝑖𝑒 𝑎𝑥𝑙𝑒 𝑙𝑜𝑎𝑑

For case 2: 7in * 9in *8ft

𝐿2 = 𝐿− 𝐿1

2=

8∗12−60

2 = 18 in

𝑀𝑐 = −𝑝

4 𝐿1 − 2𝐿2 = −

𝑝

4 60 − 36 = −6𝑝

𝑀𝑟 =𝑝𝐿2

2

𝐿= 𝑝 ∗

182

96= 3.375𝑝

Mc govern the design

1200 = 𝜎𝑚𝑎𝑥 . = 6𝑀

𝑏𝑑2 = 6(6𝑝)

9(7)2

p = 14.7 kips

Max. allowable tie axle load = 2p = 2*14.7 = 29.4 kips Increase in axle load using 8.5 ft tie (long tie) increase (39.2-29.4)/29.4 = 33% compared with 8 ft tie. Concrete ties: In U.S.A concrete ties were used experimentally as early as 1893. Early concrete ties were reinforced. Monolithic cast and weighted over 400 Ib each 4. Most of these were removed within a few years because:

1- Of corroded fastenings. 2- Concrete cracking and deterioration (shattering under impact and crushing under the rail). 3- Current leakage in bonded territory.

Pre-stress Concrete Ties 1) Pre tensioned ties. 2) Post tensioned ties.

Compressive force transmitted to concrete high compressive strength in concrete ties Dow-Mac ( ( ششمح داك

Hungarian type.

4.32 P 4.32 P

4.5 P

+ +

-

P P

214 mm

264 mm

212 mm

231 mm

264 mm

140 mm

At the end At the center

4 strand 4 strand

each strand contains 7 wires with diameter = 3.3 mm , area of each strand = 0.523 cm2

8 * 0.523 = 4.184 cm2 total area of pre-stressing steel.

174 mm

280 mm

125 mm

202 mm

260 mm

115 mm

At the end At the center

18 strand 18 strand

each strand contains 36 wires with diameter = 3 mm , area of each strand = (3/2)2 *π cm2

36 * (3/2)2*π = 2.54 cm2 total area of pre-stressing steel.




42

Important Features of Pre-stress Concrete Sleeper in Iraq

Feature Sleeper according to Hungarian

Design Sleeper to Dow-Mac Design

for 60 u/c Rail

Length 2420 mm 2515 ± 5 mm

Bottom width at ends 280 mm 246 ± 5 mm

Bottom width at rail scat 280 246

Bottom width at center 260.4 246

Top width at ends 174 mm 214 ± 3 mm

Top width at rail seat 180 216

Top width at center 203 231

Height at ends 125 mm 212 ± 5/3

Height at rail scat 160 203

Height at center 115 140

Maximum height 170 212

Volume of sleeper 0.079 m3 0.111 m3

Weight 196 kg 266 kg

Conc. Strength at transfer ------------ 350 kg / cm2

At 28 days 500 kg / cm2 525 kg/cm2

Steel strength Fs (yield point)

19000 kg/cm2 18000 kg/cm2

Total transfer applied to wires

28.3 tones 52718 tones

Two cases of design of pre-stressed concrete ties 1- Freshly packed tie. 2- Conciliated sleeper

251.5 cm

Distance from center to center or rail seat for standard gauge railway

track =152 cm

Distance from center of rail seat to end of

sleeper = (251.5-152)/2= 49.75 cm

52.5 cm

99.5 cm 99.5 cm P

Conciliated sleepers

P

99.5 cm

Freshly packed tie

49.75 cm

α P




43

Typical design calculation for pre-stressed concrete sleeper (Dow-Mac type) 1) Design Load Locomotive is most critical (in the static condition), axle load 20.2 ton. Max. axle load = 20.2 ton static wheel load =(20.2/2)=10.1 ton Max. static wheel load on any rail seat = 0.6*10.1 = 6.06 ton. Field measurement carried out on various railway system in the world have shown that the max. share coming to any individual rail seat is about 60% of the wheel load. Dynamic increment = 1.5*static=1.5*6.06 = 9.09 ton. Due to the dynamic effect of speed track / vehicle irregularities , etc. , the static wheel load is assumed to increase by 150% under the worst conditions. Max. dynamic load /rail seat = 6.06 + 9.09 = 15.2 ton. Strength of concrete

28 days cube strength Fc = 525 kg/cm2 (min). Safe compressive strength = 0.4*Fc = 210 kg/cm2. Safe tensile strength =0.04*Fc = 21 kg /cm2. Pre-stressing steel. Ultimate tensile strength = 180 kg/cm2. Initial pre-stressed applied = 70% of the ultimate tensile strength = 126 kg/mm2

Max. losses in the pre-stressing in service = 30% losses to of creep of service pre-stress = 126*0.7 =88.2 Sectional Properties Section at center - area of concrete =[(23.1+26.4)/2]*14 = 346.5 cm2 - distance of center of gravity from base = 6.84 cm - distance of center of gravity from top = 7.16 cm - moment of inertia of section about center of gravity Icg = 5651 cm4 - section modulus Z bottom = Icg / 6.84 = 826 cm3

- Z top = Icg / 7.16 = 789 cm3 - center of gravity from base of steel = 7.2 cm - eccentricity = 7.2-6.84 = 0.36 cm above center of gravity of concrete. Section at rail seat

- area of content = [(21.6+ 26.4)/2]*20.3 = 487.2 cm2 - distance of center of gravity from base = 9.81 cm - distance of center of gravity from top = 10.49 cm - moment of inertia Icg = 16675 cm4 - section modulus Z bottom = 16675 /9.81 = 1699 cm3

- Z top = 16675 / 10.49 = 1589cm3 - center of gravity from base of steel = 7.2 cm - eccentricity = 9.81 – 7.2 = 2.61 cm below center of gravity of concrete.

At the center

Center gravity of section

46.5 mm

214 mm

264 mm

140 mm

Center of steel

68.4 mm 72 mm

At the end

231 mm

264 mm

203 mm

Center gravity of section

Center of steel

72 mm 98.1 mm




44

Calculation of stresses due to tire load (wheel loads) For freshly packed tie Ballast pressure ( p) = 15.2 / (26.4*99.5) = 0.00578 t / cm2 = 5.786 kg / m2 Distributed load = 15.2 / 99.5 = kg/cm Bending moment at rail seat Mr

𝑀𝑟 = 15.2 ∗ 26.4

26.4 ∗ 99.5∗ 49.75 ∗

49.75

2−

15.2

35.2∗

35.3

2∗

35.3

4= 121.98 𝑡. 𝑐𝑚

Bending moment at center of tie Mc=0 For consildated sleeper P* (live load) P* (99.5) + 0.4 P* (52.5/2) =15.2 P* = 0.1382 t/cm Ballast pressure =[0.1382/26.4] t/cm2 Bending moment at rail seat Mr = (0.138 *99.5/2*99.5/4)-15.2/2*35.3/4 =1103.93 t .cm Bending moment at center of tie;

𝑀𝑐 = 0.1382 ∗ 99.5 99.5

2+

52.2

2 + 0.4 ∗ 0.1382

52.2

2

52.2

2 ∗ 2 − 15.2

99.5

2+

52.5

2 = −91.1 𝑡. 𝑐𝑚

2*49.75 cm

49.75 cm

99.5 cm

15.2 ton

N.A

35.5 cm

Base of rail

15 cm

h/2 h; height of tie at rail seat = 20.3 cm for Dow- Mc sleeper

N.A

45o

15 + h

121.58 cm

+ +

Mc

15.2 ton

99.5 cm 52.5/2 cm

P P 0.4 P

15.2 ton 15.2 ton

99.5 cm 99.5 cm 52.5 cm

α ; centric binding coefficient = 0.4

91.1 ton/cm

+ +

-




45

Stresses in tie due to live loads:

5. Freshing packed tie at rail seat

𝜎𝑡𝑜𝑝 =121.98

1589= 0.07676

𝑡

𝑚2= 76.7

𝑘𝑔

𝑚2

𝜎𝑏𝑜𝑡𝑡𝑜𝑚 =− 121.98

1699= −0.07179 = −71.79

kg/m2 compression

− 𝑐𝑜𝑛𝑠𝑖𝑙𝑑𝑎𝑡𝑒𝑑 𝑠𝑙𝑒𝑒𝑝𝑒𝑟 𝑎𝑡 𝑟𝑎𝑖𝑙 𝑠𝑒𝑎𝑡 ∶

𝜎𝑡𝑜𝑝 =103.93

1589= 0.00654 = 65.4

𝑘𝑔

𝑚2

𝜎𝑏𝑜𝑡𝑡𝑜𝑚 =− 103.93

1699= −0.06117 = −61.17

kg/m2

6. Center Section

𝜎𝑡𝑜𝑝 =−91.1

789= −115.46

𝑘𝑔

𝑚2 𝑡𝑒𝑛𝑠𝑖𝑜𝑛

𝜎𝑏𝑜𝑡𝑡𝑜𝑚 =91.1∗100

826= 110.3

kg/m2 compression

Stresses due to Pre-stressing Each strand 0.523 cm2 8 strand = 0.523*8=4.184 cm2 Ultimate tensile strength =180 kg/mm2 = 18000 kg/cm2 Applied pre-stress initially = 0.7 * 18000 = 12600 kg/m2 Minimum expected pre-stressing in service = o.7 *12600 =8820 kg/m2 Minimum expected pre-stressing force in service =4.184 *8820 =36903 kg = 36.903 ton

𝜎𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 =𝐹

𝐴 ±

𝐹𝑒

𝑍

F=P

F = Normal force. e = Eccentricity. Z= Section modulus. A =Cross sectional area.

Stresses at rail seat (due to pre-stressing). Mr = - P . e

𝜎𝑡𝑜𝑝 = 36903

4872−

36903 ∗ 2.61

1589= 15.1

𝑘𝑔

𝑚2

𝜎𝑏𝑜𝑡𝑡𝑜𝑚 =36905

4872+

36903 ∗ 2.61

1589= 132.4 𝑘𝑔/𝑚2

Stress at center of tie:

𝜎𝑡𝑜𝑝 = 36903

346.3+

36903 ∗ 0.36

789= 123.2

𝑘𝑔

𝑚2

𝜎𝑏𝑜𝑡𝑡𝑜𝑚 =36905

346.3+

36903 ∗ 0.36

826= 90.4 𝑘𝑔/𝑚2

Combined stresses due to pre-stress + live load (external loading)

F

P

N.A

Center gravity of steel e= 2.61 cm Mr=-P*e

15.1

132.4 Rail sat

123.2

90.4 Rail sat




46

Rail seat Allowable compression < 210 kg/m2

+ =

Center of Tie Allowable compression = 200.7 kg/m2

+ = The tie spacing:- The allowable tie spacing in Iraq are;

1) For (mainlines and branch lines) with high speed ; 60 cm. 2) For branch lines with low speed ; 75 cm.

15.1

132.4

_

+

-

+ Comp

91.8

60.61

76.7

71.79

Stresses due to pre-stress Stresses due to live load

+ Comp

7.8

200.7

_ +

-

+ Comp

123.2

90.4

113.5

110.3

+ Comp

Rail

Ballast Tie Spacing Tie Spacing




47

Beams on Elastic Foundation

.(دو تغػ ترج" و ن ذثو اىرشتح ساظا )مفح ذثو الاجاداخ اىجدج تاىرشتح ذحد اىرثيػ

Winkler Foundation ( Dense Liquid Model 1876). Represented the subgrade soil by menus of indicial closely spaced discrete springs for this winkler model pressure at any point is proportional to deflection at that point.

)ثيح تغثشل ق اىغثشل ذراعة غشدا ع قذاس اىرش (غثقح ا ذحد الأعاط )ىنش تفشض تا اىننشد جيظ عي اىرشتح

أ أح ثو اىرشتح تغثشناخ غرقيح تععا ع تعط رشاتا ف خاصا رقاستح قذاس اىقج اىجدج ف اىغشل ذراعة (اىثغ

ز قطح ظعف اىدو (و ق اىقص- ذجذ ق قص لا )ذؤثش تععا عي تعط ز اىغثشناخ لا.غشدا ع الإصاحح اىاذجح

.لأ عذ ذغيػ اىحو عف تحذز ثغ ف اىغثشل ذحد اىحو

Subgrade reaction is proportional to deflection

Ps=Kw , K=P/W === psi/in or N/ cm2. cm K=Modules of subgrade reaction or spring constant or Dense liquid constant K ; should be determined from plate bearing test.

↑ 𝐹𝑦 = 0 −𝑞(𝑥) ∗ 𝑑𝑥 + 𝑉(𝑥) − 𝑉(𝑥) ∗ 𝑑𝑥 = 0 𝑑𝑣

𝑑𝑥=−𝑞 𝑥

𝑀 = 0

𝑀 𝑥 − 𝑀 𝑥 + 𝑑𝑀 + 𝑉(𝑥) − 𝑞 𝑥 ∗ 𝑑𝑥 ∗𝑑𝑥

2− 𝑑𝑀 + 𝑉 𝑥 ∗ 𝑑𝑥 − 𝑞 𝑥 ∗

(𝑑𝑥 )2

2

𝑉(𝑥) =𝑑𝑀

𝑑𝑥

𝑑𝑀

𝑑𝑥=

𝑑2𝑀

𝑑𝑥2 =−𝑞(𝑥)

𝑦" =𝑀(𝑥)

𝐸𝐼(𝑥)

𝑑4𝑦

𝑑𝑥2=1

𝐸𝐼∗

𝑑2

𝑑𝑥2

q(x)=Po(x) – Ps(x)

𝑑4𝑦

𝑑𝑥4= −

1

𝐸𝐼 𝑃𝑜 𝑥 − 𝑃𝑠 𝑥 =

−𝑃𝑜(𝑥)

𝐸𝐼+

𝑃𝑠(𝑥)

𝐸𝐼=

−𝑃𝑜(𝑥)

𝐸𝐼+

𝑘𝑤

𝐸𝐼

4 Lecture No.

K

P

Subgrade reaction

Ps (pressure under needed subgrade (psi)

W(x,y)=deflection

q

Ps =kw

dx

Mx Mx*dM Vx Vx*dv

qx*dx




48

𝑖𝑓 𝑤 = −𝑦 𝑡𝑕𝑒𝑛 𝑑4𝑦

𝑑𝑥4=

−𝑃𝑜(𝑥)

𝐸𝐼−

𝑘𝑦

𝐸𝐼

𝑑4𝑦

𝑑𝑥 4 +𝑘𝑦

𝐸𝐼=

−𝑃𝑜 (𝑥)

𝐸𝐼(𝑥) differential equation for the deflection curve of a uniform beam on Winkler springs

𝑑4𝑦

𝑑𝑥 4 +𝑘𝑦

𝐸𝐼= 0 For concentrated load

fourth order differential equation with constant coefficient

𝑚 =𝑑𝑦

𝑑𝑥 ∴ 𝑚4 +

𝑘

𝐸𝐼= 0 𝑚4 = −

𝑘

𝐸𝐼

characteristics equation

𝑚1,2,3,4 = 𝑘

𝐸𝐼

4

−14

= 𝛽∗ −14

→ 𝛽∗ = 𝑘

𝐸𝐼

4

De Movine theorem Z = r [cosθ + i sinθ]

𝑍1/𝑛 = 𝑟1/𝑛 [cos𝜃 + 2𝑘𝜋

𝑛+ 𝑖 𝑠𝑖𝑛

𝜃 + 2𝑘𝜋

𝑛]

K = 0, 1, 2, 3, ----------, n-1 n = 4

𝑍14 = 𝑟

14[cos

𝜃 + 2𝑘𝜋

4+ 𝑖 𝑠𝑖𝑛

𝜃 + 2𝑘𝜋

4]

r=1, θ=π

𝑚1 = 𝛽∗ 𝑐𝑜𝑠𝜋


𝜋

4 = 𝛽∗

1

2+

𝑖

2

𝑚2 = 𝛽∗ 𝑐𝑜𝑠𝜋 + 2𝜋


𝜋 + 2𝜋

4 = 𝛽∗ −

1

2+

𝑖

2



𝜋 + 4𝜋

4 = 𝛽∗ −

1

2−

𝑖

2



𝜋 + 6𝜋

4 = 𝛽∗

1

2−

𝑖

2

two conjugate complex route

𝑦 𝑥 = 𝑒𝛽∗

2𝑥 𝑐1𝑐𝑜𝑠

𝛽∗

2𝑥 + 𝑐2 𝑠𝑖𝑛

𝛽∗

2𝑥 + 𝑒

−𝛽∗

2𝑥 𝑐3𝑐𝑜𝑠

𝛽∗

2𝑥 + 𝑐4 𝑠𝑖𝑛

𝛽∗

2𝑥

For x>0 Y (at x = ∞) =0 then c1 = c2 = 0 Y(x) =

𝛽 =𝛽∗

2=

𝑘

4𝐸𝐼

4

𝑦 , 𝑥 = 𝛽 𝑒−𝛽𝑥 𝑐3𝑐𝑜𝑠𝛽𝑥 + 𝑐4 𝑠𝑖𝑛𝛽𝑥 + 𝑒−𝛽𝑥 −𝑐3𝛽𝑠𝑖𝑛𝛽𝑥 + 𝑐4 𝛽𝑐𝑜𝑠𝛽𝑥 = 𝛽 𝑒−𝛽𝑥 𝑐3𝑐𝑜𝑠𝛽𝑥 + 𝑐4 𝑠𝑖𝑛𝛽𝑥 + 𝑐3𝑠𝑖𝑛𝛽𝑥 − 𝑐4 𝑐𝑜𝑠𝛽𝑥 2- 𝑦 , 0 = 0 = −𝛽 𝑐3 − 𝑐4 → 𝑐3 = 𝑐4

𝑦 𝑥 = 𝑐3𝑒−𝛽𝑥 [𝑐𝑜𝑠𝛽𝑥 + 𝑠𝑖𝑛𝛽𝑥] 3- Boundary condition at x=0

𝑦 , 𝑥 = −𝛽𝑐3𝑒−𝛽𝑥 𝑐𝑜𝑠𝛽𝑥 + 𝑠𝑖𝑛𝛽𝑥 + 𝑐3𝛽𝑒−𝛽𝑥 −𝑠𝑖𝑛𝛽𝑥 + 𝑐𝑜𝑠𝛽𝑥 𝑦 , 𝑥 = −2𝑐3𝛽𝑒−𝛽𝑥 𝑠𝑖𝑛𝛽𝑥 𝑦 ,, 𝑥 = 2𝑐3𝛽2𝑒−𝛽𝑥 𝑠𝑖𝑛𝛽𝑥 − 2𝑐3𝛽2𝑒−𝛽𝑥 𝑐𝑜𝑠𝛽𝑥 = 2𝑐3𝛽2𝑒−𝛽𝑥 [𝑠𝑖𝑛𝛽𝑥 − 𝑐𝑜𝑠𝛽𝑥]

𝑦 ′ ′ ′ ′ ′′′′ 𝑥 = −2𝑐3𝛽3𝑒−𝛽𝑥 𝑠𝑖𝑛𝛽𝑥 − 𝑐𝑜𝑠𝛽𝑥 + 2𝑐3𝛽3𝑒−𝛽𝑥 𝑐𝑜𝑠𝛽𝑥 + 𝑠𝑖𝑛𝛽𝑥 = 4𝑐3𝛽3𝑒−𝛽𝑥 𝑐𝑜𝑠𝛽𝑥

y

x

-1 Real part −𝟏 Imaginer part

Z=x+iy Z= r *cosθ + i sinθ+




49

Shear force 𝑉 𝑥 =𝑑𝑀

𝑑𝑥= 𝐸𝐼𝑦 ,,,

𝑉 𝑥 = 0 = −𝑝

2= 𝐸𝐼 4𝑐3𝛽3

𝑐3 =−𝑝

8𝛽3𝐸𝐼

𝑦 𝑥 = −𝑝

8𝛽3𝐸𝐼𝑒−𝛽𝑥 𝑐𝑜𝑠𝛽𝑥 + 𝑠𝑖𝑛𝛽𝑥

𝛽 =𝛽∗

2=

1

2

𝑘

𝐸𝐼

4

= 𝑘

4𝐸𝐼

4

𝑦 𝑥 = −𝑝𝛽

2𝑘𝑒−𝛽𝑥 𝑐𝑜𝑠𝛽𝑥 + 𝑠𝑖𝑛𝛽𝑥

W(x) = - y(x)

𝑤 𝑥 =𝑝𝛽

2𝑘𝑒−𝛽𝑥 𝑐𝑜𝑠𝛽𝑥 + 𝑠𝑖𝑛𝛽𝑥

𝑤 𝑥 =𝑝𝛽

2𝑘 ∅𝑎𝑡 𝛽𝑥

∅𝑎𝑡 𝛽𝑥 = 𝑒−𝛽𝑥 [𝑐𝑜𝑠𝛽𝑥 + 𝑠𝑖𝑛𝛽𝑥]

𝑀 𝑥 = 𝑦 ,, 𝑥 𝐸𝐼 = −𝑝

4𝛽 𝑒−𝛽𝑥 𝑠𝑖𝑛𝛽𝑥 − 𝑐𝑜𝑠𝛽𝑥 =

𝑝

4𝛽𝜑𝑎𝑡 𝛽𝑥 − 𝜑 𝛽𝑥

𝑀 𝑥 =𝑝

4𝛽𝜑𝑎𝑡 𝛽𝑥 𝑓𝑜𝑟 𝑥 ≥ 0

𝜑𝑎𝑡 𝛽𝑥 = −𝑒−𝛽𝑥 𝑠𝑖𝑛𝛽𝑥 − 𝑐𝑜𝑠𝛽

𝑉 𝑥 = −𝑝

2𝑒−𝛽𝑥 𝑐𝑜𝑠𝛽𝑥

𝑉 𝑥 = −𝑝

2 𝜃𝑎𝑡 𝛽𝑥 ∴ 𝜃 𝛽𝑥 = 𝑒−𝛽𝑥 𝑐𝑜𝑠𝛽𝑥

βx W M V

Φ ψ θ

0 1 1 1

0.1 0.9907 0.81 0.9003

0.2 0.9651 0.6398 0.8024

……

1.5 0.2184 0.2068 0.0158

7.0 0.0013 0.0061 0.0007

Rule after Heteny:- Separate elastic supports may be replaced by an imaginary continuous foundation if we have at least four of

the supports in the characteristics wave length L= 𝜋 4𝐸𝐼

𝑘

4 of the deflection curve.

Examples, how many ties are in the characteristic wave length of a railway traffic with E = 2.1 * 106 kg/m2 modulus of elasticity I = 1830 cm4 moment of inertia K = 200 kg/cm2 modulus of foundation a = tie spacing = 50 cm




50

characteristic wave length = L= 𝜋 4𝐸𝐼

𝑘

4 = π

4∗2.1∗106∗1830

200

4= 294.15 𝑐𝑚

No. of ties within L =[294.15/50]=5.88 = 6 > 4 The analysis as a beam on elastic foundation is of (the assumption of imaginary continuous foundation is accept) Determination of the modulus of the foundation:

1- After Timoshenko 2- After Hetengi 1- After Timoshenko

K=( load required to produce unit deflection of a tie) / tie spacing K= *p / δ a+ a: tie spacing k = [10000/22*0.3] = 1500 psi

2- After Hetengi D= spacing constant for tie P=D ∑ yi D = *p / ∑ yi] K = D /a a= tie spacing

Ch 4 Bending of a Rail Due to the Wheel Pressure of a Train Method of Supperposition for Various Concentrated Load J=point at which bending moment (and or the deflection) is required Xi = distance of the load No.i from point j (i.e origin of x is at )

𝑀𝑗 =1

4𝛽 𝑝𝑖 𝜑𝑖 𝛽𝑥𝑖 , 𝑀 𝑥 =

𝑝

4𝛽 𝜑𝑎𝑡

𝑛

𝑖=1

𝛽𝑥

𝑤 𝑥 =𝛽

2𝑘 𝑝𝑖 ∅𝑖 𝑎𝑡

𝑛𝑖=1 𝛽𝑥𝑖 𝑤 𝑥 =

𝑝𝛽

2𝑘∅𝑎𝑡 𝛽𝑥𝑖

For equal concentrated loads pi, pi+1, ….., p

𝑀𝑗 =𝑝

4𝛽 𝜑𝑖 𝑎𝑡

𝑛

𝑖=1

𝛽𝑥𝑖 , 𝑤 𝑥 =𝑝𝛽

2𝑘 ∅𝑖 𝑎𝑡

𝑛

𝑖=1

𝛽𝑥𝑖 , 𝛽 = 𝑘

4𝐸𝐼

4

Example :- For a system of four equal wheel loads 66,, apart moving on 100 Ib rail section with I = 49 in4 calculate the maximum bending moment and the deflection under the first and second load knowing that k=1500 psi, E=30*106 psi

𝛽 = 𝑘

4𝐸𝐼

4

= 1500

4 ∗ 30 ∗ 106 ∗ 49

4

=1

43.3 𝑖𝑛−1

Calculating bending moment and deflection under the first load

Loads 1 2 3 4 ∑

Xi (inch) 0 66 132 198

βxi 0 1.52 3.05 4.57

φ 1 - 0.207 -0.051 0.008 0.75




51

∅ 1 0.23 -0.042 -0.012 1.18

𝑀1 =𝑝

4𝛽∗ φ =

𝑝

4𝛽∗ 0.75 = 0.75 𝑀0 𝑤𝑕𝑒𝑟𝑒 𝑀0 =

𝑝

4𝛽

w=δ = 𝑝𝛽

2𝑘 ∅ = 1.18

𝑝𝛽

2𝑘 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 1

𝑘 =𝑝

𝛿𝑎 𝑜𝑟 𝑝 = 𝑘𝛿𝑎

Under load section x at 2=0

Load 1 2 3 4 ∑

Xi (inch) -66 0 66 132

βxi -1.52 0 1.52 3.05

φ -0.207 1 -0.207 -0.051 0.535

∅ 0.23 1 0.23 -0.042 1.418

𝑀2 =𝑝

4𝛽∗ 0.535 = 𝑀3

δ2 = 1.418 𝑝𝛽

2𝑘 deflection at point 2 = δ3

calculating the bending moment at the midpoint between point 2 and 3

Load No. 3 4

Xi 33 99

βxi 0.702 2.286

φi 0.017 -0.1416

Mat point 2/3 = 𝑝

4𝛽 0.017 − 0.1418 ∗ 2 = −

0.25𝑝

4𝛽

K=500 , I=1830 cm4, a=60 cm, E= 2.1 *106 kg/cm2 , δ2=1.418 𝑝𝛽

2𝑘

𝛽 = 𝑘

4𝐸𝐼

4=

500

4∗2.1∗106∗1830

4= 0.01329 cm-1

Subgrade reaction = k.w=k.δ Pressure force on a tie at point 2

Tie load at rail seat at point 2 = p2 a= k δa= k (1.418𝑝𝛽

2𝑘) a =

1.418(60)

2 (0.0134

𝑝𝛽

2𝑘)𝑝 = 0.5713 p

Stresses in Rail Steel tie 70 or 90 The quality of steel to be used may either be grade 70 or grade 90 to decided based on traffic density and radii curve Grade 70 yeild point 70000 psi Grade 90 yield point 90000 psi Working stress = allowable stress= yield point /factor of safety Factor of safety to cover :

1- Lateral bending and torsion 2- Temperature stresses 3- Unbalanced superelevation 4- Rail wear 5- Variation from wheel loading diagram




52

2-Temperature stresses: For grade 70 steel (yield point 70 000 psi 7000 psi immediate reduction for temperature stresses is jointed rails 20000 psi immediate reduction for temperature stresses for CWR continuously welded rails The tensile stresses which are caused in winter by low temperature at worst are of the order of magnitude of 100 to 120 N/mm2 for CWR in central Europ, this reduces the load ability of the rail with respect to bend. This explains why transverse rupture increasing by occurred in winter especial in CWR. In summer high compressible forces are generated in the rail by high temperature. These reduce the bending stresses caused by operating load. This counteracts the creation of transverse rupture. However, the compressed rail attempt to decompress, it does this by lateral deflection of the track panel that can effect which must be avoided under all circumistant in order to garanti operating safety. 3-under unbalanced superelevation C = m v2 / r Where: m= mass v= velocity r= radius of horizontal curve C= centerfugl force

∆𝑣 = [𝑚 𝑣2

𝑟cos 𝛼 − 𝑤 sin 𝛼]𝑕/𝑑

4-Rail wear Guide line

Design speed km/hr categori Average wear mm/year

140 1 1

100 2 0.9

80 3 0.5

Approximate formula for estimating the section modulus of worn rails:

𝑍𝑤 = 𝑍 −∆𝑕

30 𝑤 + 0.53 𝑕 − ∆𝑕 𝑖𝑚𝑝𝑒𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎

Zw =section modulus for the worn section in cm3 Z =original section modulus in cm3 w= original weight of the rail kg/m h=intial height of the rail in mm Δh=weir in rail head in mm After Hai 1982 p.525 Common practice cubs for cascading used rails from maintains truck into secondary lines and branches. Example : A new rail section with section modulus Zmin=225 cm3, height h=155mm, weight w=52 kg/m, is subjected to a maximum bending moment Mmax=3.6 ton.m in a railway track with an average wear of 1 mm/year. Determine the tire t which the maximum bending stress in rail becomes equal to maximum allowable stress of 1500 kg/cm2




53

𝜎 =𝑀𝑦

𝐼=

𝑀

𝑍 𝑍 =

𝑀

𝜎

𝑍𝑤𝑜𝑟𝑛 =3.6 ∗ 106𝑘𝑔. 𝑐𝑚

1500 𝑘𝑔/𝑐𝑚2= 240 𝑐𝑚3

𝑍𝑤 = 𝑍 −∆𝑕

30 𝑤 + 0.53 𝑕 − ∆𝑕

240 = 255 −∆𝑕

30 52 + 0.53 155 − ∆𝑕

0.58 (∆𝑕)2 − 134.15 ∆𝑕 + 450 = 0 → ∆𝑕2 − 253.13 ∆𝑕 + 849.06 = 0

∆𝑕1,2 =−253.13 ∓ 253.132 − 4 ∗ 1 ∗ 849.06

2

∆𝑕 = 3.04 𝑚𝑚 𝑜. 𝑘 𝑜𝑟 ∆𝑕 = 249.2 𝑚𝑚 > 155 𝑛𝑒𝑔𝑙𝑒𝑐𝑡𝑒𝑑

𝑡𝑖𝑚𝑒 =3.04

1= 3.04 𝑦𝑒𝑎𝑟

Guide line:- Rail guide for estimate of rail cross section

𝑤 𝑘𝑔

𝑚 =

𝑎 ∗ 𝑤𝑚𝑎𝑥

13 𝑖𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙

a= tie spacing wmax=max. wheel load [tone] ex. : wmax=13 ton, a=60 cm, w=(13*60)/13 = 60 kg/m allowable stress for jointed rail 32000 psi (for 70 grade section) allowable stress for CWR 25000 psi (for 70 grade section) for rail design:

𝑃𝑑𝑒𝑠𝑖𝑔𝑛 = 𝛼 𝑃𝑠𝑡𝑎𝑡𝑖𝑐

𝛼 = 1 +4.5 𝑉2

105−

1.5 𝑉3

107

𝛼 = 𝑖𝑚𝑝𝑎𝑐𝑡 𝑓𝑎𝑐𝑡𝑜𝑟 V= speed km/hr Ex :- speed = 200 km/hr

𝛼 = 1 +4.5 (200)2

105−

1.5 (200)3

107= 1.6

Flat and tie plate To lessen the over turning and tie destroying tendancies cost is introduced into the tie plate to tilt the rail award and bring the rail road resultant closer to the center line of the rail. Note that in so doing a greater area of wheel that is in contact with the rail load, thereby locomotive the contact pressure. Track Stability: 1- Tracks that buckle in vertical plane 2- Tracks that buckle in horizontal plane Factors assisting buckling 1- Excessive alignment errors

- Vertical or lateral direction - Poor subgrade soil - Defects in construction 2- Local stress peaks in the track

- Lack of adjustment of the join opening in joint track




54

- Too low a laying temperature compared to the neutral temperature in the case of a length exceed 120 m

- Breaking forces in the area of falling grades especially of constrained point - Other factors 3- Reduced track lateral strength caused by track maintenance 4- Reduced panel rigidity caused by loosened or untension rail fastening

Note:- The so called classified track buckling occurs under a running train in summer at high temperature. This classified track buckling is initiated in the area of the life off wave. Contact Stress These stresses concentrated immediately beneath the wheel contract area. Contact stresses are especially susceptible to high wheel loads. Rail strength in contact is not a function of rail weight or section modulus but rather of the toughness and quality of rail steel. If wheels and rails are made out of perfectly rigid materials. They would contact at a mathematical point , infinite stress yield of rail than contact area is developed. Hertzain theory: see p. 341

- Compression of two bodies having the same modulus of elasticity.

- Let 1

𝑟1 and

1

𝑟1, denote the principle curvature at the point contact of one of the bodies.

- Let 1

𝑟2 and

1

𝑟2, denote the principle curvature at the other

- Let ∅ denote the angle between the normal plane contact curvatures 1

𝑟1 and

1

𝑟2

- The principle curvature are the maximum and minimum curve and these are in planes at right angle. - The curvature of a body is considered as positive if the comy center of curvature is within the body - The surface of contact for the general case is an ellipse, the semiaxes of which are given by the

equations:-

𝑎 = 𝛼 𝑃𝑚

𝑛

3

𝑏 = 𝛽 𝑃𝑚

𝑛

P = compressive force

𝑚 =4

1𝑟1

+1𝑟1

, +1𝑟2

+1𝑟2

,

𝑛 =4𝐸

3(1 − 𝜇)

E = modulus of elasticity μ= passion ,s ratio

𝑐𝑜𝑠𝜃 =𝐵

𝐴 𝐴 =

2

𝑚 𝐵 =

1

2

1

𝑟1−

1

𝑟1,

2

+ 1

𝑟2−

1

𝑟2,

2

+ 2(1

𝑟1−

1

𝑟1,)

Constant for the calculating the semianes of the ellipse of contact . θ function of α and β

Θ Α β




55

20 3.772 0.408

30 2.731 0.493

35 2.397 0.530

40 2.136 0.567

45 1.926 0.504

50 1.754 0.641

55 1.611 0.688

60 1.486 0.717

65 1.398 0.759

70 1.284 0.802

75 1.202 0.846

80 1.128 0.893

85 1.061 0.944

90 1 1

The contact pressure associated with the ellipse is itself of ellipsoidal form described by 𝑃 ≡ 𝜎𝑍 =

𝑃𝑜 1 − (𝑥

𝑎)2 − (

𝑦

𝑏)2

Po(x,y)

Pmax at x=0, y=0 , 𝑃𝑚𝑎𝑥 = 𝑃𝑜 =3𝑃

2𝜋𝑎𝑕

Ex :- (crossed circular cylinders) Determine the surface of contact and the maximum pressure between two circular cylinders whose axle are mutually perp Solution

1

𝑟1, = 0 ,

1

𝑟2, = 0 ∅ =

𝜋

2 𝑐𝑟𝑜𝑠𝑠𝑒𝑑 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟

𝑚 =4

1

𝑟1+

1

𝑟1, +

1

𝑟2+

1

𝑟2,

=4

1

𝑟1+

1

𝑟2

n=4𝐸

3(1−𝜇 )

𝐴 =2

𝑚=

1

2 (

1

𝑟1+

1

𝑟2)

𝐵 =1

2

1

𝑟1−

1

𝑟1,

2

+ 1

𝑟2−

1

𝑟2,

2

+ 2 1

𝑟1−

1

𝑟1,

1

𝑟1+

1

𝑟2 =

1

2

1

𝑟12 +

1

𝑟22 −

2

𝑟1𝑟2

= 1

2

1

𝑟1−

1

𝑟2

2

= ∓ 1

2

1

𝑟1−

1

𝑟2 =

1

2

1

𝑟1−

1

𝑟2

Cos θ = 𝐵

𝐴=

1

𝑟1−

1

𝑟2

1

𝑟1+

1

𝑟2

Note : for two cylinder of equal raduii 𝑟1 = 𝑟2 then cos θ =

1

𝑟1−

1

𝑟1

1

𝑟1+

1

𝑟1

= 0

Cos θ=0 𝑚 =42

𝑟

= 2𝑟




56

For θ = 90 then α=β=1

𝛼 = 1 𝑃𝑚

𝑛

3

= 𝑃2𝑟

𝑛

3

= 2𝑃

𝑛

3

𝑟3

Example :- Find the maximum pressure between a whal with a cylindrical rim of radius r1 = 15.8 in and rail with the radius of head r2 = 12 in. if Pwheel load = 1000 Ib, and poissions ratio, μ=0.25, E=30*106 psi.

1

𝑟1=

1

15.8= 0.0633 𝑖𝑛−1 ,

1

𝑟2=

1

12= 0.0833 𝑖𝑛−1

𝑚 =4

1𝑟1

+1𝑟2

=4

0.0633 + 0.0833= 27.285 𝑖𝑛

n=4𝐸

3(1−𝜇 )=

4∗30∗106

3(1−0.25) = 42.666 * 106 psi

𝐴 =2

𝑚=

2

27.285= 0.0733 𝑖𝑛−1

B = 1

2

1

𝑟1−

1

𝑟2 = 0.01 𝑖𝑛−1

Cos θ = 𝐵

𝐴 =

0.01

0.0733= 0.1364 then θ = 82.1589o

From table with linear interpolation α = 1.0991 β = 0.915

𝑎 = α 𝑃𝑚

𝑛

3

= 1.0991 ∗ 1000 ∗ 27.285

42.666 ∗ 106

3

= 0.0946 𝑖𝑛

𝑏 = β 𝑃𝑚

𝑛

3

= 0.0915 ∗ 1000 ∗ 27.285

42.666 ∗ 106

3

= 0.079 𝑖𝑛

𝑃𝑚𝑎𝑥 = 1.5𝑃

𝜋𝑎𝑏= 1.5 ∗

1000

𝜋∗0.0946∗0.079= 6388.8 𝑝𝑠𝑖

Maximum Shearing Stress After Panel p. 350: The maximum shear stress τmax. occurs at a distance h beneath the surface. Typical values of h and τmax are given below for steel (poisson s ratio, μ=0.3) for two different aspects ratio (b/a) of the contact patch (b/a=1) for circular contact area then h = 0.47a τmax = 0.31 Po

𝜏𝑚𝑎𝑥 =23500 ∗ 𝑃

13

2 ∗ (𝑅1

𝑅2)0.271 ∗ 𝑅2

2/3

τmax = maximum shear stress in rails due to vertical wheel load compression psi P = dynamic wheel load R1 = radius of the larger cylinder wheel R2 = radius of the smaller cylinder rail head

Railway Transportation Engineering

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