Rings and Balls Apparatus (Softening point test of Bitumen)

Softening point of Bitumen: It is the minimum temperature at which bitumen get soft.
It is important to find the softening point of Bitumen because bitumen can be easily applied on the road surface when it is soft and we have to use the fuel and fire to soften the available bitumen. Softening point tells us about the amount of fuel/energy that will be needed to soft it before applying on the road surface. Softening point of bitumen must not be below the average day temperature and it should also not be very much higher, because that will require much energy and fuel to soft it before applying on the road.

• Aim: To find out the softening point of the given bitumen sample.
• Apparatus: 

1. Rings and Balls Apparatus
2. Heating arrangement
3. Thermometer.

Rings and Balls Apparatus

Ring and Balls apparatus contain metallic rings and steel balls and a base metallic plate, encased inside a jar. This whole arrangement is immersed in the water inside the jar, so as to gradually and uniformly raise the temperature by heating with the help of the heating arrangement.

• Theory: Theory of this test is that as the temperature rises, bitumen starts to soft, and attains certain fluidity sufficient enough for this to be applied on the road surface. The temperature plays an important role in the hardness or the softness of the bitumen so we have to find out the minimum temperature at which bitumen attains sufficient fluidity.

• Procedure: 

1. Fill the top surface of the metallic rings with the given bitumen, forming a sheet like surface.
2.  At lower temperature surface of the bitumen will be hard enough to support the steel balls on the top surface, so put the steel balls on it just above the rings. 
3. When the whole arrangement immersed in the water is heated at the specified standard rate, the bitumen will start to soft and the steel balls under their own weight, will start to sink on the bitumen surface. 
4. Steel balls will sink in a way that they will drag the bitumen along with them in the downward direction, bitumen still supporting them but elongating at the same time. 
5. At certain rise of temperature, steel balls will touch the surface of the base plates below and that temperature is noted.  Thermometer must be put in through the slot initially to take the readings.

Result: The reading on the thermometer gives the softening point of the bitumen.

Road Materials - Soil as Sub-grade

Hi,
Road Materials are the materials which are used for the construction of the roads, commonly used road materials are, soil, Aggregates and binders.
Soil is used for the construction of the bottom most layer of the pavement, i.e. sub-grade. Here is a short details of the sub-grade and its function.:

Soil as sub-grade material

• sub-grade is the layer of the pavement whose main function is to support the upper layers of the pavement and to provide the good drainage facility to the infiltrating rain water. It has to act as a single structure along with other layers of the pavement.
• Soil is compacted to its maximum dry density which can be achieved by using the optimum moisture content and the methods of compaction control. Strength has to be ensured which is required for the given design thickness of the pavement.
• Strength analysis and the thickness of pavement are inter linked because more thickness of the pavement is needed if the soil is weak but if the soil possess a good strength then less thickness is needed.

This is ensured by using the CBR(California Bearing Ratio) Test which is produced or was first used by the California State Highway Department. Using the CBR test and the empirical charts you can find out the thickness of the flexible pavement required above the sub-grade.

Thanks for your visit!

P.S. for more about the functions of the various pavement layers please visit this article:  Functions of various pavement layers.

Westergaard's theory for rigid pavements

Hi, 

Rigid Pavements are constructed with some rigid materials like Cement Concrete(Plain, reinforced or prestressed).

Here the load is transferred  through the slab action not like in the flexible pavements. Westergaard's theory is considered good to design the rigid pavements.

He considered rigid pavement slab as a thin elastic plate resting on soil sub-grade, which is assumed to be a dense liquid. So, here the upward reaction is assumed to be proportional to the deflection, i.e. p = K.d, where K is a constant defined as modulus of subgrade reaction. Units of K are kg/cm^3.

• Westergaard's modulus of sub-grade reaction:

Modulus of sub-grade reaction is proportional to amount of deflection d. Displacement level is taken as 0.125 cm in calculating K i.e. d = 0.125 cm, so modulus of sub-grade reaction
K = p/d = p/0.125 kg/cm^2

• Radius of relative stiffness of slab to sub-grade:

Amount of deflection which will occur on the pavement surface depends on the stiffness of the slab and also on the stiffness of the sub-grade. Same amount of deflection will occur on the top surface of the sub-grade.

This means that the amount of deflection which is going to occur in the rigid pavement pavement layer depends both on relative stiffness of the pavement slab with respect to that of sub-grade.

Westergaard defined this by a term "Radius of relative stiffness" which, can be written numerically as below:
            l = [Eh^3/ (12K(1-U^2)]^(1/4)

Where,  l = radius of relative stiffness, cm
         E = Modulus of elasticity of cement concrete kg/cm^2
         U = Poisson's ratio for concrete = 0.15
         K = Modulus of Sub-grade reaction in kg/cm^2

Traffic Parameters:
(1) Design Wheel Load
(2) Traffic Intensity

• Critical Load Positions:

  

  Rigid Pavement - a 3D view

When the wheel load is applied on the pavement surface, flexural stresses are induced in the pavement. There are three critical positions which are to be checked for maximum stresses.

1. Interior loading
2. Edge Loading
3. Corner loading

Whenever loading is applied at the interior of the slab, remote than the edges and corner, this is called interior loading.

When loading is applied on the edges, remote than the corners is called edge loading.

When the loading is applied on the corner angle bisector and loading is touching the corner the edges.

• Equivalent Radius of Resisting section:

When the loading is at the interiors there is a particular area which will resist the bending moment. Westergaard assumed that the area will be circular in plan and its radius is called as Equivalent radius of Resisting section.

Numerically,

     b= (1.6.a^2 + h^2)^(1/2)  - 0.675.h

Here,

      b = equivalent radius of resisting section, cm when 'a' is less than 1.724.h

      a = radios of wheel load distribution, cm

      h = slab thickness, cm

When 'a' is greater than 1.724.h, b =a.

• In case of corner loading, maximum stresses are not produced at corner but they are produced at a certain distance X along the corner bisector. This is given by the relation:

     X = 2.58.(a.l)^1/2

 Here,    X = distance from apex of the slab corner to section of maximum stress along the corner bisector, cm.

   a= Radius of wheel load distribution, cm

    l = Radius of relative stiffness, cm.

Here is an image which shows you the formulas used to calculate the amount of stresses developed at the three critical positions due to the given wheel load P.

Rigid Pavement- Stresses at interior, edges and corners - Westergaard's theory

References:
 

Thanks for you visit!

Part -2-GATE preparation - Transportation Engg. - One liners

5. What is the value of intermediate sight distance for a highway with a design speed of 65 kmph? Assuming the data for co-efficient of friction f = 0.36 and for reaction time t = 2.5 sec.
Ans: 182.8 m

6. The speed of overtaking and overtaken vehicles are 70 and 40 kmph respectively on a two way traffic road. If acceleration of overtaking vehicle is 0.99 respectively, then safe overtaking sight distance, assuming reaction time of 2 sec, will be  278 m.

7. What is the minimum length of overtaking zone for a design speed of 96 kmph assuming acceleration as 0.69 m/sec^2, reaction time as 2 sec and traffic as one way?   Ans: 1026 m  OSD = 3(d1+d2)

8. The radius of horizontal circular curve is 100 m. If design speed is 50 kmph and design co-efficient of lateral friction is 0.15, then super elevation required if full lateral friction is assumed to develop will be 0.047

9. What is the co-efficient of friction needed if no super elevation is provided for a horizontal circular curve of radius 190 m and design speed of 65 kmph?   Ans: 0.177

10. Ruling minimum radius of horizontal curve of a national highway in plain terrain for a ruling design speed of 100 kmph with e = 0.07 and f = 0.15, is close to  360 m.

11. Design rate of super-elevation for horizontal highway curve of radius 450 m for a mixed traffic condition, having a speed of 125 km/hour, is 0.07

Refer GK Publishers for elaborated answers.

GATE 2014- Transportation Engineering - one liners - part 1

1. The safe stopping sight distance for design speed of 50 kmph two way traffic on a two lane road assuming co-efficient of friction as 0.37 and reaction time as 2.5 seconds is 61.4 m.

2. The stopping sight distance for design speed of 80 kmph for two way traffic on a single lane road. Assume co-efficient of friction as 0.35 and reaction time as 2 seconds, is 232.94 m.

3. What is the minimum sight distance required to avoid head on collision of two cars approaching from the opposite directions at 90 and 60 kmph? It is given that the reaction time is 2.5 seconds, co-efficient of friction is 0.7 and a break efficiency is 50 percent in either case.   Sol: 235.8 m.

4. What is the stopping sight distance on a highway at a ascending gradient of 5% for a design speed of 70 kmph assuming co-efficient of friction as 0.35 and reaction time as 2 seconds?  Sol: 87.14 m.

Reference: GATE 2013 by GK publishers (refer for elaborate answers)

CBR(California Bearing Ratio) Test

Hi,

CBR loading frame

• Aim: To find out the CBR(California bearing ratio) value of the given soil sub-grade.
• Apparatus: 

1. CBR Apparatus
2. soil sample with known OMC and MDD.

   

   CBR apparatus - accessories

• Theory: California Bearing Ratio test was invented by California State Highway Department. This test is used to design the thickness of the flexible pavement. 

CBR value signifies the strength characteristics of the soil sub-grade which is compacted to the MDD using the OMC. The samples used for the testing are prepared in the laboratory. If a new pavement is to be constructed then, sample are prepared by compacting it with OMC and then sample is soaked in water for four days.

If the test is done for the overlay design then the sample is prepared by compacted it to the density of the soil at the site. Four days soaking is necessary in order to achieve the worst site conditions.

Higher the CBR value more is the strength of the soil sub-grade. Empirical charts are prepared by the IRC to find out the thickness of the flexible pavement corresponding to the given CBR value and for the given design wheel loading.

If the CBR value obtained with the 5 mm penetration exceeds the value obtained with the 2.5 mm penetration then this test is performed again and if the value are still same then CBR value at 5 mm is taken as the final value.

• Procedure: 

  

  CBR apparatus

1. Take the soil from the given soil sub-grade remove the gravel of size more than 20 mm and and find out its OMC.
2. Compact the soil in three layers in the CBR mold, with 25 numbers of blows given to each layer.
3. Put the surcharge weight and then immerse the sample in the water for four days.
4. Put the sample with the surcharge weight on the loading base of the CBR apparatus.
5. Apply the loading with a standard rate 1.25 mm/min with the 5 cm plunger and note down the load values corresponding to the penetrations of 1mm, 2 mm, 3 mm, 4 mm, 5 mm, 7.5 mm, 10 mm and 12 mm.
6. Plot the graph and apply the correction if any.
7. Repeat the test for two more samples.

• Calculations: 

Check if there is any correction to be applied to the penetration and loading curve.
CBR value at 2.5 mm penetration = (Load at 2.5 mm penetration/ Standard load at 2.5 mm penetration i.e. 1370kg)*100
CBR value at 5 mm penetration = (Load at 5 mm penetration/ Standard load at 5 mm penetration i.e. 2055kg)*100

Use the CBR value at the 2.5 mm penetration if it is more than that at 5 mm penetration otherwise, repeat the test. If the value at the 5 mm penetration still comes out to be more, then use it.

• Result: CBR value of the given soil sub-grade sample = .....%

Check the relevant books:

Aggregate Impact Value Test

Hi,
AIM: to find the aggregate Impact value.

Theory:
Aggregate to be used as road materials must possess some properties or characteristics. Toughness is one of the highly required property which is necessary for the aggregates to bear the impact loads. 

Road aggregates are applied with the impact loads a millions of times in their life as the road materials. So they must be tested for their toughness before they are used in the pavement layers. 
Toughness can be tested using a Impact testing machine, which is an arrangement to apply the impact loads on the aggregates just like they are applied on the roads. 
Materials which get fractured or crushed into smaller particles are not tough. In this test we will find out the percentage of the weight of the aggregate sample which gets crushed with respect to the total weight of the sample. This percentage is known as the aggregate impact value and more the aggregate impact value less is the toughness of the road aggregates and vice verse.  

If impact value is less than 10% then aggregate is said to be exceptionally strong, if it is in range of 10 to 20% then they are good, aggregates having impact value less than 35% are considered satisfactory by the IRC(Indian Roads Congress).

Apparatus Used:  Aggregate Impact Load Testing Machine, IS Sieves of size 12.5 mm; 10 mm and 2.36 mm. Electronic weighing balance, tray.

Calculations:
Suppose weight of the Total aggregate samples in the cylindrical mould( which passed through 12.5 mm Sieve and retained on 10 mm Sieve) = W1
Let weight of the crushed aggregates(Which pass through 2.36 mm IS Sieve) = W2
Impact value of the aggregates(%) = W2/W1 *100

Results: Aggregate Impact Value =       %