पाठ- ६ सिभिल

पाठ -छ सिभिल बि.ई र डिप्लाेेेमा तहमा;wikipedia inbox
if, any more query facebook massesanger or email. This civil engineering part is available from NPI lecture PREM ADHIKARI NEC no 17717.Airport planning Airport planning

Practical subject 3 yr Diploma

Programme: civil

Yr 1^st

Sem 1^st ,

Civil I/I	Engg. Drawing	40	16
	Physics	10	4
	Workshop Practice -I	80	32
	Chemistry	10	4

sem 2^nd

Civil I/II	Engg. Drawing	40	16
	Physics	10	4
	Chemistry	10	4
	Computer Application	20	8
	Workshop Practice -II	80	32

Yr 2^nd

Sem 3^rd

*Program*	*Subjects*	*F.M.*	*P.M.*
	Construction Drawing	40	16
	Surveying I	40	16
	CAD	20	8
	Building Construction	20	8

sem 4^th

Civil II/II	Surveying II	40	16
Civil II/II	Estimating & Costing	40	16

Yr 3^rd

Sem 5^th

Civil III/I	Structural Design and Drawing	20	8
	Estimating and Costing II	40	16
	Surveying II	20	8
	Minor Project(Survey Camp)	40	16

sem 6^th

Civil III / II	Irrigation and Drainage eng.	20	8
	Estimating and Costing III	20	8
	Elective	20	8
	Entrepreneurship	10	4
	Major Project	80	32

See-थिन वालेट भेसल ,टु हय्ङग्ड एपाटर्स, थ्रि हेङग एपारेटर, स्टक्चर ,भिस्काेमिटर ,रिनाउण्ड एपाटर्स ,सफ्टर्न फाेव ,जेट ईम्प्यक्ट ,बिटुमिन टेष्ट ,जेट इमेप्यक्टर ,मार्सल स्टेविटि,पाेर्टल अपरेशन,टर अपरेटर

,पर्माबिलिटप टेष्ट 1.Structure

2. TWO HINGED ARCH APPARATUS( INSTRUCTION MANUAL)

OBJECTIVE:- “MARS” Made to determine the horizontal displacement in two hinged apparatus.

APPARATUS:-

1. 1 Digital Dial gauge ( 0-25)mm

2. 9 Hanger

3. Weight 10N × 3 5N × 8, 2N × 4, 2.5N × 2,

4. span of 1000mm and rise 250mm

5. stand dimension( height = 900mm , length = 1280mm)

THEORY:

The two hinged arch is a statically indeterminate structure of the first degree. The horizontal thrust is the redundant reaction and is obtained y the use of strain energy methods.

ffig a

Fig. (b)

Two hinged arch is made determinate by treating it as a simply supported curved beam and horizontal thrust as a redundant reaction. The arch spreads out under external load as shown in fig. (a). This results in a horizontal displacement of support B by D₁. Here, deflection due to flexure only has been considered. Since the support conditions dictate that that the final displacement at support B should be zero, horizontal reaction H should be such that displacement D₂ caused by H must satisfy the condition.

Where, f is the displacement caused by a unit force applied in the direction of H. (1)

Therefore, it is required to calculate the horizontal displacement in arch caused by external load as well as unit horizontal force.

The horizontal displacement in a curved member can be found by either Castiglano's second theorem or the unit load method.

Where, M = Bending moment on any point on the arch due to given loading.

m = moment on any point on the arch due to a unit horizontal force applied at B in the direction of H.

The expression given by Eq. (2) will become simpler provided the curve of the arch axis is parabolic and moment of inertial of curve at any section varies as I= I₀secq where q is angle between the horizontal and tangent to the arch axis at that particular point.

I₀ = moment of inertia at the crown

I = moment of inertia at any other section

m = y, ds = secq dx

it may be noted that the integration is to be carried out from 0 to L then Eq. (2) will become

(3)

and D₁ = horizontal displacement

For a concentrated load W at the crown it is found that

Horizontal displacement (4)

Where, L is the span of the arch in mm and r is the rise in mm.

The horizontal movement of the roller end can be found by this method for any position of the load on the arch. E = 210000N/mm² I₀ is the planar moment of inertial = bt³/12 where b is the width of span and t is thicknesss of span.

The ordinate for the influence line diagram for H at any distance z = aL form L.H.S. can be obtained as follows:

Now H can be evaluated using Eq. (3)

Now substituting z = aL

We have

Taking W=1kg

\ Influence line ordinate are given by...
PROCEDURE:-

1: Fix the dial gauge to measure the movement of the roller end of the model and keep the lever out of contact.

2: Place a load of 0.5kg on the central hanger of the arch to remove any slackness and taking this as the initial position, set the reading on the dial gauge to zero.

3: Now add 1kg weights to the hanger and tabulated the horizontal movement of the roller end with increase in the load in steps of 1kg. Take the reading up to 5kg load. Dial gauge reading should be noted at the time of unloading also.

4: Plot a graph between the load and displacement (theoretical and experimental) compare. Theoretical values should be computed by Eq. (4).

5 for infulance of line:-

6 now move the lever in contact with 200 gm load. And place the 1kg .5 kg load at the

Centre. For removing any slackness in the wire.set the initial reading of dial gauge to

Zero.

7 place the 2.5 kg load on to the centrel hanger and observe the dial gauge reading.

8 restore the dial gauge reaing to zero by adding load to the lever hanger.

Required load to restore dial gauge reading to zero is equal to = 4w/5 , w is the load

Applied at the centre. As shown in fig.below.

1 2 3 4

Central hanger ( applying weight = 2.5kg

Lever

Required weight to zero the dial gauge = 4w/5

W = apply weight at central hanger

Similarly for 3^rd position( when applied weight = 2.5kg)

Required weight to zero dial gauge reading at lever = 3w/5

Similarly for 2^rd position( when applied weight = 2.5kg)

Required weight to zero dial gauge reading at lever = 2w/5

Similarly for 1^rd position( when applied weight = 2.5kg)

Required weight to zero dial gauge reading at lever = 1w/5

9 This experiment is perform for any other loads applied at different position.

OBSERVATION DATA SHEET:

Table-1

Horizontal displacement

Sl. No.	Central load (kg)	0.0	0.5	1.0	1.5	2.0	2.5	3.0
	Observed horizontal displacement (mm)
	Calculated horizontal displacement Eq.(4)

FORMULA:-

Horizontal displacement (4)

Where, L is the span of the arch =900mm

r is the rise = 250mm.

The horizontal movement of the roller end can be found by this method for any position of the load on the arch.

W is the applied weigth in N ( 100gm = 1N)

E = 210000N/mm² is young modulus of elasticity.

I₀ is the planar moment of inertial = bt³/12

where b is the width of span = 40mm

t is thickness of span. = 10mm

FOR EXAMPLE:-

let 1kg weight is the applied into the central hanger, so that

w= 10 N

L is the span of arch = 900mm

r is the rise = 250mm

E= 210000 N/mm²

I₀ = planar moment of inertia.= bt³/12

Where b = 40mm

t = 10mm

I₀ = 40 × 10³

= 3333mm⁴

Horizontal displacement

= 5 × 10 ×900² × 250

48 × 210000 × 3333

= .30mm

RESULTS AND DISCUSSIONS:

Compare the horizontal displacement results.( obtained theoretically and experimentally.)PRECAUTIONS:

1 Apply the loads without jerk.

2 Perform the experiment away from vibration and other disturbances.

3 Dial gauge must be fitted in proper manner.( if dial gauge reading is not zero on pressing zero position button. Please reset the dial gauge by on/ off switch (२)(2)

3. THREE HINGED ARCH APPARATUS (INSTRUCTION MANUAL)

OBJECTIVE:- “MARS” Made to determine the horizontal displacement in three hinged apparatus.

APPARATUS:-

6. 1 Digital Dial gauge

7. 8 Hangers.

8. Weight Set different type.

9. Span of 900mm and rise 210mm.

1.0 THEORY:

A three hinged arch is a statically determinate structure with the axial thrust assisting in maintaining the stability. The horizontal thrust H in the arch for a number of loads can be obtained as follows:

W₁

W₂

W₃

a₁

a₂

a₃

Taking moment about A

(1)

Taking moment about B

Taking the moment of all the forces on left hand side about C, we get

(2)

The value of horizontal reaction can be evaluated by Eq. (1).

2.0 OBJECTIVE:

(1) To determine the horizontal thrust in a three hinged arch for a given system of loads experimentally and verify the same with calculated values.

3.0 APPARATUS:

The model has a span of 900mm and rise 210mm, with hinges at supports and crown. One of the end rests on rollers. Along the horizontal span of the arch various points are marked at equidistant for the application of load. A dial gauge with magnetic base is supplied with the apparatus.

4.0 SUGGESTED EXPERIMENTAL WORK:

Step 1: Use lubricating oil at the roller end of the arch so as to have a free movement of the roller end..

Step 2: Apply the weight on to the lever hanger to obtain the proper unbalancing position.

apply weight at position a1,a2,a3 for initial position of 3hinged arc.

Step 3: Place a few loads on the arch in chosen positions.put the weight in small steps in order of 100gm, now note down the w1 ,w2 and w3 weight value where arc in balance or reach initial position.

Step 4: Put the value of distance a1,a2 a3 and respectively value of weight and find horizontal force for it. Compare theoretical value to experimental value and also find % error.

RESULTS AND DISCUSSIONS:

1. Find the horizontal thrust for a given set of load experimentally and theoretically..

2. Plot the observed and calculated values of influence line ordinates on the same graph and comment on the accuracy obtained in the two cases.

5.0 SAMPLE DATA SHEET:

Span of the arch, L = 900mm

Central rise, h =

Initial load on the thrust hanger for balancing, kg =

Table-1

Sl. No.	Load on hanger member from roller end		Total load on thrust hanger (kg)	Distance from left hand support (cm)	Calculated value of H (kg
	Set No.	Load (kg)
Set I	W₁ W₂ W₃			a₁ =225mm a₂ =337.5mm a₃ =562.5mm

6.0 PRECAUTIONS:

· Put the weights in thrust hanger very gently without a jerk.

· Measure the distance of loaded points from left hand support accurately.

· Perform the experiment away from vibration and other disturbances.

Sample result-

In three hinged apparatus total L= 900mm

Ist position =112.5

2^nd position =225

3^rd position =337.5

4^th position =450

5^th position = 562.5

6^th position = 675

7^th position =787.5

Hanger position where bridge is gone upward direction

3.500kg

W1=1.3kg

W2=1.3kg

W3=1.3kg

Rb= 1.3X225+ 1.3X337.5+ 1.3X562.5 / 900

=292.5+438.75+731.25

=1462.5/900 = 1.625kg

= 1/180( 1.625*900/2 - 1.3( 562.5-450) )

= 0.00555(731.25 – 146.25)

= 3.24KG

Note -percentage error from 10-20%

4. BUCKLING

When a structure (subjected usually to compression) undergoes visibly large displacements transverse to the load then it is said to buckle. Buckling may be demonstrated by pressing the opposite edges of a flat sheet of cardboard towards one another. For small loads the process is elastic since buckling displacements disappear when the load is removed.
Local buckling of plates or shells is indicated by the growth of bulges, waves or ripples, and is commonly encountered in the component plates of thin structural members.

Buckling proceeds in manner which may be either:

	*stable* -	in which case displacements increase in a controlled fashion as loads are increased, ie. the structure's ability to sustain loads is maintained, or
	*unstable* -	in which case deformations increase instantaneously, the load carrying capacity nose- dives and the structure collapses catastrophically.

Neutral equilibrium is also a theoretical possibility during buckling - this is characterized by deformation increase without change in load.
Buckling and bending are similar in that they both involve bending moments. In bending these moments are substantially independent of the resulting deflections, whereas in buckling the moments and deflections are mutually inter-dependent - so moments, deflections and stresses are not proportional to loads.
If buckling deflections become too large then the structure fails - this is a geometric consideration, completely divorced from any material strength consideration. If a component or part thereof is prone to buckling then its design must satisfy both strength and buckling safety constraints - that is why we now examine the subject of buckling.

Buckling has become more of a problem in recent years since the use of high strength material requires less material for load support - structures and components have become generally slenderer and buckle- prone. This trend has continued throughout technological history, as is demonstrated by bridges in the following sequence :

The Pont du Gard in Provence was completed by the Romans in the first century AD as part of a 50km aqueduct to convey water from a spring at Uzès to the garrison town of Nemausus (Nimes). The bridge is constructed from limestone blocks fitted together without mortar and secured with iron clamps. The three tieredstructure avoids the need for long compressive members. ( source Art images for College Teaching )

The Royal Border Bridge, Berwick upon Tweed, was built by Robert Stephenson whose father George built the Stockton and Darlington Railway (the first public railway) in 1825. Opened in 1850, the bridge continues today as an important link in the busy King's Cross (London) - Edinburgh line. The increased slenderness of the columns compared to the Pont du Gard reflect technological improvements over many centuries. ( source FreeFoto.com )

The Crymlyn Viaduct over the Ebbw Alley opened in 1857 as Welsh coal mining expanded. It was constructed of wrought and cast iron, and remained the highest railway viaduct in the UK until its closure in 1964 due to increased locomotive weights (1908 photo). The advance from masonry to the slender metal compressive members which make up each column requires substantial bracing to prevent buckling ( source John Croeso )

The Humber road bridge, opened in 1981, comprises a continuously welded closed box road deck suspended from catenary cables supported on reinforced concrete towers. Suspension bridges eliminate the need for struts other than the two towers, however avoiding buckles in other slender components becomes an issue ( source FreeFoto.com )

The dangers associated with over-slender build were tragically driven home by the collapse of the Tacoma Narrows road bridge over the Puget Sound in 1940. Although this failure was apparently due to wind- structure aerodynamic coupling rather than buckling as such, this film clip demonstrates graphically the ability of large structures to undergo significant elastic deflections. ( MoviePlayer or similar is required to view this .mov video ) ( source CamGuys.com )

6.Buckling of thin-walled structures

A thin-walled structure is made from a material whose thickness is much less than other structural dimensions. Into this category fall plate assemblies, common hot- and cold- formed structural sections, tubes and cylinders, and many bridges and aeroplanestructures.
Cold- formed sections such as those illustrated are increasingly supplanting traditional hot- rolled I-beams and channels. They are particularly prone to buckling and in general must be designed against several different types of buckling.
It is not difficult to visualise what can happen if a beam is made from such a cold- rolled channel section. One flange is in substantial compression and may therefore buckle locally at a low stress ( ie. much less than yield ) thus reducing the load capacity of the beam as a whole. Buckling rather than strength considerations thus dictate the beam's performance.

Let us now look at typical examples of buckling.

The slender elastic pin-ended column is the protoype for most buckling studies. It was examined first by Euler in the 18th century. The model assumes perfection - the column is perfectly straight prior to loading, and the load when applied is perfectly coaxial with the column.

The behaviour of a buckling system is reflected in the shape of its load- displacement curve - referred to as the equilibrium path. The lateral or 'out-of-plane' displacement, δ, is preferred to the load displacement, q, in this context since it is more descriptive of buckling.

Nothing is visible when the load on a perfect column first increases from zero - the column is stable, there is no buckling, and no out- of- plane displacement. The P-δ equilibrium path is thus characterised by a vertical segment - the primary path - which lasts until the increasing load reaches the critical Euler load P_c = π²EI_min/L² a constant characteristic of the column ( for a derivation of this, see below or Timoshenko & Gere op cit. for example ).

When the load reaches the Euler load, buckling suddenly takes place without any further load increase, and lateral deflections δ grow instanteously in either equally probable direction. After buckling therefore, the equilibrium path bifurcates into two symmetric secondary paths as illustrated. Clearly the critical Euler load limits the column's safe load capacity.

Local buckling of an edge-supported thin plate does not necessarily lead to total collapse as in the case of columns, since plates can generally withstand loads greater than critical. However the P-q curve illustrates plates' greatly reduced stiffness after buckling, so plates cannot be used in the post- buckling region unless the behaviour in that region is known with confidence.
It should be emphasised that the knee in the P-q curve is unrelated to any elastic- plastic yield transition; the systems being discussed are totally elastic. The knee is an effect of overall geometric rather than material instability.This photograph illustrates local buckling of a model box girder constructed from thin plates, not unlike the road deck of the Humber bridge above.

Inclined striations are caused by shear loading in the web of a beam or in a torqued tube giving rise to compressive buckling stresses at 45^o to the longitudinal direction as predicted by Mohr's circle.

The behaviour of a compressed shell after buckling is quite different to that of a plate; in this case an unstable ( negative ) stiffness is accompanied by a sudden reduction of load capacity.

Since the displacements are uncontrolled in most practical systems, shells behave in a snap- buckling mode - ie. as an increasing load reaches the bifurcation point, the cylinder must undergo an instantaneous increase in deflection ( "snap" ) to the point 1 in order to accomodate the increasing load. A subsequent decrease in load is accomodated by a corresponding decrease in buckling deflection until the point 2 is reached whereupon the structure again snaps instantaneously - this time back to the point 3 on the primary path.
Clearly this behaviour makes it imperative in design to apply large safety factors to the theoretical buckling loads of compressed cylinders.

It has been noted that a pressure vessel head is subjected to a compressive hoop stress at its junction with the cylinder.

The two photographs here (from Ramm op cit) show both inward and outward buckles arising from this compression in the torispherical heads of internally pressurised 3 m diameter stainless steel vessels.

Longitudinal stresses in a vertical cylinder may also promote buckling as these two photographs illustrate (from Rhodes & Walker op cit).
Warning of impending failure of the 7.3 m diameter vitreous enamelled silo on the left is provided by the visible buckles.
Grain pours out of the buckled bin on the right - the ladder gives an idea of the bin size.

Torsional buckling of columns can arise when a section under compression is very weak in torsion, and leads to the column rotating about the force axis.

More commonly, where the section does not possess two axes of symmetry as in the case of an angle section, this rotation is accompanied by bending and is known as flexural torsional buckling.
Lateral buckling of beams is possible when a beam is stiff in the bending plane but weak in the transverse plane and in torsion, as is the I-beam of the sketch.

It often happens that a system is prone to buckling in various modes. These usually interact to reduce the load capacity of the system compared to that under the buckling modes individually. An example of mode interaction is the thin box section which develops local buckles at an early stage of loading, as shown greatly exaggerated here.
The behaviour of the column is influenced by these local buckles, and gross column buckle will occur at a load much less than the ideal Euler load. The Steel Structures Code, AS 1250 op cit. sets out rules for the avoidance of mode interaction in large components, and its guidelines should be followed in design.

Buckling has mixed blessings in automotive applications.
The photograph on the left illustrates how local buckling of a car's thin-walled A-pillar dramatically reduces passenger cell integrity in the event of roll-over.
Conversely, the energy absorbed by plastic buckling can reduce significantly the injuries suffered by a vehicle's occupants in the event of a crash. The energy absorption capability of thin- walled sections is demonstrated clearly by the experiment photographed on the right. (from Murray op cit)

The detailed analysis of most practical buckle-prone structures is too complex mathematically to attempt here. We therefore examine instead some mechanisms which demonstrate (un)stable behaviour similar to that of structures. The mechanisms allow us to appreciate buckling behaviour and the tools used to analyse it, and to introduce the concept of imperfections which must occur in practical components and which have a relatively large effect on buckling behaviour and safety.
This work leads to the derivation of a design equation for practical columns, in which the twin failure modes of strength and geometric instability invariably interact. This interaction is apparent also in the behaviour of cracks - the subject of a later chapter.
Prediction of the plastic collapse of sub-sea pipelines is also addressed.

7.applied mechaniesअप्लाईड मेकानिक्स

Principle of transmissibility and equivalent forces

Principle of transmissibility states that “The condition of equilibrium or motion of a rigid body will not be affected by transmitting a force (F) along its line of action”.

In above figure, the force F and F’ are equivalent forces. (Newton mechanics)

We observe that the line of action of force F is horizontal line passing through the front and rear bumper of the truck. Using the principle of transmissibility, we can therefore replace F by equivalent force F’ acting on the rear bumper. In other words, conditions of motion are unchanged and all the other external forces acting on a truck (W, R₁, R₂) remain unchanged if the people push on the rear bumper instead of pulling on the front bumper.

Moving the point of the force F to the rear bumper does not affect the motion of the other forces acting on the truck.

In general, “the condition of equilibrium or of motion of a rigid body will remain unchanged if a force F acting at a given point if the rigid body replaced by a force F’, of the same magnitude and direction, but acting at a different point, provided that the two forces have the same line of action” as shown in above figure.

Varignon’s theorem

Statement: - “The moment about a given point O of the resultant of several

concurrent forces is equal to the sum of the moments of the various

forces about the same point O.”

It is also known as the distributed property of vector product

Illustration

Considering a force and its moment about O. Let F be resolved into a number of forces,

F_1, F_2, F_3, ---------- F_n_,etc. with position vectors r₁, r₂, r₃, ------ r_n,

Therefore, individual moments will be r_1×F_1, r_2×F₂, r_3×F₃ ------ r_n×F_n

Now, in vector form,

= ×

= ₁ + ₂ + ₃ ---------------------------- + _n )

= ₁× ₁ + ₂× ₂ + ₃× ₃ +-----------------+ _n× _n

This property, which was originallyestablished by the French mathematician Varignon (1654–1722) long before the introduction of vector algebra, is known as Varignon’stheorem. The above relation makes it possible to replace the direct determinationof the moment of a force F by the determination of the moments of two or more component forces. As you will see in the next section, F will generally be resolved into components parallel to the coordinate axes. However, it may be more expeditious in some instances to resolve F into components which are not parallel to the coordinate axes.

8. STUDY OF ORIFICEMETER AND VENTURIMETER FOR DISCHARGE MEASUREMENT

OBJECTIVE- Iभेन्चुराेमिटर

To Study the Cd (coefficient of discharge )of venturimeter & orificemeter.

THEORY

An orificemeter consists of a flat circular plate with a hole called an orifice,which may be circular or a sector of a circle ,but generally it is circular and is concentric with the pipe axis.

By applying the bernoulli’s equations between section 1 and 2 and using the continuity equation,the discharge Q through the orificemeter is given by the following expression:

Where A is the area of the pipe ,A is the cross- sectional area of the orifice .∆h is known as the differential head or simply the head loss,Cd is the coefficient and is given by the expression.

For the mercury –water column manometer ,the value of the differential head (∆h)is given by the expression.

∆h=12.6X∆x equ-(iii)

Where ∆x is the difference of mercury levels in the manometer.

EXPERIMENTAL SET-UP

The Set-up consists of an orifice meter fitted in the same horizontal pipeline.Water is supplied to the orifice through an inlet valve provided in the supply pipeline connected to a constant overhead water tank .A regulating valve is provided at the exit to regulate discharge through the orifice meter. The difference of pressure between the inlet and throat section is measured by a U-tube differential manometer containing mercury as manometer liquid.

PROCEDURE

1. Open the pressure tappings of the orifice meter and close the valve of venturimeter

2. Attach the manometer tube with orifice meter for pressure measurement. Remove air in the manometer.

3. Open the inlet valve fully and wait for sometimes so that flow becomes steady.

4. Note the difference of the levels in the manometer.

5. Measure the discharge, in the discharge measurement tank.

6. Repeat steps (4) and (5) for difference discharges by regulating the flow with the help of an exit valve.

7. Above steps 2-6 is same for experiment of venturimeter, after opening the valve of venturimeter and closing the valve of orificemeter.

8. Note the temperature of water used in the experiment.

OBSERVATIONS AND CALCULATIONS FOR ORIFICE-

Diameter of the main pipe,D = 26mm

Diameter of an orifce ,d = 13mm

Area of pipe ,A =

Area of an orifice ,A =

Temperature of water ,Tc =

Kinematic viscosity of water at Tc , V =

OBSERVATIONS AND CALCULATIONS FOR VENTURIMETER-

Diameter of the main pipe,D = 26mm

Diameter of an venturimeter ,d = 13mm

Area of pipe ,A =

Area of an orifice ,A =

Temperature of water ,Tc =

Kinematic viscosity of water at Tc , V =

DISCUSSION

Diameter at an inlet section,D1 =

Diameter at the throat section,D2 =

Plan Area of the measuring tank,Am =

Temperature of water ,Tc =

Kinematic viscosity of water at Tc , V =

Operating Instructions

9.INSTRUCTION MANUALFOR BITUMEN EXTRACTOR

INTRODUCTION: The Centrifuge Extractor is used for the quantitative determination of bitumen in hot mixed paving mixtures and pavement samples. The bitumen content is calculated by difference from the weight of the extracted aggregate, moisture content and ash from an aliquot part of the extract.

DESCRIPTION: The centrifuge extractor consists of a revolving bowl inside a housing (5). The bowl is provided with a cover plate and it is secured in position by tightening a nut.

The bowl housing (5) is provided with an outlet (4). The housing is mounted on an enclosed gear box (1). A cover plate (2) is clamped to the housing.

The gears are flash lubricated and the system is operated manually with a detachable handle (3).

SPECIMEN PREPARATION: To separate with a spatula or trowel, place it in a large, flat pan and warm to 240^oF, only until it can be handled. Separate the particles of the mixture as uniformly as possible using are not to fracture the mineral particles. Normally 1000 g samples are used for extraction. For routine testing smaller samples may be used when the maximum size aggregate therein is less than ¼ in. the precision f the method becomes less as the aggregate size increases due to variation in samples. This procedure may be used on mixtures containing aggregate larger than 1 in by using samples weighing at least 3000 g. They may be tested by extracting 1000 g at a time.

TEST PROCEDURE: Determine the moisture content of the sample using a standard procedure.

Weigh a 1000 g sample into the bowl. Cover the sample in the bowl with benzene or trichloroethylene and allow sufficient time for the solvent to disintegrate the sample (Not over one hour).

Place the bowl containing the sample and the solvent inside the housing. Dry and weigh filter disc and fit, it around the edge of the bowl. Clamp the cover plate tightly on the bowl. If necessary the tommy pin may be a used for loosening the nut for removing the cover plate. Position the housing cover and clamp it to the housing. Place a beaker under the outlet (4).

Start the centrifuge revolving slowly and gradually increase the speed until the solvent ceases to flow from the outlet. Allow the centrifuge to stop, add 200 ml of trichloroethylene of benzene and repeat the procedure. Use sufficient 200 ml solvent additions (not less than three) so that the extract is clear and not darker than a light straw color. Collect the extract and washings in a suitable graduated vessel.

Remove the filter disc from the bowl and dry in air. Remove the as much of the mineral matter adhering to the disc as possible and add to the aggregate in the bowl. Dry the disc to constant weight in an oven at 210 to 220F. Dry the contents of the bowl on a steam bath and then to constant weight in an oven at 210 to 220 F.

Record the volume of the total extract in the graduated vessel. Agitate the extract thoroughly and immediately measure approximately 100 ml into a previously weighed ignition disc. Evaporate to dryness on a steam bath. Ash residue at a dull red heat. (500 to 600 C). Cool and add 5 ml of saturated ammonium carbonate (NH)2 Co3 solution per gram of ash. Digest at room temperature for one hour. Dry in an oven at 100 C to constant weight, cool in desiccator, and weight.

Calculate the weight of ash in the total volume of extract.

Total ash, g = G (V1 )

V1-V2

Where

G = ash in aliquot in grams.

V1 = total volume in milliliters

V2 = volume after removing aliquot in milliliters.

Calculate the percentage of bitumen in the sample. Bitumen content of dry sample, percent.

= (W₁– W₂) - (W₃ + W4) _ X 100

W₁ – W₂

Where

W1 = Weight of sample

W2 = Weight of water in sample,

W3 = Weight of extracted mineral matter

W4 = Weight of ash in extract.

Add the increase in weight of filter disc to the weight of the recovered aggregate and the ash in the recovered bitumen.

MAINTENANCE: Keep all parts clean.

10.BITUMEN EXTRACTOR HAND OPERATED

TRODUCTION: The apparatus meats the essential requirements of ASTM S-2172 – 72 for the determination of percentage of bitumen in having mixtures by cold solvent and centrifugal force method.

DESCRIPTION: The apparatus consists consists of a bowl with cover fitted on a shaft and enclose in housing with removable cover. The cover (1) can be clamped to the housing with clamp (3) and has lifting handles (2) the drain nipple (4) is attached to the housing and helps to drain out the extracted solution. The rotor in driven by handle (6) by rotating it clock vise. The hand brake (5) is provided to stop the bowl.

SPECIMEN PREPARATION: If the mixture is not sufficiently soft to separate with a spatula or trowel, place it in a large flat pan and warm to 240^oF only until it can be handled. Separate the particular of the mixture as uniform ally as possible taking care not to fracture the mineral particles. Normally 1000gms sample are used for extraction. For routine testing, smaller sample may be used when the maximum size of aggregate therein is less than 26mm. this procedure may be used on mixture containing aggregate larger than 25mm by using samples at least 3000gms. They may be tested by extraction of 1000gms at a time.

TEST PROCEDURE: Determine the moisture contents of the sample using a standard procedure. Weight a 1000gms sample into the bowl. Cover the sample in the bowl with benzene or trichloroethylene and allow sufficient time for the sample to disintegrate the sample (Not more than one hour) place the bowl containing the sample and the solvent inside the housing fitting it over the shaft. Dry and weigh the filter paper disc and fit around the edge of the bowl and position the housing cover clamping it to the housing. Place a beaker under the outlet drain on the side of housing.

Rotate the bowl with the help of (6) and gradually increase the speed of the bowl until the solvent ceases to flow the nipple (4). Then stop the bowl by pulling brake (5). Add 200ml of benzene and repeat the procedure. Use sufficient 200ml solvent additions (Not less than three) so that the extract is clear or not dark than a light straw color.

Remove the filter disc from the bowl and dry in air. Remove as much mineral matter adhering to the disc as possible and add to the aggregate in the bowl. Dry the disc to constant weight in an oven at 210^oF. Dry the contents of the bowl on a steam bath and than constant weight in an oven at 210^o to 220^oF. Record the volume of the total extract in the graduate cylinder. Agitate the extract thoroughly and immediately measure approximately 100ml into a previously weight ignition disc. Evaporate to dryness on a steam bath. Ash the residue at a dull red heat (500^o to 600^oc), cool and add 5 ml of saturated ammonium carbonate solution per gram of ash. Digest at room temperature for one hour. Dry in an oven 100⁰c to constant weight. Cool in a desiccators and weight.

Calculate the weight of ash in the total volume of extract.

Total ASH: -G – V1

Vi – V2

Where G: - ash in aliquot in games.

Vl: - Total volume in ml

V2:- Volume after removing aliquot in milliliters

Calculate the percentage of bitumen in the sample as below:-

% :- (Wl-W2) – (W3-W4) _{X 100}

W1-W2

Where W1 Weight of sample

W2 Weight of water in sample

W3 Weight of extracted mineral matter

W4 Weight of ASH in extract.

Add the increase in weight of filter disc to the weight of the recovered aggregate and the ash in the recovered bitumen.

MAINTENANCE: Keep the apparatus clean.

Grease the bearing of the shaft occasionally.

11.RAPID MOISTURE METER

CAUTION: The absorbent with the outfit is highly susceptible to absorbent of moisture and so must not be exposed to atmosphere. Replace the lid firmly as soon as the required amount of the absorbent for a test is taken from the bottle. The absorbent suffer deterioration with the result that the instrument will always give lower results. Please ensure that the lid is screwed firmly after use.

OPERATING INSTRUCTIONS: With this portable unit, the moisture contents of soils, sand and other materials can be determined in a few minutes. It can also be used to determine the moisture content of various types of pastes powders and mixes.

The instrument operates on the principle of a reagent being introduced to the free moisture in the sample. This forms a gas, the amount being dependent upon the amount of free moisture in contact with the reagent. By confirming the resultant gas in sealed chamber it is possible to measure the created gas pressure. The gauge in the instrument is calibrated to interpret this pressure as the percentage of moisture in the sample on the wet weight basis.

The Rapid Moisture Meter consists of an aluminum body fitted with a gauge at one end and a sealing at the other end held to the body by a u-clamp and a clamp screw with an ‘o’ ring between the body and the cup. The cup can be removed from the body by just unscrewing the clamping screw and by swinging the u-clamp to the side. The gauge is calibrated in percentage of moisture content on the wet weight basis and a balance id mounted on a base which is fixed in the carrying case and swivels into position when required, when the catch to the left of it is moved back. And the base is rotated about the swivel pins. When the balance base is thus set horizontal the same catch engages to keep it in that position.

The balance arm is prc-loaded to indicate level when the correct weight (5 Gms) sample is placed in the pan. The arm sticrup and the pan are all detachable and are camped separately inside the carrying case. A scoop one bottle containing absorbent are also supplied. The whole unit with the accessories is packed in a wooden carrying case.

Keep the box on a firm level base. Open the carrying case. Move the balance base catch back and rotate the base into horizontal position. Detach the balance arm stirrup and pan from the spring clips holding them and assemble.

The balance arm knife supports have two side covers which can be rotated into position after placing the knife edges in their support hearings.

SAMPLE PREPARATION: Fine powders and send coarse: No preparation grind or

Powders cohesive soils. Pulverize. Run test as

Describe in note under

Test procedure using dry

Pieces of 1” size gravel.

TEST PROCEDURE: Weight of samples. Set up the balance as described above. Place sample in pan till the mark on the balance arm are at same level.

Unclamp the clamping screw of the instrument sufficiently to move the u-clamp off the cup. Life off the cup. Check cup and body are clean, clean it properly. Hold the body horizontal and gently one level scoopful of absorbent (calcium carbide) halfway inside the chamber. Then lay the chamber down without disturbing the absorbent charge. Transfer the soil weighed out as above from the pan to the cup.

Holding cup and chamber approximately horizontal bring them together without disturbing sample or absorbent, bring the u-clamp round and clamp the cup tightly in to plate.

NOTE: If the sample is bulky reverse the above placements i.e. put the sample in the chamber and the absorbent in the cup.

With gauge downwards (except when the steel balls are used) shake the moisture meter up and down vigorously for 5 seconds, then quickly turn it so that the gauge is upwards, give a tap to the body of the moisture meter to ensure that all the contents fall into the cup. Hold the rapid moisture meter in this position for one minute. Repeat this for the third time. Once more ingest the R.M.M. and shake up and down to cool the gas. Turn the R.M.M. with the gauge upwards and dial facing you at eye level. When the needle comes to re take the reading.

NOTE:2: In the case of clayey soils plaid 6 steel balls in the absorbent and the material in the cup and seal up the unit as usual. Mix as follow. Hold the rapid moisture meter vertical so that the material in the cup falls into the body. Now holding the unit horizontal rotate it for 10 seconds so that the balls are rolled round the inside circumference of the body. Rest for 20 seconds fall into the body. Now holding the unit horizontal rotate it for 10 seconds so that the balls are rolled round the inside circumference of the body. Rest 20 second. Repeat the rotate-rest cycle unit the gauge reading is constant (usually this takes 4 to 8 minutes). Note the reading as usual.

NOTE:3: Where sticky or post samples are to be tested quicker, results may be obtained if after weight the sample, it is spread on a small piece of paper with a spatula and the test conducted in the following manner.

Introduce the piece of paper with the sample into the chamber. Place 8 to 10 sharp edged, dry pieces of gravel of 1” size along with the sample. Take two scoopful of absorbent in the cup and assemble 2 the R.M.M. as above in test procedure. The remaining test procedure is the same as in Note 2 of the test procedure.

In the case heavy, the test results can be obtained by shaking the R.M.M. off and on for about 30 minutes.

Finally, release the pressure slowly away from the opening the clamp screw and taking the cup out, empty the contents and clean the instrument with brush etc. the readings obtained on the R.M.M. are the percentage of moisture on the weight basis. To calculate the percentage of moisture on the dry weight basis as in the conversion in soil mechanicals, follow the calculations given below:-

Let the reading on the R.M.M. for a particular sample be w%

Weight of dry material in the sample (100-w) %

The moisture content of the sample on dry weight basis w x 10%

100w

MAINTENANCE: Keep all the parts clean always.

12.Torsionटाेरिजन

1) A solid circular shaft and a hollow circular shaft whose inside diameter is ¾ of the outside diameter, are of the same materials, of equal length and are required to transmit a given torque T. compare the weights of these two shafts if the maximum stresses developed in the two shafts are equal. Ans: D_H = 1.135 Ds, Ws=1.772W_H

2) A solid shaft transmits 250 kw at 100 rpm. If the shear stress is not to exceed 75 N/mm². What should be the diameter of the shaft? If this shaft is to be replaced by a hollow one whose internal diameter = 0.6 times the outer diameter, determine the size and percentage saving in weight, the maximum shearing stress being the same. Ans: d=117.473 mm, D_H=123.036mm, d_H= 73.882mm.

3) A solid shaft 6.50m long is fixed at each end. A torque if 91 Nm is applied to the shaft at a section 2.5m from one end. Find the fixing torques set up at the end of the shaft. If the shaft diameter is 35mm, find the maximum shear stresses in the two portions. Also find the angle of twist for the section where the torque is applied. Take G= 8.4 * 10⁴N/mm2.

Ans: T₁= 56Nm, T₂= 35Nm, ᴓ= 0.01113 radian, τ_s1=6.65 N/mm², τ_s2= 4.16 N/mm².

4) Two solid shaft AC and BC of aluminium and steel are rigidly fastened together at C and attached to rigid supports at A and B shaft AC is 75 mm in diameter and 2m in length. Shaft BC is 55mm in diameter and 1m in length. A torque of 200 Nm is applied at the junction C. Compute the maximum shearing stresses in each material. What is the angle of twist at junction? Take moduli of rigidity of materials as G for aluminium = 3*10⁴N/mm², G for steel =3*10⁴N/mm^2.

Ans: τ_s1=0.88 N/mm², τ_s2= 3.89 N/mm², ᴓ= 0.0015radian.

5) For the shaft shown in fig, determine the end fixing couples diameter of the shaft if the maximum shearing stress is not to exceed 85 MN/mm² and the position of the section where the shaft suffers no angular twist. Ans: D = 93.8mm, x = 0.404m

6) In a tensile test, a test piece 25mm in diameter, 200mm gauge length stretched 0.0975mm under a pull of 50 KN. In a torsion test, the same rod twisted 0.025m radian over a length of 200mm, when a torque of 400 Nm was applied. Evaluate the Poisson’s ratio and the three elastic moduli for the material. Ans: E= 2.089*10⁵, G= 0.834*10⁵ N/mm². µ= 0.252, k= 1.404*10⁵ N/mm^2.

7) A hollow shaft of diameter ration 3/8 is to transmit the 375kw power at 100 rpm. The maximum torque being 20% greater than the mean, the shear stress is not to exceed 60 N/mm² and the twist in a length of 4m is not to exceed 2 degree. Calculate its external and internal diameter which would satisfy both the above conditions. Take G= 8.5*10⁴ N/mm². Ans: T= 35810 Nm, when τ_s=60 N/mm^2, D= 155mm & d= 58.11mm, when ᴓ= 2 °, D= 157mm & d= 59mm.

8) A hollow shaft having internal diameter 40% of its diameter, transmits 562.5 KW power at 100 rpm. Determine the external diameter of the shaft if the shear stress is not to exceed 60 N/mm² and the twist in length of 2.5 m should not exceed 1.3°. assume maximum torque is equal to 1.25 times of mean torque and modulus of rigidity as 9*10⁴ N/mm². Ans: T_max= 67143.5 mm, T= 53714 Nm, when τ_s=60 N/mm^2, D_H= 180.18mm, when ᴓ= 2 °, D_H= 180mm.

9) Shaft BC is hollow with inner and outer diameter of 90 mm and 120 mm respectively. Shafts AB and CD are solid and of diameter d. for the loading shown in fig. Determine

a) The maximum and minimum shearing stress in shaft BC

b) The required diameter d of shaft AB and CD if the allowable shearing stress in these shafts is 65 MPa.

Ans: a) τ_max= 86.2 MPa, τ_min= 64.65 MPa & b) d= 77.8 mm.

10) A brass tube of external diameter 80 mm and internal diameter 50 mm is closely fitted to a steel rod of 50mm diameter to form a composite shaft. If a torque of 6 KNm is to be resisted by this shaft, find the maximum stress developed in each material and the angle of twist in 2m length. Ans: (τ_{s) max}= 64.701 N/mm², (τ_{b) max}= 64.701 N/mm², ᴓ= 0.0647 radian.

11) A 30 mm diameter circular steel shaft is provided with enlarged portions A and B as shown in fig. On to this enlarged portion, a steel tube 2mm thick is shrunk. While the shrinkage process is going on the 30 mm diameter shaft is held twisted by a couple of magnitude 80 Nm. When the tube is firmly set on the shaft, this twisting couple is removed. Calculate what twisting couple is left on the shaft, the shaft and the tube being made of the same material.

Ans: Ts= 60 Nm

12) A stepped shaft is subjected to a torque as shown in fig.Determine the angle of twist at the free end. Take G = 80 KN/mm². Also find the maximum shear stress in any step.

Ans: (τ_{s) AB}= 17.25 N/mm², (τ_{s) BC}= 9.95 N/mm²& (τ_{s) CD}= 23.579 N/mm²

13) A solid shaft of 200 mm diameter has the same cross-sectional area as that of a hollow shaft of the same material with inside diameter 150 mm. Find the ratio of power transmitted by the two shafts at the same speed. Ans: P_H/Ps= 1.7

14) A solid circular shaft, 1.5m long and 60 mm in diameter is subjected to a 7.20 kN-m torque at the free end. Assuming that the shaft is made of an elasto-plastic material with yield strength in shear of 150 MPa and a modulus of rigidity of 80 GPa, determine a) the radius of the elastic core, b) the angle of twist of the shaft.

Ans: 25.28mm, 0.1115 radian

15) A steel shaft and an aluminium tube are connected to a fixed support and to a rigid disk as shown in the cross-section in fig. Knowing that the initial stresses are zero, determine the maximum torque T_o which may be applied to the disk if the allowable stresses are 120 MPa in the steel shaft and 70 MPa in the aluminum tube. Take Gs= 80 GPa for steel and G_Al = 27 GPa for aluminum. Ans: 6.188 KN-m

13.Marshall Stability Apparatus

Operating Instruction For

Resistance in to plastic flow of bituminous matured using

Marshall Stability Apparatus

1. Scope: This method coves the measurement of the resistance to plastic flow of cylindrical specimens of bituminous paving mixture loaded on the lateral surface by means of the Marshall apparatus. This method is for use with hot mixture containing asphalt of tar and aggregate unit in maximum size.

2. Apparatus

a) Specimen Mould Assembly - Mould cylindrical 4 inch dia meter and 3

Inch in height base plates and extension collar shall conform to the details. Three mould cylinders are recommended.

b) Specimen Extractor - A specimen extractor the compacted from the specimen mould is recommended. A suitable bar is required to transfer load from the extension collar to the upper proving ring attachment while extracting the specimen.

c) Compaction Hammer - The Compaction Hammer shall have a flat, circular tamping face and a 10 Lb sliding weight with a free fall of 18 inch two compaction Hammer are a recommended.

d) The Compaction pedestal shall consist of a wooden post capped with a 30 cm by 30cm by 25mm (Aprox) in steel plate. The wooden post shall be oak, yellow pine, or other wood having a dry. The wooden post shall be secured by four angle brackets to a solid concrete stab. The steel cap shall be firmly fastened to the post. The pedestal assembly shall be installed so that the post is plumb and cap is level.

e) Specimen Holder - The specimen mould holder shall consist of a semicircular base and circular ring to hold the specimen mould in place during compaction of the specimen the top section shall be flanged to fit over the collar of the specimen mould and shall be attached to the base by means of a fulcrum on one side. Holder shall be provided in the base for mounting on the compaction pedestal. The specimen mould holder shall be mounted on the pedestal cap so that the center of the mould is over the post.

f) Breaking Head - The breaking head shall consist of upper and lower cylindrical segments or test heads having an inside radium of N curvature of 2^nd accurately machined. The lower segment is mounted on a base having two perpendicular guide rods or posts extending upward. Guide sleeves in the upper segment are in such a position as to direct the two segments together without appreciable binding or less of motion on the guide rods.

g) Loading Frame - The loading frame consists of a Hand crank operated screw jack mounted in steel testing frame and shall produce a uniform vertical movement of 2^nd per min. An Electric motor may be attached to the jacking mechanism, by fixing an additional base attachment with suitable pulleys for obtaining required speed of travel. A circular plate is mounted on top of the traversing screw for supporting.

h) Proving Ring Assembly - One proving Ring of 5000 kg capacity shall be equipped with a dial gauge 0.002mm / travel upper and lower proving Ring attachments are required for fastening the proving ring to the testing frame and transmitting the load to the breaking head.

i) Flow Gauge - One dial gauge also supplied 0.01mm and 25mm travel for flow. This dial gauge is fitted with Braking Head (f).

j) Oven Or Hot Plates - Oven or hot plates shall be provided for heating aggregate bituminous material, specimen mould, compaction hammers, and other equipment to the required mixing and molding temperatures. It is recommended that the heating units be thermostatically controlled so as to maintain the required temperature within 5F (2.8 C) suitable shields baffle plates bath shall be used on the surfaces of the hot plates to minimize localized over heating.

k) Mixing Apparatus - Mechanical mixing is recommended. Any type of mechanical mixer may be used provided it can be maintained at the required mixing temperature bad will produce a well coated, homogeneous mixture of the required in the allow able time, and further provided the essentially all the bath can be recovered. A metal pan or bowl of sufficient capacity and hand mixing may also be used.

l) Water Bath - The Water bath shall be at least 6 in deep and shall be thermostatically controlled so as to maintain the bath at IS 140-1-8F (60) (10) The tank shall have a perforated false bottom or be equipped with a shelf for supporting specimen 2^nd above the bottom of the bath.

3. Miscellaneous Equipments

1) Containers for heating aggregate flat-bottom metal pans or an other suitable

Containers.

2) Container for heating bituminous material, either gill type tins, beakers pouring posts, or saucepans may be used.

3) Mixing tool either steel trowel (Garden type) or Spatula, or spading and hard mixing.

4) Thermometer of determining temperatures of aggregates, bitumen and bituminous mixtures. Armored glass thermometers or dial type with metal stem are recommended. A range of 50 to 400F.

5) Thermometer for water bath with a range of 20 to 70 deg. Cent.

6) Balance 5kg capacity, sensitive to 0.1gm for weighing moulded specimens.

7) Balance 5kg capacity, sensitive to 1.0gm for bathing mixtures.

8) Gloves for handing hot equipments.

9) Rubber Gloves for removing specimens from water bath.

10) Marking pencils for identifying specimens.

11) Scoop flat bottom for batching aggregates.

12) Spoon Large for placing mixture in specimen moulds.

Test Specimens –

a) Number of Specimens - Prepare at least three specimens for each combination of aggregates and bitumen contents.

b) Preparation of Aggregates - Dry to constant weight at 221 to 231 F (105 to 110 deg. Cent.) Hand operates the aggregates by dry sieving into the desired size fractions. The following size fractions are recommended. 1 to /3¹/2 inch, ¹/2 to 3/8 inch, 3/8 inch to 4 (4.75mm) No. 4 (4.75mm to 8 (2.36mm) passing No. 2.36mm. (8).

c) Preparation of Mixtures - Weight into separate pans for each test specimen, the amount of each size fraction required to produce a batch that will result in compact specimen 2.5 = 0.05 in, in height (about1200gm). Place the pen son the hot plate or in the oven and best to a temperature of approximately 27.8 deg. Cent. Above the mixing temperature established in paragraph (0) charge the mixing bowl with the heated aggregate and dry mix thoroughly, from a crater in the dry blended aggregate and weight the required amount of bituminous material an into the mixture. At this point the temperature of the mixing temperature established in paragraph (e) Mix the aggregate and bituminous material rapidly until thoroughly coated.

d) Compaction of Specimen - Thoroughly clean the specimen mould assembly and the face of the compaction hammer and best than either in boiling water or on the hot plate to temperature between 200 and300F (93.3 and 148.9C) place a piece of filter paper or paper to woling out to size in the bottom of the mould spade the mixture vigorously with a heated spatula. Remove the collar and smooth the surface of the mixtures immediately prior to compaction shall be within the limits of the compacting temperature established in paragraph.

e) Replace the Collar - Place the mould assembly on the compaction pedestal in the mould hold and unless otherwise specified, apply 50 blows with a hammer perpendicular to the base of the mould assembly during compaction. Remove the base plate and collar, and reverse and reassembly the mould. App the same number of compaction blows to the face of the reversed specimen after compaction; remove the base plate and the place the sample extractor on that end of the specimen. Place assemblies with the extension collar up the testing machine apply pressure to the collar by of the load transfer bar and force the specimen into the extension collar. Lift the collar from specimen. Carefully transfer the specimen to smooth flat surface and allow it to stand overnight at room temperature. Weight cooled as specified in paragraph (f). When more rapid cooling is desired table fans may be used. Mixtures that lack suffidder sufficient cohesion to result in the required cylindrical shape shape on removal from the mould immediately after compaction may be cool. In the mould in air until sufficient cohesion has developed to result in the proper cylinder shape.

Procedure:-

a) Bring the specimens to the desired temperature by immersing them in the bath and place in the lower segment of the breaking head. Place the upper segment of braking head on the specimens and place the dial gauge to zero while holding it firmly against the upper segment.

b) Apply load to the specimen by means of the constant rate of movement of the load jack or testing machine head of 2 inch per minute until the maximum is read reached and the load jack or testing machine head of 2 inch indicated by dial. Record the maximum is reached and the load noted on the testing machine or converted from the micrometer dial reading. Note the micrometer dial reading where the instant the maximum load begions to decrease. Note an record the indicated flow value or equivalent units in hundred the of inch if a micrometer dill is used measure the flow the elapsed time for the test from recovery of the test specimen from the water bath to the maximum load determination shall not exceed 30 deg. Cent.

Note – For core specimens correct the load when thickness is other than 2 inch by using the proper multiplying factor from table I.

Report –The report shall include the following the information for each specimen tested,

i. Weight of the test specimen.

ii. Maximum load in kgs corrected when required.

iii. Flow value in hundred of an inch.

iv. Mixing temperature.

v. Compacting temperature, and

vi. Test temperature.

14.Road Pavement

1.1 Introduction

Road pavement can be defined as a relatively stable layer constructed over the natural soil for the purpose of supporting and distributing the wheel loads and providing an adequate surface for the movements of vehicles with certain speed safely, comfortably and economically. One main objective/function of a well-designed pavement is to keep temporary deformation of the pavement with in the permissible limits so that the pavement can sustain a large number of repeated load applications during design life. In short, road pavement is a stable layer constructed over the natural soil.

Pavement layers

Pavement consist of one or more layers of subgrade, sub base, base, surface or wearing course.

Surfacing course is the top course and is provided to provide smooth, abrasion resistant, dust free, reasonably water proof and string layer. Base course is the medium through which the stresses imposed are distributed evenly to the underlying layers. Sub base layer provides additional help in distributing the loads. Subgrade course is the compacted natural earth and top of the sub grade level is also known as the formation level.

Surface

Binder

Surface

Base Base

Sub Base Sub Base

Sub Grade Sub Grade

Indian Practice American Practice

Wearing

Base

Road Base

Sub Base

Sub Grade

British Practice

Types of Pavement:

Based up on the structural behavior of the materials used in the construction, pavements are classified as

a) Flexible pavement

b) Rigid pavement

And also, c) Semi rigid pavement

d) composite pavement

Flexible Pavement:

• The pavements which have very low flexural strength and are flexible in their structural behaviour under the load are called flexible pavements. The flexible pavement layers reflect the deformation of the lower layers on to the surface of the layer. Thus, if lower layer somehow gets deformed, then the surface of the pavements also gets deformed.

• Different layers used in flexible pavement are- Soil subgrade, sub base layer, base layer, surface layer.

• The flexible pavement layers transmit the vertical loads to the lower layers by grain to grain transfer through their point of contact in the granular structure. The load spreading capacity of the flexible layer depends on the type of material and the mix design factors. The materials which fall in the category of flexible pavement layers are soil aggregate mix, crushed aggregate, WBM, granular materials with bituminous binder, bituminous concrete. The top layer has to be the strongest as the highest stresses are to be sustained by this layer. They also have to withstand the wear and tear due to traffic. As the lower layers have to sustain lesser intensity of stresses and there is no direct wearing action, therefore inferior materials with lower cost can be used.

• The pavement thickness is so designed that the stresses on the sub grade soil are kept within its bearing power and the sub grade is prevented from excessive deformations.

Rigid Pavement:

• Rigid pavements are those which possess considerable flexural strength.

• The rigid pavements are made of cement concrete which may be either plain, reinforced or pre stressed.

• The rigid pavements have a slab action and are capable of transmitting the wheel loads stresses through a wider area below.

• The main difference between rigid and flexible pavements in the structural behaviour is that the critical condition of stress in rigid pavement is the maximum flexural strength occurring in the slab due to the wheel load and the temperature changes where as in flexible pavement it is the distribution of compressive stress to the lower layers and lastly over the soil subgrade. The rigid pavement doesn't get deformed to the shape of the lower surface as it can bridge the minor variations of lower layers.

• Usually the rigid pavement structure consists of a cement concrete slab, below which a granular base or sub-base course may be provided.

• A good base or sub base course under the cement concrete slab increases the pavement life considerably and therefore works out more economical in the long run.

Semi rigid pavement:

• When bonded materials like pozzolanic concrete (lime-fly ash-aggregate mix), lean cement concrete or soil cement are used in base or sub-base course layer the pavement layer has considerably higher flexural strength than the common flexible pavement layers. However, these bonded materials do not possess as much flexural strength as the cement concrete pavements. These pavements are called semi rigid pavements. These semi rigid pavement materials have low resistance to impact and abrasion and therefore are usually provided with flexible pavement surface course.

• Intermediate between the flexible and the rigid pavement.

• Much lower flexural strength compared to concrete slabs but derives the support by the lateral distribution of loads through the pavement depth as in a flexible pavement.

Composite pavement

• A composite pavement comprises of multiple, structurally significant layers of different composition.

• In its widely used form, composite pavement consists of PCC as bottom layer and bituminous layer as a top layer resulting in an ideal pavement with most desirable characteristics.

• The bottom layer (PCC) provided a string base and the bituminous layer (top) provides a smooth and non-reflective surface.

• This type of pavement is very expensive and is rarely used as a new construction

Differences between flexible and rigid pavement

Flexible and rigid pavements differ in many characteristics which can be described as below

Rigid pavement	Flexible pavement
1. Flexible pavements under heavy loads yield to excessive stresses resulting in the local depression of the surface.	A rigid pavement under heavy load ruptures thereby producing a crack to the surface
2. A flexible pavement with subgrade of varying thickness will adjust itself to the irregularities due to different settlements	A rigid pavement with the subgrade of varying strength will not adjust the irregularities due to different settlement but acts as a beam or cantilever.
3. flexible pavements under load worsened condition in subgrade will get depression in the pavement	A rigid pavement instead is capable of bridging the small weakness and depressions in the subgrade
4. temperature variations due to atmospheric conditions do not produce stresses	Temperature variations produce heavy temperature stresses
5. the flexible pavement has self-healing properties (recoming to shape)	The rigid pavement doesn't have self-healing properties
6. strength of flexible layer is a result of building up thick layers and thereby distributing the load over subgrade	Strength of rigid layer is rather by bending action

Design precision—Cement concrete pavement design is much more precise structural analysis because flexural strength of concrete is well understood. But Flexible pavement designs are mainly empirical.

Design life—Well-designed concrete slab has a life of about 40 years whereas Flexible pavements has a design life of 10~20 years (with extra maintenance input)

Maintenance—Awell-designed cement concrete pavement needs very little maintenance (mainly of joints) and in other hand, Bituminous surfaces need great inputs in maintenance

Initial cost—Argument is made that cement concrete slab is much costlier than flexible pavement. But If higher specification of bituminous pavement is selected, the argument that a cement concrete specification is costlier than a flexible pavement should no longer be valid.

Stage construction—stage construction is possible on bituminous pavement—initial outlay is minimum and additional outlays are in keeping with traffic growth thus at no stage the investment made in advance of the actual requirement. In other hand, Cement concrete slab do not fit such scheme of stage construction.

Availability of material— Bitumen is a scarce resource and should be imported involving hard earn foreign exchangewhereas Cement can be manufactured with in the country

Surface characteristics— Cement concrete provides smooth pavement surface, free from rutting, potholes and corrugations with good riding quality.

Asphalt concrete pavement provides comparable riding quality

Well-constructed cement concrete pavement gives permanent non-skid surface but faulty design may become very smooth which is extremely costly to restore the non-skid characteristics.

Penetration of water— Cement concrete is practically impermeable except at joints where the problem of mud pumping exists. However bituminous surface is mot impervious in which water enters through pores and cracks which impairs stability of pavement.

Utility location— No digging up the pavement for water supply pipes, telephone lines, electric poles etc. is possible for rigid pavement but the same be accomplished in flexible pavement though the practice is not good one.

Glare and night visibility— Cement concrete is grey in color which produces glare on the sun lights whereas Bituminous roads being black in color, need more street lighting for night driving conditions.

Traffic dislocation during construction— Cement concrete pavement needs around 28 days for setting and curing to gain its strength but bituminous surface can be opened to traffic shortly after it is rolled and, in this case, traffic will facilitate its compaction. Concrete pavement causes longer dislocation of traffic if the work is done on existing pavement.

Environmental considerations during construction— Heating of aggregates and bitumen in hot mix plants can prove to be much more hazardous. Use of cutbacks can also prove to be environmentally hazardous due to evaporation of volatile constituents into the atmosphere.

Overall economy on a life cycle basis—on overall economic considerations, rigid pavement is far more economical than flexible pavement in the long run.

Advantages and disadvantages of Rigid pavement

Advantages

- High strength: compressive, abrasion, compression-tension

- Good stability: water, heat stability, strength increases with increase in time

- Durability: 20~40 years

- Low maintenance cost, large economic gain: big initial investment but long design period therefore maintenance cost per year is low.

- Suitable for night driving

Disadvantages

- Necessity of cement and water is large: for 20cm depth, 7 m wide cement concrete pavement for every 1000 m needs about 400~500t cement and 250t water. Not included water necessary for curing. Difficult where these materials are very hardly available

- Have joints: increases difficulty in construction and maintenance, easily causes vibration on vehicle. If not handled properly, damages may occur

- Pavement is quite lately available for vehicle operation: needs of 15~20 days

- Difficult to repair: repairing work big, influence the traffic movement

1.2 Functions of different layers in pavement structure

v Soil subgrade

The soil subgrade is a layer of natural soil prepared to receive the layers of pavement materials placed over it. Traffic load moving on the surface of the road is ultimately transferred to the subgrade through intermediate layers. The pressure transmitted on the top of the subgrade should be within the allowable limit so as not to cause excessive stresses condition or deform the same beyond the elastic limit. The top layer of the subgrade soil should be well-compacted under controlled condition of optimum moisture content and maximum dry density. It is necessary to evaluate the strength properties of soil subgrade. If the strength properties are inferior, suitable treatment should be given to impart improvements in the performance of soil subgrade.

v Sub base and Base layer:

These layers are made of broken stones bounded or unbounded. Sub base layer may sometimes be constructed by stabilized soil or selected granular soil. At sub-base course, it is desirable to use smaller size graded aggregates or soil aggregate mixes or soft aggregates instead of large boulder stone. Sub base course primarily has the similar function as that of the base course and is provided with inferior materials than of the base course.

Base and sub base courses are used under flexible pavement primarily to improve the load supporting capacity by distributing the load through a finite thickness. Base courses are used under rigid pavement for preventing the mud pumping, protecting the subgrade against frost action.Thus, the fundamental purpose of a base and sub base course is to provide a load transmitting medium to spread the surface wheel loads in such a manner as to prevent shear and consolidation deformation.

Ø Wearing course:

Wearing course performs the following functions:

- Provides smooth and dense riding surface

- Resists pressure and takes up wear and tear duo to traffic

- Provides water tight layer against the filtration of surface water

- Provides hard surface which can withstand the pressure exerted by tyres of vehicle

In flexible pavement, wearing surface is generally made of bituminous material. In cement concrete pavement, the cement concrete slab is used as wearing course. There are many types of surface treatments employed as wearing course. The type of surface depends upon the availability of materials, plants and equipment and upon the magnitude of surface loads.

There is no test for evaluating the structural stability of the wearing course. However, the bituminous mixes used in the wearing courses are tested for their suitability (Marshall Stability test--optimum content of bitumen binder is worked out based on stability, density….)

Chapter 2 Hill Roads

1.1 Definition and importance

A hill road can be defined as one which passes through the area with a cross slope of 25% or more. A hilly or mountainous area is characterized by a highly broken relief with widely differing elevations, steep slopes, deep gorges and great number of water courses.

Living in hill area depends upon the agriculture and other products of that area which are not sufficient for them where as in plain terrain the agricultural and industrial products are always surplus. This unbalanced productivity of land makes the people of hilly area underprivileged. For the overall economic development of the nation this unbalanced distribution of national product should be curtailed down. Surplus products in one area should be served in deficit area. Construction of hill roads is important not only from economic consideration but it is associated with the social justification for providing facilities to underprivileged mass of the country.

The hilly regions generally have extremes of climatic conditions, difficult and hazardous terrain, topography and vast high-altitude areas. The region is sparsely populated and the basic infrastructural facilities that are available in plain areas are mostly absent. The roads in these areas are affected by floods, landslides, snowfall etc. compelling certain roads to be closed in part of the year especially in winter months. But these areas are rich in natural resources, flora and fauna and are important to launch development projects, tourism etc.

The population

A terrain can be classified into four groups based on the ground cross slope i.e. the slope approximately perpendicular to the C.L. of the highway alignment.

Terrain type	Cross slope, %
Level/Plain Rolling Mountainous Steep	0 – 9.90 10 – 24.9 25 – 60 Above 60

2.2 Design and construction problems in hill roads

Hilly area has broken relief with widely differing elevation in a short distance. It requires considerable length of road

ØGeological condition varies in a short distance. It makes difficult to design and construct embankment and for other structures.

ØCross slope may become unstable after road construction due to removal of vegetation and other activities.

ØDifficult to investigate and forecast the hydro-geological condition of hill slopes which may cause land slide

ØRequires more numbers of cross drainage structures and other special structures

ØSteep slope needs careful arrangement of erosion control measures.

ØSurface run-off become very high in a short period after heavy rain fall, it requires the big opening for cross drainage.

ØVariation of climatic condition causes the road damage

vTemperature

vAtmospheric pressure

vPrecipitation

vWind

ØSpecial safety measures should be taken or construction works.

2.3 Special consideration of hill road geometric design

We have to consider following things,

a) Road location and alignment

b) Geometric design

c) Typical cross section

d) Special structures (retaining walls, drainage, slope protection works etc.)

Alignment of hill roads:

Special consideration should be made in respect to variation in:

vTemperature

vPrecipitation/rainfall

vAtmospheric pressure and wind

vGeological condition

Temperature:

- Temperature decreases per 100m increase in elevation (0.50C)

- Hill slopes facing south-west and south receive enough solar heat; snow disappears quickly and rain water evaporates rapidly.

- North and north east; rain water and snow remain for long period.

- Sharp temperature variation of south and south-west slopes results the fast weathering process, which causes deposition of alluvial fan on that side, mud flow and avalanches occur frequently.

Rainfall:

- In general, the increase of rainfall for every 100m of elevation averages 40 – 60 mm.

- Hill/ mountain slopes which face winds, receive more rainfall.

- Heavy rainfall may occur in hills, which may cause serious damages to hill roads

- Openings of the cross drainage are requiring very large.

Atmospheric pressure and wind:

- Atm. pressure decreases with the elevation

- At low atmospheric pressure engine effort become less

- At high altitude wind blows with high velocities cause erosion of pavement material

- Wind may damage the road pavement: by blowing away the binding particles (in dry season), and by the erosive action of surface run off (in wet season).

Geological condition:

- Sedimentary rocks often occur as folds which may concave and convex with inclinations.

- Degree of stability of hill slopes depends upon the rock type, strata, inclination and presence of ground water.

Alignment of hill road (Route location)

Selection of alignment in hilly region is a complex problem. The designer should attempt the shortest, most economical and safe route. Hill roads tend to follow routes with large number of curves.

Route location may be:

Ø River route

Ø Ridge route

River route:

The location of a route along the river valley is the most frequent ease of hill road alignment owing to the distinct advantage of running the road at a comparatively gentle gradient. The route along the river route serves rural settlements situated next to the water course. It has the advantage of low operating cost, availability of water and other construction materials in the vicinity. However, a valley run may involve numerous horizontal curves, construction of large bridges over tributaries and on stretches along steeply sloping hill sides, which in some places may be insufficiently stable.

Fig. typical example of river route

Advantages:

ü Minimum gradient

ü Availability of water and other construction materials

ü Road can serve settlements, situated along river valley

ü Low vehicle operation cost

Disadvantages:

ü River alignment has many horizontal curves

ü Route cross large number of tributaries. It means large number of cross drainages works.

ü Need of protection works against the washout, or toe cutting of foundation of road bed or other retaining structure.

Basic consideration in locating the river route:

ü Road bed should be away and above from the maximum water level in the river so that the risk of erosion or seepage during high water level.

ü Embankment slope facing the river should be protected and stabilized

ü It needs good geological and hydro-geological surveys of local conditions

ü When crossing water courses, several route alternatives have to be investigated.

Ridge route:

Ø Steep gradient

Ø Large number of curves (hair pin bends)

Ø Expensive rock work, many other structures

Ø It climbs continuously up from the valley till it reaches a mountain pass and descends down to follow another hill system

Ø Mountain pass should have the least elevation to the direction of destination

Ø If land slope is steeper than the permissible maximum gradient, the route can’t be laid along the shortest direction. So, length has to be increased to gain the height with permissible gradient.

Ø The route is traced out in the map by following more or less the line of equal gradient, slightly lower than the ruling gradient.

Alignment survey of hill roads:

The alignment of hill roads is fixed in the three stages:

Ø Reconnaissance

Ø Trace cut-1 to 1.2 m wide track

Ø Detailed survey-B/M Fixation, L- section, X-section, Center line marking, Hydrological and soil investigations

Geometric design of hill Roads: Main reasons for the difference in design are complexity of terrain, high altitude factors and other problems in the design and construction of hill roads. Special consideration should be given for the selection of gradient and the design of hair pin bends.

2.4 Typical cross sections of hill roads

The cross section of a road in hilly terrain is determined by the original ground slope of the site, the slope of the road formation, width of roadway, side drain size and shape and so on. Various configurations of road cross sections include.

1. Cut and fill

2. Bench type

3. Box cutting

4. Embankment with retaining walls

5. Semi bridge

6. Semi tunnel

7. Platforms

Hill road cross section example 1

Hill road cross section example 2

Different cross section used are

1. Cut and fill:

When roadbed slope has a gradient 2% or more a cut and fills road bed is cheaper and environmentally friendly as well. The fill mass is generally balanced by the cut mass. For adequate stability, benches are made on the surface of the hill side with a height of 0.5 m and length varying from 1.5 to 3.0 m depending upon the slope.

2. Bench type

A cross section of the bench type although entails some increase in earthwork but ensures the complete stability of the road bed, if hill side is itself stable.

3. Box cutting

When the location of road bed is unstable or unsuitable along the hillside due to one or other reasons, the road bed is designed as trench type of cross section. Itincreases earthwork to a large extent. It is introduced to meet the geometric design standards for a given category road.

It is introduced in order to meet the geometric design standard for given category of road. When a road is ascending up the grade is reduced substantially by raising formation line at the beginning with fill and lowering the same at the above section with box cutting. This way, the length of road may be substantially reduced.

4. Embankment with retaining walls

On steep slopes of about 30-35°, the earthwork involved in constructing the embankment increases substantially. The retaining wall is sometimes provided to reduce earthwork’s cost and to increase

stability. Also, the retaining wall is provided when embankment soil on steep grounds itself need support. They may also be constructed on a less steep ground slope to increase the stability of road bed.

5. Semi bridge

If the road is located on a hill slope the retaining wall needs to be at a substantial height. In such

cases, to reduce quantities of work, road bed with a semi-bridge type of structure may be

constructed. Part of the roadway is accommodated on bench cut and part on the semi bridge.

6. Semi tunnel

When inscribing is to be cut into steep hills in stable rock faces, the rock may be permitted to overhang the road to reduce rock works. Such a cross section is called a semi-tunnel.

Fig: With Accommodating Road-Way Only and With Retaining and Breast Walls4

7. Platform

On the precipitous slopes, where shifting of the route into the hillside will lead to enormous rock

works which eventually increases the cost and where semi-tunnel cannot be constructed,platforms are usually cantilevered out of the rock on which road way is partially located.

2.5 Special structures in hill road

When constructing hill roads, a lot of special structures are required owing to harsh geological and hydrological conditions as well as highly broken relief.

Following are the objectives of providing special structures in hill road,

Ø To retain soil mass

Ø To increase stability of road embankment

Ø To accommodate road bed in steep slope

Ø To dissipate energy of surface water

Ø To provide snow avalanche control and protection system

Ø To river training and erosion control

Ø To prevent bed scouring

The following types of structures are mostly used in the hill roads for strength durability and stability:

1. Retaining structures

2. Drainage structures

3. Slope protection structures

1. Retaining structures

A retaining structure is usually a wallconstructed for supporting vertical ornearly vertical earth bank. Retaining wallsare constructed on the valley side on thecut hill side to preventthe slide towardsthe roadway.

Situations whereconstruction of retaining walls is required:

Ø Places where the valley sidesurface gets saturated in themonsoons and is likely to result inslip taking a part of the roadwith it.

Ø Places where undercutting by astream or other water course causesdamage to the valleyside and theroad.

Ø In valley point where water flowsover the road

Ø To achieve roadway width, wherecutting into the hill is noteconomical or has to be restricted due to other reasons.

Major types of retaining structures are as follows

Fig(a). crib wall

Fig (b). buttressed walls fig(c) counterfort walls

Fig(d). gravity retaining wall Fig(e). cantilever retaining wall

Design of retaining walls

1. Assemble the general information about topographical and physical surveys.

2. Analyze the subsoil condition.

3. Establish surcharge load- highway, building, and other loads

4. Select the type and tentative proportion of the wall.

5. Compute the earth pressure and surcharge pressure.

6. Analyze the structural stability.

7. Analyze foundation stability.

8. Design structural elements.

9. Select drainage in backfill.

10. Predict settlement and movement of the wall.

2. Drainage Structures

The main problems that hill roads face are the harmful effect of water. Water may come from

different sources to the parts of the road. This water must be drained using any means necessary.

Drainage of hill roads can be studied under following sub-topics.

a) Drainage of waterfrom hill slope

Surface water flowing fromthe hill towards the roadwayis one of the main problemsin the drainage of hill roads.Since a large amount ofwater flows along withdebris from the hill slopesduring heavy storms, a catchdrain is generally providedto catch the water in themiddle of the slope. Waterintercepted in catch waterdrains are then diverted bysloping drains and carried tothe nearest watercourse or toculvert to cross the roadway.The figure below shows alayout for drainage from hillslopes.

b) Roadside Surface Drainage

Side drains are provided all along the hill side of the road. Due to the limitation in theformationwidth side drains are usually constructed to such a shape that at emergency thevehicles couldutilize this space for crossing. The shapes may be angular, saucer or kerb and channel drains.

c) Cross drainage

A cross drainage is always required on a hill road. The drainage must be taken under the road as

far as possible. At the heads of the small cross drains, catch pits must be provided to collect debris and to prevent scouring.

d) Subsurface drainage

Seepage flow is one of the major problems in hill road. Ground water may seep across hillside

above or below the road level depending upon several factors such as nature and depth of hard

stratum, its inclination, the quantity of ground water etc.

3. Slope protection structures

In hill roads, landslides are very common due to steep slopes. The basic cause of landslide is thedevelopment of shear stresses more than the shear strength of the soil. Fresh unturfedembankmentand cut slopes are the least stable part of the road bed since the soil on the surface of the slopes issubjected to the direct action of sun, rain, and wind.

Causes of landslides

- Increase in moisture content of the soil in hill slopes which increases the pore water

- pressure.

- Alternate swelling and contracting of the soil mass.

- Seepage pressure of percolating groundwater.

- Steeper slopes.

- Human activities like blasting and using heavy vehicles at unstable zones.

Preventive measures

- The highway may be realigned at areas more prone to landslides.

- Construction of retaining walls must be done at places where required

- Adopting easy slopes during design and construction of the road.

- Treatment of slopes to increase stability conditions.

Reinforced retaining walls

This is a type of retaining wall of composite construction material in which strength of fill is

enhanced through the addition of inextensible tensile reinforcement in the form of strips, sheets,

grids, or geotextiles.

It is suitable for hill roads because:

- The fill material is readily available at cheaper cost.

- The land required for embankment is less.

- Cost effective, easy to construct and environmentally friendly.

- It causes less alteration in natural slope.

15.Road Machineries

3.1 Methods of road construction

There are two methods in which road construction works are carried out.

a) Labor based method of construction

b) Machine based method of construction

The methods are selected according to the availability and suitability of equipment, plants and tools in the field. Also, it differs according to size of construction and budget available too.

Labor based method is used when the large plants and advanced technologies are not available in the field. For the rural areas, and very steep slope areas such as hill roads, and mountain areas, we cannot bring the very large mixer, dozers, and compacting equipment’s. So, we must apply the labor-based method in road construction. Generally single lane roads, water bound macadam’s, and green roads are suitable to be constructed by this method.

On the other hand, for the very large projects and in the conditions where the plants, materials,equipment is available then we use the machine-based methods of construction. Also, in some situations, there is difficulties in finding labor, also sometimes labors can not work in adverse conditions like mining, golf countries, underwater tunnels etc. then we must use machines and apply technologies in road construction.

The initial cost of machine-based work is high, however over the life of road, labor-based work also seems to cost more due to necessity of maintenance, repair and future forecast.

3.2 Different types of Tools, Equipment and Plants used in Road Construction

Highway construction can be carried out either by using mechanical appliances or by manual labor. Although adoption of mechanical method involves heavy initial investment but it results superior and economical than those conducted by manual labor. Machines that can be used in road construction can be classified into following heads.

ü Earthwork machinery

ü WBM road machinery

ü Bituminous road machinery

ü Cement concrete road machinery

In road construction, earthwork has to be done to obtain necessary formation level.Tractor, dozer, scrapper, grader, shovel, dragline, power rammers, rooters, trucksetc. are the usual mechanical equipment used for earthwork.

Road metal machineries are primary crusher, secondary crusher and tertiary crusher. Bitumen road machinery consists of bitumen boiler, bitumen sprayer, bitumen mixer and sprayer, spreader, gritting machine, hot mix plant, bitumen plane etc.

Machinery required for cement concrete road construction is as follows--concrete batching plant, concrete mixer, concrete pavers, concrete screens, concrete vibrators, concrete finishers etc.

Ø Small projects—labor intensive works

Ø Big projects—almost impossible without construction equipments.

Types of road pavement

1. Earth and gravel roads

2. WBM roads

3. Soil stabilized roads

4. Bituminous or black top roads

a. Surface dressing

b. Seal coat/prime coat/tack coat

c. Grouted or semi grouted macadam

d. Premix— bituminous bound macadam

Bituminous carpet

Bituminous concrete

Sheet asphalt or rolled asphalt

Mastic asphalt

5. Cement concrete roads

a. cement grouted layers

b. rolled concrete layer

c. cement concrete slab

Construction equipments

1. Tools

a. hand shovel

b. chisel

c. peak

d. spade

e. hand rammer

f. brushes

g. trowel

h. wheel barrows etc.

2. Equipments

a.Earth moving equipments

i. dozer (bull dozer, angle dozer, tree dozer)

ii. scrapper

iii. loader

iv. excavator (back hoe)

v. drag line

vi. clamshell

vii. trench digger

b.Compaction equipment

i. smooth wheel rollers

ii. vibrating rollers iii.pneumatic rollers iv.sheep foot rollers

v.rammers

c.Leveling equipment

i.grader

d.Paving equipment

i. binder spreader

ii. heating kettle for binder

iii. aggregate spreader

iv. cement concrete mixer

v. bituminous paver

vi. cement concrete paver etc.

e.lifting equipment

i. backhoe (for low load)

ii. crane (different capacity)

f.transporting equipment

i. dumping trucks (tipper)

ii. trucks (flat body)

iii. mini dumpers

g.plants

i. cement concrete plant

ii. asphalt concrete plant

iii. cold premix mixing plant

iv. aggregate crusher plant

v. screening plant

vi. washing plant

vii. sand blowing plant

3.3 Different types of Compacting Equipment

Soil compaction can be achieved in the field either by rolling, ramming or by vibration. Hence the compacting equipment may also be classified as rollers, rammers and vibrators. Compaction of sands is also achieved by watering, pounding and jetting. Trucks and heavy equipments do compaction of loose materials to some extent.

Rollers:

The principle of roller is the application of pressure, which is slowly increased and then decreased. The various types of rollers which are used for compaction are smooth wheel, pneumatic tired and sheep-foot rollers.

Smooth wheel rollers

Ø Two types

1. Three wheel or macadam rollers with gross weight of 4~18t

2. Tandem rollers (twowheel)with gross weight of 1~14t

· The compacting efficiency of the smooth wheel rollers depends on the weight, width and diameter of each roller

· Useful for finishing operations after compaction of fills and for compacting granular base course of highways

· Used to seal the surface of the fill to provide a smooth surface to quickly drain off the rainwater.

· They are suitable for compacting gravel, sand, crushed rock and any material where crushing action is required.

Pneumatic tired rollers

· Number of pneumatic wheels (9~11 wheels fixed on two axles) are mounted on two or more axles under a loading platform. Sandbags or some other weights can be placed over the platform to provide the effective compaction.

· Pneumatic tires are so spaced that a complete coverage is obtained with each pass of the roller.

· Compacts the soil by kneading action

· Effective for compacting both cohesive soils and cohesion less soils

· The weight of such roller may be as large as 50t and 2~4 passes are generally sufficient to achieve compaction of 60cm thick soil layer.

· Light rollers (weight up to 20t) for soil layers of small thickness up to 15 cm, heavy rollers useful for layers of thickness up to 30 cm.

Sheep foot rollers

· Consists of hollow steel cylinder with projecting feet.

· The weight of the roller can be increased by filling water.

· The weight, diameter and width of the roller may be varied and also the shape and size of the feet.

· Efficiency of the sheep foot rollers depends on the weight of the roller and the number of feet in contact with the ground at a time.

· Suitable to compact clayey soils

· Combine the soil by the combined action of tamping and kneading

· About 24 or more number of passes of the roller may be necessary to obtain adequate compaction

Rammers

· Block of iron or stone attached to a wooden rod

1. Hand operated of weight about 3.5 Kg

2. Mechanical

· Useful to compact relatively small areas and where the rollers cannot operate due to space limit such as trenches, foundation and slopes.

· The output of the rammer is much lower than that of rollers.

Vibrators

They are most suitable for compacting dry cohesionless granular material. There is also vibrator mounted roller (vibratory roller) to give combined effects of rolling and vibration. They are advantageously used in compacting a wide range of materials.

Watering (Jetting and Pounding)

· Is considered to be an efficient method of compacting cohesionless sands.

· Watering heavily and rolling by smooth wheel of pneumatic tyred roller may also give adequate compaction of cohesionless sands.

The compaction of roller depends up on the following factors:

- Contact pressure

- Number of passes

- Layer thickness

- Speed of roller

16.Road Construction Technology

4.1 Introduction

Road construction technology is that branch ofhighway engineering when deals with all kinds ofactivities and technology or operations for changingexisting ground to the designed shape, slope and toprovide all necessary facilities for smooth, safe andefficient traffic operation and also includes thereconstruction of existing roads.

As per nature and type of works and elements of road to be constructed various activities can broadly divided into several works.

Highway construction project may be broadly divided into two phases:

ü Earthwork and preparation of Sub-grade

ü Pavement construction

The selection of base course and the surface course depends upon:

- Type and intensity of the traffic.

- Funds available for the construction project

- Sub-grade soil and drainage condition

- Availability of construction materials at site.

- Climatic condition.

- Plants and equipment available.

- Time available for completing the project.

4.2 Activities involved in road construction

1. Earthwork and site clearance

· site clearance

· earthwork in filling for embankment

· excavation for cutting

· excavation for borrow pit

· excavation for structural foundation

· disposal of surplus earth

2. Drainage works

· minor bridges

· culverts

· causeways

· side drains

· other surface and sub surface drainage works

3. Protection works

· earth retaining structures

· river training works

· gully control works

· land slide stabilization

· bridge protection works

4. Pavement works

· sub grade works

· sub base works

· base works

· surface works

5. Miscellaneous works

· road ancillaries

· Traffic signs/signals/markings etc.

· bio-engineering works

Preparation of road bed

a) site clearance

b) preparation of subgrade

c) earthwork i) earthwork in excavation

ii) earthwork in embankment

Site Clearance

Site clearance is the first operation to be startedjust after completion of survey works for fixing theroad alignment and before the beginning of anyearth works for the road construction. Major worksto be done under this heading along the alignmentare as follows:

- Clearing hedges and shrubs at least covering toe width.

- Removal of existing tree stump, and roots along the alignment

- Removal of existing structures along the alignment.

Preparation of subgrade

Subgrade preparation includes all operations before the pavement structures could be laid over it and compacted. Subgrade may be situated on embankment, excavation or at the existing ground surface. Therefore, the sites should be cleared off and grading is necessary to bring the vertical profile of the sub grade to the designed grade and camber. The top of the sub grade should be well compacted before placing the pavement layer.

4.3 Earthwork

It is the process to prepare the sub-grade level bringing it to the desired grade and camber by compacting adequately. The earth work may be either in embankment (filling) or in excavation(cutting) depending on the topography.

It includes all construction operations required to convert the road land from its natural condition and configuration to the sections and grades prescribed in the plans.

Earthwork, which may be excavation or filling, can be performed manually or using machines. In order to reduce the cost of construction it is necessary to plan the movement of materials from cuts to the nearest fills; therefore, it is necessary to decode the limits of economical haul and lift.

Earthwork in Excavation:

· Process of cutting or loosening and removing the earth including rock from its original position transporting and dumping it to the site as a fill or spoil bank.

· May be needed before preparing the sub grade

· Done when the natural ground level is higher the designed grade line level.

· The depth of cutting depends up on the height of grade line below natural ground level and can be calculated from L-section and cross section of the road.

The slope to be provided for excavation depends upon the nature and type of soil and depth of cutting, construction of side drains also requires excavation along roadside.

· The selection of excavating equipment and cost analysis is made based on the stiffness of the materials to be excavated. The excavation equipments are bulldozer, drag line, scrappers, clam shell, hoe etc. The selection of particular type of equipment depends upon the types of soil, availability of equipment and cost benefit analysis of the project.

· The design elements of highway excavation works are

- Depth

- Stability of foundation

- Stability of slopes

- Accommodation of road side drains

Earthwork in Embankment:

Is the filling of earth or soil to achieve the desired grade line with the consideration of vertical alignment. It is necessary when natural ground level is below the grade line level or formation level. The grade line may be raised due to any of the following reasons.

· To keep the subgrade above the high ground water table.

· To prevent damage to pavement due to surface water and capillary water.

· To maintain the design standards of the highway with respect to the vertical alignment.

The design elements of highway embankments are

i. Height of fill: Depends on the formation level and location of natural ground. In case of weak soils, its bearing capacity and stability control the height of embankment.

ii. Fill materials: Generally, granular soil is preferred as highway embankment material. As far as possible organic soils, silts should be avoided. If the foundation is very weak then light soil as cinder nay also be used as fill material. iii.settlement of embankment:

The settlement of fill material i.e. embankment may be caused due to

- settlement of fill material itself

- settlement of foundation

- both of the above

To reduce the settlement of foundation at high moisture content sometimes following remedy is taken into account.

iii. The use of vertical drains and sand blanket will reduce the path of flow so that there is no danger from settlement point of view. Sand blanker is extended beyond the bottom width. Whatever is the type of settlement it is desirable that the settlement is almost complete before the construction of pavement.

iv. Stability of foundation: The foundation stability is evaluated and the factor of safety is estimated by any of the following approaches.

· Assuming a certain failure surface such as a circular arc or any other composite shape and analyzing it with Swedish circular arc analysis or method of wedges as the case may be.

· Estimating the average shear stress and strength at the foundation layers by approximate methods and estimating the factor of safety. ™Using theoretical analysis based on elastic theory.

v. Stability of slopes: Embankment slopes should be stable enough to eliminate the possibility of a failure under adverse moisture and other conditions. The stability of the slope should be checked by providing minimum factor of safety of 1.5. Flatter slopes are preferred than in cutting.

Construction of Embankment

The embankment may be constructed either by rolling in relatively thin layers called rolled earth method or by hydraulic fills. In rolled earth method each layer is compacted by rolling to a satisfactory degree or to a desired density before next layer is placed. While rolling the layers of the soils are maintained at optimum moisture content. Compaction at optimum moisture content provides the maximum dry density. The thickness of the layers may vary between 10~30 cm. depending on various factors such as soil type, equipment specification etc.

The practice of dumping the earth without compacting properly and allowing the fill to get consolidated under weather during few subsequent seasons should be avoided as the settlement will continue for a very long period. If pavement is constructed before the settlement of the fill is almost complete, the pavement is likely to become uneven and also fail later.

Soil Compaction

Compaction of soil is the process by which the soil particles are constrained to pack more closely together through a reduction in air voids generally by mechanical means. The object of compacting soil is to improve its properties and to increase its strength and bearing capacity reduce its compressibility and decrease its ability to absorb water due to reduction in volume of voids. The various factors influencing soil compaction include moisture content, amount and type of compaction, soil type and stone content. There is optimum moisture content for a soil, which would give maximum dry density for a particular type and amount of compaction. Hence, it is desirable to compact the soil at the OMC after deciding the compacting equipment. The moisture content during compaction must also be specified and carefully controlled during construction to achieve the maximum density by the selected method of compaction.

Field Control of Compaction

For adequate quality control in construction, it is necessary to have proper field control in construction. The two field control tests needed are:

· Measurement of moisture content

· Measurement of dry density

Before compaction of earthwork is undertaken, it is always preferred to know the optimum moisture content for the soil, which can be determined by Proctor's Field control method. If the moisture content of the soil during compaction is controlled at OMC then the next control needed is the dry density, the desired value of which may be achieved by increasing the number of passes for the selected equipment and the thickness of each later (sand replacement method is widely used.)

In field, it is not possible to achieve 100% results in comparison to standard results obtained in the laboratory. However, by field checks it is possible to control the compaction to achieve the best possible results.

Relation of optimum moisture content and maximum dry density

To assess the degree of compaction, it is necessary to use the dry unit weight, which is an indicator of compactness of solid soil particles in a given volume. The laboratory testing is meant to establish the maximum dry density that can be attained for a given soil with a standard amount of compactive effort.
In the test, the dry density cannot be determined directly, and as such the bulk density and the moisture content are obtained first to calculate the dry density as

A series of samples of the soil are compacted at different water contents, and a curve is drawn with axes of dry density and water content. The resulting plot usually has a distinct peak as shown. Such inverted “V” curves are obtained for cohesive soils (or soils with fines), and are known as compaction curves.

Dry density can be related to water content and degree of saturation (S) as

The relation between moisture content and dry unit weight for a saturated soil is the zero air-voids line. It is not feasible to expel air completely by compaction, no matter how much compactive effort is used and in whatever manner.

Effect of Increasing Water Content
As water is added to a soil at low moisture contents, it becomes easier for the particles to move past one another during the application of compacting force. The particles come closer, the voids are reduced and this causes the dry density to increase. As the water content increases, the soil particles develop larger water films around them.

This increase in dry density continues till a stage is reached where water starts occupying the space that could have been occupied by the soil grains. Thus, the water at this stage hinders the closer packing of grains and reduces the dry unit weight. The maximum drydensity (MDD) occurs at an optimum water content (OMC), and their values can be obtained from the plot.

Effect of Increasing Compactive Effort
The effect of increasing compactive effort is shown. Different curves are obtained for different compactive efforts. A greater compactive effort reduces the optimum moisture content and increases the maximum dry density.

An increase in compactive effort produces a very large increase in dry density for soil when it is compacted at water contents drier than the optimum moisture content. It should be noted that for moisture contents greater than the optimum, the use of heavier compaction effort will have only a small effect on increasing dry unit weights.

It can be seen that the compaction curve is not a unique soil characteristic. It depends on the compaction effort. For this reason, it is important to specify the compaction procedure (light or heavy) when giving values of MDD and OMC.

Factors Affecting Compaction
The factors that influence the achieved degree of compaction in the laboratory are:

ü Plasticity of the soil

ü Water content

ü Compactive effort

Mass Haul Diagram

It is the Graphical representation of the amount of earthwork involved in road construction and the manner in which the earth to be hauled economically.

Characteristics and principle of diagram are as follows

• Is plotted below the longitudinal profile

• Horizontal distances are the chainage along the centre line

• The ordinate at any station along the curve indicates the earthwork quantity accumulated up to that point and is the summation of the differences between cut and fill.

• The maximum ordinate (+) indicates a change from cut to fill

• The minimum ordinate (-) indicates a change from fill to cut

• A rising curve at any point indicates an excess of excavation over till at that point. A falling curve indicates the reverse.

• If the curve has steep slopes it indicates heavy cuts or high fills. Flat slopes indicate small earthwork quantities.

• A convex loop of the mass diagram indicates that the haul from cut to fill is from left to right. A concave loop indicates that the haul from cut to fill is from right to left.

• Balance point—a point where the volume in excavation balances the volume in embankment.

• Any line drawn parallel to the base line and intersecting two point within the same curve indicates a balance of cut and fill between these two points

• The area between a balancing line and the mass diagram is a measure of the haul between the balance points. This area divided by the maximum ordinate between the balance line and the curve gives the average distance of haulage of the cut material to make the fill.

• When the earth excavation and embankment quantities balance at the end of the section, the mass diagram curve would end at the base line at the zero point.

• Free haul—it is the distance to which the contractor is supposed to move the earth without any additional charge. The charge for free haul is covered by the unit rate of earthwork and it is generally 50m

• Overhaul is the distance in excess of free haul for which the contactor will be paid extra for each unit of haulage

• Economic haul: -

When the haul distance area large it may be economical to waste excavated material and borrow from a more convenient source that pay for overhauling.

§ Economic haul distance is a distance to which material from excavation to embankment can be moved more economically than to get material from borrow opening

§ The economic over haul distance can be determined by equating the cost of roadway excavation plus overhaul and tipping in embankment with the cost of borrow pit material (including original cost as well as cost of excavation, hauling and tipping from borrow pit to embankment) plus excavation, haul and wasting of roadway material within the free haul distance.

Thus if,a= cost of roadway excavation per m³

b= cost of overhaul and tipping per m³ per station

c= cost of borrow materials per m³

L= economic overhaul distance in stations

a+b*L = c+a

If the free haul distance is denoted by F stations, then the economic haul distance is given by

F+L = F + c/b

Shrinkage: when earth is excavated from borrows area and deposited on the embankment its volume increases. But as compaction is done, the final volume of the compacted bank becomes less than the borrow area volume. This is known as shrinkage. Actual shrinkage factor depends on the soil deposit and may vary from 10~20%.

Swell—when rock is excavated and deposited in the bank, the volume of material may occupy a larger volume. 20~40%.

Construction of low-cost roads

A low-cost road is a road constructed at a low cost and is capable of being maintained at a low cost. In villages and underdeveloped areas, the immediate need is not of good roads but the access which may serve the traffic needs. As the traffic increases on the road, as a result of the development existing roads may be improved upon to serve the need of the increased traffic. This enables economical use of the funds. Construction of low-cost roads is very preferred in developing countries like Nepal where large length of roads are to be constructed in the rural area with the limited available funds. Earth roads, Gravel roads and Soil stabilized roads are the examples of such roads.

fig
17.Thin Walled Vessel

1) A cylindrical shell is 3m long, and is having 1m internal diameter and 15mm thickness. Calculate the maximum intensity of shear stress induced and also the changes in the dimensions of the shell, if it is subjected to an internal fluid pressure of 1.5 N/mm².

Ans: - ɛ_c = 2.125*10^-4Ϩ_d = 0.2125 mm, Ϩ_l = 0.15 mm, Ϩ_v= 1119192.4 mm²

2) A thin cylindrical shell, 2m long has 200 mm diameter and thickness of metal 10 mm. it is filled completely with a fluid at atmospheric pressure. If an additional 25000 mm³ fluid is pumped in, find the pressure developed and hoop stress developed. Also find the changes in diameter and length. Take E = 2*10⁵N/mm² and µ= 0.3.

Ans: - p= 4.188 N/mm², σ_c = 41.88 N/mm², σ_l = 20.54 N/mm², Ϩ_d = 0.0356 mm, Ϩ_l = 0.08376 mm

3) The diameter of a city water supply pipe is 750mm. It has to withstand a water head of 60m. Find the thickness of the seamless pipe, if the permissible stress is 20 N/mm². Take unit weight of water as 9810 N/mm³.

Ans: - t= 9.197 mm

4) A thin cylindrical shell made up of a copper plate 6 mm thick is filled with water under a pressure of 6 N/mm². The internal diameter of the cylinder is 250 mm and its length is 1.0 mm. Determine the additional volume of the water pumped inside the cylinder to develop the required pressure. E_Copper= 1.04*10⁵ N/mm²_,µ= 0.32, K _water = 2100 N/mm².

Ans: -250.72*10³ mm³

5) A 100mm diameter 5mm thick cylindrical shell is provided with hemispherical ends, and subjected to internal fluid pressure. Find the thickness of the hemispherical part of the condition of no distortion of the junction. Take µ= 0.25.

Ans:- t₂= 2.14mm

6) What is the largest diameter of the boiler that can be manufactured with 16mm plates to resist internal pressure of 2 N/mm². If the permissible stress is 150 N/mm² and efficiencies of longitudinal and circumferential joints are 75% and 45% respectively.

Ans: - d= 1800mm, 2160 mm

7) A cylindrical water tank having vertical axis is open at the top. Its inside diameter is 500cm and the height is 20m. the tank is full of water and is made of structural steel having yield stress of 2200 kg/cm². Determine the thickness of the steel sheet used if the efficiency of longitudinal joint is 80% and the factor of safety is 3.

Ans: - t ≥ 0.85 cm

8) A thin cylinder having 12 cm internal diameter and 2 mm wall thickness is closed at its ends. The cylinder is subjected to an internal pressure of 3 N/mm² and torque of 800 Nm rotating the cylinder in the clockwise direction for a point on the outside surface of the cylinder. Determine the principal stresses and maximum shear stress developed.

Ans: - τ = 17.145 N/m², σ₁ = 95.934 N/mm², σ₂ = 39.066 N/mm²

16.Marshall Stability Apparatus

Operating Instruction For

Resistance in to plastic flow of bituminous matured using

Marshall Stability Apparatus

2. Apparatus

a) Specimen Mould Assembly - Mould cylindrical 4 inch dia meter and 3

Inch in height base plates and extension collar shall conform to the details. Three mould cylinders are recommended.

b) Specimen Extractor - A specimen extractor the compacted from the specimen mould is recommended. A suitable bar is required to transfer load from the extension collar to the upper proving ring attachment while extracting the specimen.

c) Compaction Hammer - The Compaction Hammer shall have a flat, circular tamping face and a 10 Lb sliding weight with a free fall of 18 inch two compaction Hammer are a recommended.

i) Flow Gauge - One dial gauge also supplied 0.01mm and 25mm travel for flow. This dial gauge is fitted with Braking Head (f).

1) Containers for heating aggregate flat-bottom metal pans or an other suitable

Containers.

2) Container for heating bituminous material, either gill type tins, beakers pouring posts, or saucepans may be used.

3) Mixing tool either steel trowel (Garden type) or Spatula, or spading and hard mixing.

4) Thermometer of determining temperatures of aggregates, bitumen and bituminous mixtures. Armored glass thermometers or dial type with metal stem are recommended. A range of 50 to 400F.

5) Thermometer for water bath with a range of 20 to 70 deg. Cent.

6) Balance 5kg capacity, sensitive to 0.1gm for weighing moulded specimens.

7) Balance 5kg capacity, sensitive to 1.0gm for bathing mixtures.

8) Gloves for handing hot equipments.

9) Rubber Gloves for removing specimens from water bath.

10) Marking pencils for identifying specimens.

11) Scoop flat bottom for batching aggregates.

12) Spoon Large for placing mixture in specimen moulds.

Test Specimens –

a) Number of Specimens - Prepare at least three specimens for each combination of aggregates and bitumen contents.

Procedure:-

Note – For core specimens correct the load when thickness is other than 2 inch by using the proper multiplying factor from table I.

Report –The report shall include the following the information for each specimen tested,

i. Weight of the test specimen.

ii. Maximum load in kgs corrected when required.

iii. Flow value in hundred of an inch.

iv. Mixing temperature.

v. Compacting temperature, and

vi. Test temperature.

cont….

professional

पाठ- ६ सिभिल

Compare the horizontal displacement results.( obtained theoretically and experimentally.)PRECAUTIONS:

The model has a span of 900mm and rise 210mm, with hinges at supports and crown. One of the end rests on rollers. Along the horizontal span of the arch various points are marked at equidistant for the application of load. A dial gauge with magnetic base is supplied with the apparatus.

Principle of transmissibility and equivalent forces

Varignon’s theorem

14.Road Pavement

1.1 Introduction

Pavement layers

Types of Pavement:

Flexible Pavement:

Rigid Pavement:

Differences between flexible and rigid pavement

Advantages and disadvantages of Rigid pavement

1.2 Functions of different layers in pavement structure

Chapter 2 Hill Roads

1.1 Definition and importance

2.2 Design and construction problems in hill roads

2.3 Special consideration of hill road geometric design

Alignment of hill road (Route location)

River route:

Ridge route:

2.4 Typical cross sections of hill roads

1. Cut and fill:

2. Bench type

3. Box cutting

4. Embankment with retaining walls

5. Semi bridge

6. Semi tunnel

7. Platform

2.5 Special structures in hill road

c) Cross drainage

d) Subsurface drainage

3. Slope protection structures

15.Road Machineries

3.1 Methods of road construction

3.2 Different types of Tools, Equipment and Plants used in Road Construction

3.3 Different types of Compacting Equipment

16.Road Construction Technology

4.1 Introduction

4.2 Activities involved in road construction

Preparation of road bed

4.3 Earthwork

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