# ELEMENTS OF CIVIL ENGINEERING BOOK PDF

Elements of Civil Engineering and Engineering Mechanics - Free ebook download as PDF File .pdf), Text File .txt) or read book online for free. ELEMENTS OF. Buy Elements of Civil Engineering and Engineering Mechanics Notes eBook by PDF Online from VTU eLearning. Related Engineering 1st SEM Books. Elements of Civil Engineering () - Teaching and Examination Scheme, Content, Reference Books, Course Outcome, Study Material.

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Elements of Civil Engg. & Engg. Mechanics (10CIV13). Sl. No. The American Society of Civil Engineers defines civil engineering as “ the PDF = KN. Hence all important aspects of civil engineering are taught as elements of civil engineering in all Author hopes that students and faculty will receive this book Basic Civil Engineering Ebook & Lecture Notes PDF Download. The book has been divided into 5 sections namely Engineering Materials, subjects ELEMENTS OF CIVIL ENGINEERING/BASIC CIVIL ENGINEERING and .

Let us consider a man climbing a ladder. The direction of a force can be represented by an arrowhead. The unit of mass is the kilogram. The weight of the man is not actually concentrated at a fixed point but for the purpose of analysis it is assumed to be concentrated at a particular point. This causes an acceleration directed towards the centre of earth. It is the line along which the force acts. When a force is applied to a body which is at rest.

A continuous distribution of molecules in a body without intermolecular space is called the continuum. The total amount of matter present in a body is known as its mass. It is the external agency which tends to change the state of a body or a particle. A physical quantity which has only magnitude. It is called acceleration due to gravity and is denoted by g.

Coplanar force system 2. Non-coplanar force system 3. See Figure 2. If two or more forces are acting in a single plane and their lines of action do not meet at a common point.

Coplanar force system If two or more forces are acting in a single plane. Types of Force System The types of force system are: The types of coplanar force system are: Collinear force system. If two or more forces are acting in a single plane and their lines of action pass through a single point. Such a system of forces can be i Non-coplanar concurrent force system ii Non-coplanar non-concurrent force system iii Non-coplanar parallel force system.

The coplanar parallel force system is of two types: All the forces act parallel to one another and are in the same direction.

If two or more forces are acting in a single plane with their lines of action parallel to one another. The forces act parallel to another. Non-coplanar force system If two or more forces are acting in different planes. If two or more forces are acting in different planes and are parallel to one another.

If two or more forces are acting on different planes but do not pass through the same point. Principle of Superposition of Forces This principle states that the net effect of a system of forces on a body is same as that of the combined effect of individual forces on the body.

Principle of Transmissibility of Forces This principle states that a force can be transmitted from one point to another point along the same line of action such that the effect produced by the force on a body remains unchanged.

According to the principle of transmissibility. Non-collinear force system If the lines of action of the forces do not coincide with one another. Let us consider a rigid body subjected to a force of F at point O as shown in Figure 2. F is the force which makes an angle q with the horizontal axis.

If the force F makes an angle of q with the horizontal. Composition of Forces It is the process of combining a number of forces into a single force such that the net effect produced by the single force is equal to the algebraic sum of the effects produced by the individual forces. In this figure. In D CAD. The single force in this case is called the resultant force which produces the same effect on the body as that produced by the individual forces acting together.

Resolution of a Force The process of splitting of a force into its two rectangular components horizontal and vertical is known as resolution of the force. The positive and negative convention of forces used in the resolution of forces in Figure 2.

Parallelogram law: If two forces are acting simultaneously on a particle and away from the particle. To find the magnitude R of the resultant. If two forces acting simultaneously on a particle can be represented both in magnitude and direction by the two sides of a triangle taken in order.

This is illustrated in Figure 2. Let P and Q be the two forces. Let R be the resultant of forces P1and P2 and B be the moment centre. If a number of forces acting on a particle can be represented in both magnitude and direction by the sides of the polygon taken in order. Let d. The magnitude of the moment is given by the product of the magnitude of the force and the perpendicular distance between the line of action of the force and the point or axis of rotation.

P1 and P2. Types of moments i If the tendency of a force is to rotate the body in the clockwise direction. Moment of a Force The turning effect produced by a force on a body is known as the moment of the force. This is shown in Figure 2. Let q be the angle made by R with the x-axis and note that the same angle is formed with the y-axis by the perpendicular to R from B and note this point as B1. Join AB and consider it as the y-axis and draw the x-axis at right angles to it at A.

We know that. Define the following terms: What are the two divisions of dynamics? What are the four characteristics of a force? What are the three types of force systems? What is a couple? What are two types of couple? What are the various methods of finding the resultant of a system of coplanar concurrent forces?

Define the principle of superposition of forces. Types of couple i Clockwise couple. Define the moment of a force.

## Re: Basic Of Civil Engineering PDF Free Ebook Download

SFx and also in the y-direction i. The position. Determine the magnitude of the resultant using the formula. The magnitude of resultant. Calculate the algebraic sum of all the forces acting in the x-direction i.

Determine the direction of the resultant using the formula. In a coplanar concurrent force system. Solution Here: Determine the components of this force along the x and y directions. Example 3. Component along the y-direction. Determine the resultant in magnitude and direction. VTU March Figure 3.

## Civil Engineering Books

Figure 3. Find the resultant of the force system shown in Figure 3. Determine the unknown force P and its inclination with the x-axis. The resultant has a magnitude of N and is acting along the x-axis. One of the forces is unknown and its magnitude is shown by P. Determine the other force. Solution Let P be the unknown force. VTU August Figure 3. Also determine the magnitude of the resultant. Determine the resultant force in the x-direction. Determine the direction of the force N such that the hook is pulled in the x-direction.

Determine the magnitude and direction of the resultant force. Determine q between the forces and the direction of the resultant. The resultant of the two forces is N. Solution Magnitude of the resultant. Take the co-ordinate directions as shown in the figure. Solution In triangle OAB. VTU January Figure 3. Find the components of the force along the horizontal and vertical axes. The effect produced as shown in Figure 3. Solution Apply equal and opposite forces of N at O as shown in Figure 3.

Determine the effect of N force at O. See Figure 3. Couple at point O. Solution Apply equal and opposite forces of 40 kN at O parallel to the given force. Determine the resultant of the force system shown in Figure 3. Determine the resultant of the coplanar concurrent force system shown in Figure 3. Compute the value of force P and its inclination with x-axis. Determine the resultant of the four forces acting on a particle as shown in Figure 3.

The force system as shown in Figure The following forces. Determine the resultant. If the force in cable AB is N. The two forces P and Q are acting on a bolt at A. The car is moving along AC. Determine the magnitude and direction of the resultant. A collar which may slide on a vertical rod is subjected to three forces as shown in Figure 3.

Magnitude of resultant. The position of the resultant means the calculation of d. R and x-intercept: In a coplanar non-concurrent force system. VTU January Figure 4. Compute the magnitude. The line of action of kN is directly passing through the point A.

Example 4. Figure 4. VTU August Figure 4. Solution Convert the uniformly distributed load UDL into point load. Determine the magnitude. Taking moments about A. Determine the magnitude and line of action of single resultant of the system.

If the resultant is to pass through B. When the resultant is passing through B. A couple N-m is acting on a bracket when the resultant is at a distance of 1. Determine the x and y intercepts also. VTU February Figure 4. Find the resultant of these forces.

Locate the resultant force with respect to point D. Direction of resultant a: VTU July Figure 4. Direction of resultant q with reference to A: Out of them. Direction of resultant q: Find the missing force and its position. Find the magnitude. Find the resultant of the force shown in Figure 4. Locate the two points where the line of action of the resultant intersects the edge of the plate.: Find the resultant of these forces.: Find the distance of D from C.

Find the resultant of the system. Three forces 60 N. Find the resultant of the force system shown in Figure 4. First find the resultant and then apply a vertical force of P at A assume down or up. Find the resultant of the three forces. Three forces act on a vertical pole as shown in Figure 4. Locate the point where the resultant cuts the pole.

Various forces to be considered for the stability analysis of a dam are shown in Figure 4. The dam is safe if the resultant force passes through middle-third of base. Find the point of intersection of the resultant with the base of the dam. Neglecting the weight of the dam.: Forces acting on 1m length of a dam are shown in Figure 4. Conditions of equilibrium for different force systems 1. Figure 5. Principle of equilibrium According to this principle.

See Figure 5. P and R. The force which is required to keep the body in equilibrium. P and Q respectively. See figure 5. That means the body is not in equilibrium. R be the three forces acting at a point O and let a.

Let P. In Figure 5. Let us consider a spherical ball of mass m.

In the coplanar concurrent force system. VTU August Figure 5. Solution Apply two conditions of equilibrium to calculate the magnitude and direction of the unknown force. Determine the magnitude and the direction of the minimum force F that should be exerted at the free end of the rope. Determine the minimum required magnitude of force F.

VTU February Figure 5. Free-body diagram. Example 5.

Its free-body diagram is shown in Figure 5. Solution or Again. A force of 30 kN is applied at C. Determine the forces in the cables CA and CB. Solution or Now. Find the tensions in the portions AB. FBD at B. At D Figure 5. Determine the tensions in the various segments of the cable. BC and CD as shown in Figure 5.

At B Figure 5. VTU January Figure 5. Compute the tensions in the strings AB.

Suppose the two planes are perpendicular to each other. If all contact surfaces are smooth.

Solution At contact points. Consider the FBD of sphere 2. Therefore it is necessary to consider the FBD of sphere 2 first. The weights are N and N respectively. Find the reactions at all the points of contact. FBD of sphere 1. FBD of sphere 2. From Figure 5. Solution Consider the FBD of sphere 2: Assuming that all contact surfaces are smooth.

Free body diagram of sphere 2. The radii of spheres 1 and 2 are. RQ RP sin Free body diagram of sphere 1. Calculate the reactions at all the points of contact. Write the free body diagram of identical spheres A and B shown in Figure 5. Using equilibrium conditions. Free body diagrams of spheres A and B. It is given that: Free body diagram of sphere C. Free body diagram of sphere B. Free body diagram of sphere A. The radii of spheres. Solution Two equal and opposite reactions will be developed between the contact surfaces of two bodies, as shown in Figure 5.

R1 1 sin Neglecting the weight of the connecting bar, find the force P applied such that the system is in equilibrium.

A ball of weight N rests in a right-angled groove as shown in Figure 5. If the entire surface is smooth, determine the reactions RA and RC at the points of contact. A circular roller of radius 5 cm and weight N rests on a smooth horizontal surface and is held in position by an inclined bar AB of length 10 cm as shown in Figure 5. A horizontal force of N is acting at B. Find the tension force in the bar AB and reaction at C.

A pendulum ball weighs 50 N. It is pulled sideways by a horizontal force of 20 N. Calculate the angle of the rope of the pendulum to the vertical and tension in the string as shown in Figure 5.

Two spheres each of weight N and of radius 25 cm rest in a horizontal channel of width 90 cm as shown in Figure 5. Find the reactions on the points of contact A, B, C and D. Two identical rollers each of weight N are supported by an inclined plane and vertical wall as shown in Figure 5. Find the reaction exerted by the wall and the inclined plane at C, D, and F. Two spheres are resting in a trench as shown in Figure 5. Determine the reactions acting on the spheres at A, B, C, and D, assuming the surface to be smooth.

Two smooth spheres rest between two vertical columns as shown in Figure 5. The larger and smaller spheres weigh respectively N and N. The diameters of the larger and smaller spheres are 36 cm and 16 cm respectively. Find the reactions at 1, 2, 3 and 4. Two cylinders are placed as shown in Figure 5. Neglecting friction, find the reactions at all contact surfaces. Find the force S in the string AB and the pressure produced on the floor at the point of contacts D and E.

Determine the magnitude and direction of force P such that the force system is in equilibrium. A beam 20 m long supports a load of 12 kN, the cable AC is horizontal and 10 m long.

Find the stress in the cable for the beam shown in Figure 5. What axial forces does the vertical load 16 kN induce in the tie rod and in the jib of the crane shown in the Figure 5.

Neglect the weight of the member. The tie rod will be under tension and the jib under compression. A body weighing 16 kN is supported from a fixed point by a string 10 cm and is kept at rest by a horizontal force P at a distance of 6 cm from the vertical line drawn through the point of suspension. Find the tension in the string and the value of P Figure 5. A body of weight 70 kN is supported by two strings whose lengths are 6 cm and 8 cm from two points in the same horizontal level.

The horizontal distance between the two points is 10 cm. Determine the tensions in the string Figure 5. At C, weight of N is a suspended. The pulley at E is frictionless. The structural members exert forces on supports known as action. Simple supports 2. A beam is a horizontal member. Roller supports 3. They restrict translation of the body in one direction only. Supports exert forces. Fixed supports Simple supports Simple supports Figure 6. Hinged or pinned supports 4.

The beam is subjected to vertical forces known as action. Types of Supports The following types of supports are found in practice: Figure 6.

They restrict translation of the body along one direction only. But rotation is possible. Such a beam can resist forces normal to the axis of the beam.

Fixed supports develop an internal moment known as restraint moment to prevent the rotation of the body. Types of Beams Simply supported beam It is a beam which consists of simple supports Figure 6. Roller supports Roller supports Figure 6.

Hinged or pinned supports Hinged supports Figure 6. Fixed supports Fixed supports Figure 6. Overhanging beam It is a beam which extends beyond support s.

Cantilever beam It is a beam whose one end is fixed and the other end is free Figure 6. Propped cantilever beam It is a beam whose one end is fixed and the other end is simply supported Figure 6. Types of Loads Concentrated load A load which is concentrated at a point in a beam is known as concentrated load Figure 6. The overhang portion is BC. In Figure 6. The magnitude of the point load corresponding to a uniformly varying load such as that shown in Figure 6.

Draw the free body diagram of the given beam by showing all the forces and reactions acting on the beam. Example 6. Solution It is a coplanar non-concurrent force system. Apply the three conditions of equilibrium to calculate the unknown reactions at the supports. Find the reactions at A and B. Determine the reactions at A and B. VTU January Figure 6. VTU February Figure 6. VTU July Figure 6. Using Figure 6. Solution Ans. Calculate the support reactions for the cantilever beam shown in Figure 6.

The moment at A is anticlockwise in nature. Using the conditions of equilibrium shown in Figure 6. Determine the reactions at the supports A and B for a beam loaded as shown in Figure 6. Solution Using the conditions of equilibrium shown in Figure 6. Solution Using the equilibrium conditions shown in Figure 6. Solution First.

## Elements of Civil Engineering (2110004)

Using the conditions of equilibrium. Determine the support reactions for the compound beam shown in Figure 6. Consider the beam BE Figure 6. Consider the beam AC Figure 6. TBC sin Solution or Applying the conditions of equilibrium to Figure 6.

Substituting the value of TBC in i. Solution Applying the conditions of equilibrium to Figure 6. Determine the support reactions for a beam loaded as shown in Figure 6. Determine the support reactions for the beam supported and loaded as shown in Figure 6. Calculate the support reactions for the beam loaded and supported as shown in Figure 6. A horizontal beam 6 m long is supported on a knife edge at its end B and the end A.

Also find the reaction at support B of beam as shown in Figure 6. Find the reactions at the supports A and B. Find the magnitude of anticlockwise couple M to be applied at D so that reaction at F will be 35 kN upwards.

Find the support reactions for a beam loaded and supported as shown in Figure 6. Also compute the reaction at the F support. Find the direction and magnitude of the couple to be applied at D so that the reaction at F will be 50 kN upwards.

Compute the reactions at the supports of beam ABCD which is loaded and supported as shown in Figure 6. A beam ABF is loaded and supported as shown in Figure 6. Individual members of a truss is a two force member subjected to either tension or compression. They are: Plane trusses are made of several bars or members connected together at the joints by riveting or welding to form a rigid formwork.

Rigid truss or perfect truss A rigid truss is one in which the number of members are sufficient to resist the external loads. Classification of Trusses The trusses are classified into three types which are depicted in Figure 7.

The forces are acting only on the joints. The members of trusses are straight. Non-rigid truss or deficient truss A non-rigid truss is one in which the number of members are less than that required for a perfect truss. The following are the assumptions made in the analysis of statically determinate trusses: All members are rigid.

Water Resources: Hydrological cycle, water use and its conservation, Introduction to dams, weirs, barrages and check dams. Transportation Engineering: Role of transportation in national development, Modes of transportation, Introduction to road traffic and traffic control, Introduction to mass transportation system. More contact details. Elements of Civil Engineering Computer Engineering Courses Subject Detail. Teaching Scheme in Hours.

Examination Scheme in marks. A water resource is such a vast subject that it includes in itself hydrology, irrigation, hydraulics and water supply. Tremendous volume of water is stored in the earths crust. According to one estimate the total volume stored under the surface of earth may be about 80 million km cube, half of which may be at depths less than m.

The use of surface water for irrigation is likely to create problems like waterlogging in certain areas. Groundwater is obtainable all the year round and its use along with the surface water keeps the subsoil water level within reasonable limits.

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The judicious use of water for purposes of irrigation has gained such importance in the recent times that water management has become a science in itself. Engineers have learnt to tame the water resources by construction of dams, construction of bore wells and construction of hydroelectric plants for the benefit of mankind.

Environmental Engineering This is an important branch of civil engineering which covers both water supply and sanitary engineering. The importance of clean environment was felt with the rapid growth in population, and the growth in urbanization and industries.

Environment is polluted through the mediums of air, water or such other agents. The science of civil engineering deals with the subject of tapping water from different sources, testing its quality, purification processes and distribution of water to the consumers.

Similarly the environmental engineering encompasses the subject of treatment of wastes which originate from different sources and deals with the removal of harmful substances in these wastes by different processes. The impact of wastes originating from industries is felt by living organisms if such wastes contain toxic substances. The Central Government as well as state governments have enacted laws for the protection of environment needed for the safe living of human beings.

Geotechnical Engineering This branch of civil engineering is also called soil mechanics. It is a discipline of civil engineering in which the study of soil, its behaviour on the application of load and its use as an engineering material in the construction of earth dams, is done. The properties and strength characteristics of different types of soil are studied in this subject.A diversion dam raises the water level slightly in the river.

Example 4.

Coefficient of Friction m It is the constant ratio which the limiting friction F bears to the normal reaction N, i. The laws of dynamic friction are: Such dams are also known as spillways. Diaphragm type dams.

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