Mechanical Engineering

Motion of a Car Under Uniform Acceleration A car starts from rest and accelerates uniformly to a speed of 72 km/h over a distance of 500 m . Calculate the acceleration and the time taken to attain this speed. If a further acceleration raises the speed to 90 km/h in 10 seconds , find this acceleration and the further distance moved. The brakes are now applied to bring the car to rest under uniform retardation in 5 seconds . Find the distance travelled during braking. Solution Given: u = 0, v = 72 km/h = 20 m/s, s = 500 m Acceleration of the Car Let a = acceleration of the car. Using the equation: v² = u² + 2as (20)² = 0 + 2a × 500 = 1000a a = (20)² / 1000 = 0.4 m/s² Time Taken to Reach the Speed Let t = time taken. v = u + at 20 = 0 + 0.4 × t t = 20 / 0.4 = 50 s Further Acceleration from 72 km/h to 90 km/h Given: u = 20 m/s, v = 96 km/h = 25 m/s, t = 10 s Acceleration v = u + at 25 = 20 + a × 10 a = (25 − 20) / 10 = 0.5 m/s² D...

Graphical Representation of Displacement with Respect to Time The displacement of a moving body at different instants of time can be represented using a graph. Such a graph is drawn by taking displacement on the Y-axis and time on the X-axis. This curve is called the s–t curve . We consider two important cases: 1. When the body moves with uniform velocity When a body moves with constant velocity, it covers equal distances in equal intervals of time. Plotting displacement on the Y-axis and time on the X-axis produces a straight-line s–t graph , as shown in Fig. 2.1 (a). The equation of motion for uniform velocity is: s = u × t Thus, velocity at any instant is: Velocity at t₁ = s₁ / t₁ Velocity at t₂ = s₂ / t₂ Since velocity remains constant: s₂ – s₁ / t₂ – t₁ = tan θ Here, tan θ is the slope of the s–t graph. Therefore, the slope of the displacement-time curve gives the velocity of the body.              ...

Kinematics of Motion Introduction In the previous chapter, we studied that the Theory of Machines deals with the motion of machine parts and the forces acting on them. In this chapter, we focus only on the kinematics of motion , which covers the relative motion of bodies without considering the forces responsible for that motion. In simple terms, kinematics explains the geometry of motion and important concepts such as displacement, velocity, and acceleration. Plane Motion When a body moves in such a way that its motion is restricted to a single plane, it is known as plane motion . Such motion may be rectilinear or curvilinear . Rectilinear Motion Rectilinear motion is the simplest form of motion and occurs along a straight line. It is often referred to as translatory motion . Curvilinear Motion Curvilinear motion occurs along a curved path. When the motion is restricted to a plane but follows a curved trajectory, it is called plane curvilinear motion...

Force Force is a very important concept in engineering science. It may be described as an agent that produces motion, stops motion, or tends to change the existing state of motion of a body. In simple words, force can create, destroy, or modify motion. Resultant Force When multiple forces such as P, Q, R act on a particle at the same time, a single force that can replace all of them and produce the same overall effect is called the resultant force . The individual forces are known as component forces . The method of determining the resultant of these forces is known as the composition of forces . The resultant may be found analytically, graphically, or by using the following laws: Parallelogram law of forces: “If two forces act simultaneously on a particle, their resultant can be represented in magnitude and direction by the diagonal of a parallelogram formed using the two forces as adjacent sides.” Triangle law of forces: “If tw...

Fundamental Units Measuring physical quantities is one of the most essential tasks in science and engineering. Every measurement is expressed using standard and internationally accepted units known as fundamental units . In this chapter, all quantities are expressed using these three basic units: Length (L or l) Mass (M or m) Time (t) Derived Units Certain units are formed by combining fundamental units. These are called derived units . Examples include units of area, velocity, pressure, acceleration, and many more. Systems of Units Four systems of units are commonly used and globally recognized: C.G.S. System F.P.S. System M.K.S. System S.I. System C.G.S. Units In this system, the fundamental units are: centimetre for length, gram for mass, and second for time. This system is often referred to as the physicist’s unit system or absolute system. F.P.S. Units In the F.P.S. system, the basic units are: foot (length), pound (mass), and ...

Definition The subject Theory of Machines can be described as that branch of engineering science which focuses on the study of relative motion between different components of a machine and the forces acting on them . The understanding of this subject is extremely important for engineers while designing the various machine parts. Note: A machine is a device that receives energy in a usable form and converts it to perform a specific kind of work. Sub-divisions of Theory of Machines The Theory of Machines is generally divided into the following four major branches: Kinematics: This branch deals with the motion of machine parts without considering the forces causing that motion. It focuses only on relative movement between components. Dynamics: This part of Theory of Machines studies the forces and their effects acting on machine elements while they are in motion. Kinetics: This branch examines the inertia forces that arise due to the ...

ANALYSIS OF BARS OF COMPOSITE SECTIONS A composite bar is made up of two or more bars of different materials, of equal lengths, fixed rigidly so that they behave as one single unit. When such a bar is subjected to axial tensile or compressive load , the load is shared between each material according to its stiffness. Important Points Strain (extension per unit length) in each bar is the same , because bars are connected rigidly. Total load on the composite bar = Sum of loads carried by each material. Consider a Composite Bar Let the bar consist of two materials: L = Length of both bars A₁, A₂ = Cross-sectional areas of bar 1 and bar 2 E₁, E₂ = Young’s Modulus of materials 1 and 2 P = Total load P₁, P₂ = Load shared by bar 1 and bar 2 σ₁, σ₂ = Stress in bar 1 and bar 2 Total Load P = P₁ + P₂     ...(i) Stress in Bar 1 σ₁ = P₁ / A₁ Stress in Bar 2 σ₂ = P₂ / A₂ Strain Condition Because bars deform equally: σ₁ / ...