Mechanical Engineering

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 / ...

1- Extension of a Tapered Rectangular Steel Bar A rectangular steel bar is 2.8 m long and 15 mm thick . It carries an axial tensile load of 40 kN . The bar tapers in width from 75 mm at one end to 30 mm at the other end. If the modulus of elasticity of steel is E = 2 × 10 5 N/mm² , find the extension of the bar. Given: Length, L = 2.8 m = 2800 mm Thickness, t = 15 mm Load, P = 40 kN = 40,000 N Width at bigger end, a = 75 mm Width at smaller end, b = 30 mm Modulus of Elasticity, E = 2 × 10 5 N/mm² Formula Used: For a tapered rectangular bar, dL = (P × L) / (E × t × (a − b)) × ln(a/b) Substituting Values: dL = (40,000 × 2800) / (2 × 10 5 × 15 × (75 − 30)) × ln(75/30) = 0.8296 × 0.9163 Extension, dL = 0.76 mm ✔ Final Answer: 0.76 mm 2- Find Axial Load from Extension in a Tapered Bar A rectangular steel bar 400 mm long and 10 mm thick extends by 0.21 mm when loaded. The width tapers uniformly from 100 mm to 50 mm . Given E = 2...

Analysis of Uniformly Tapering Circular Rod A circular bar whose diameter gradually reduces from D₁ at one end to D₂ at the other end is shown in Fig. 1.13. Such a bar is said to be uniformly tapering . Let: P = Axial tensile load acting on the bar L = Total length of the bar E = Young’s Modulus of the material Diameter at any Section Consider a very small element of the rod at a distance x from the left end. The diameter at this section may be written as: Dₓ = D₁ − kx where k = (D₁ − D₂) / L Area of Cross-Section at Distance x Aₓ = (π/4) × (D₁ − kx)² Stress at Distance x The stress in the infinitesimal section is: σₓ = P / Aₓ σₓ = 4P / [π (D₁ − kx)²] Strain at Distance x eₓ = σₓ / E eₓ = 4P / [πE (D₁ − kx)²] Extension of a Small Element dx dL = eₓ × dx dL = [4P dx] / [πE (D₁ − kx)²] Total Extension of the Rod To get the overall elongation, integrate from x = 0 to x = L : L_total = ∫ (4P / πE (D₁ − kx)² ) dx Performing the inte...

Analysis of Bars of Varying Sections Many practical bars are made of different lengths and with different diameters, so their cross-sectional areas are not the same along the length. Figure 1.6(a) shows a bar composed of three segments, each with its own area and length. The entire bar is subjected to an axial load P . Even though the same load acts through all sections, the stress, strain and extension in each part will be different because the areas (and possibly material properties) are different. The total change in length of the bar is obtained by adding the change in length of each segment. Let: P = axial load on the bar L₁, L₂, L₃ = lengths of sections 1, 2 and 3 A₁, A₂, A₃ = cross-sectional areas of sections 1, 2 and 3 E = Young’s Modulus for the material of the bar (same for all sections here) Stress in each section: σ₁ = P / A₁     (section 1) σ₂ = P / A₂     (section 2) σ₃ = P / A₃     (section 3...

Problem 1.1 – Stress, Strain and Elongation of a Rod A rod of 150 cm length and 2.0 cm diameter is subjected to an axial tensile force of 20 kN . The modulus of elasticity of the rod material is: E = 2 × 10 5 N/mm² Determine: (i) The stress (ii) The strain (iii) The elongation of the rod Given Data Length , L = 150 cm Diameter , D = 2.0 cm = 20 mm Load , P = 20 kN = 20,000 N Modulus of Elasticity , E = 2 × 10 5 N/mm² 1. Calculate Cross-Sectional Area For a circular rod: A = (π/4) × D² A = (π/4) × (20)² = 100π mm² 2. Calculate Stress (σ) Using the formula: σ = P / A σ = 20,000 / 100π = 63.662 N/mm² Answer: Stress = 63.662 N/mm² 3. Calculate Strain (e) Using Hooke’s Law ratio: e = σ / E e = 63.662 / (2 × 10 5 ) = 0.000318 Answer: Strain = 0.000318 4. Calculate Elongation (ΔL) Using the strain formula: e = ΔL / L Therefore: ΔL = e × L ΔL = 0.000318 × 150 = 0.0477 cm Answer: Elongation = 0.0477 cm ...

Constitutive Relationship Between Stress and Strain For One-Dimensional Stress System For a uniaxial or one-directional stress condition (i.e., normal stress acting only in one direction), the relationship between stress and strain follows Hooke’s Law . According to this law, when a material is loaded within its elastic range, the normal stress developed in the body is directly proportional to the strain it produces. This implies that the ratio of normal stress to corresponding strain remains a constant as long as the material stays within the elastic limit. This constant is known as the Modulus of Elasticity or Young’s Modulus . Normal stress / Corresponding strain = Constant or σ / e = E Where: σ = Normal stress e = Strain E = Young’s Modulus e = σ / E     ...(1.7 A) The above expression provides the stress–strain relation for normal stress acting in a single direction. For Two-Dimensional Stress System Before establishing th...

Elasticity and Elastic Limit Whenever an external load acts on a material, the body experiences deformation. If the applied load is removed and the body fully regains its original size and shape (with all deformation disappearing), the material is called an elastic body . This characteristic — the ability of a material to return to its original configuration after the external force is withdrawn — is known as elasticity . A body will return to its initial form only when the deformation produced by the force remains within a certain allowable range. There exists a specific maximum force up to which the deformation will completely vanish after unloading. This limiting value of force is termed the elastic limit . If the applied stress goes beyond this elastic limit, the material will lose part of its elastic behavior. Even after removing the force, the body will not fully return to its original shape — a permanent deformation will remain. Hooke’s Law and Elastic...

Types of Stresses In engineering mechanics, a body may experience two primary types of stresses: normal stress or shear stress . Normal stress is the stress that acts perpendicular to the surface area. It is commonly represented by the symbol σ (sigma) . Normal stress is further divided into tensile and compressive stresses. Tensile Stress Tensile stress develops in a material when it is pulled by two equal and opposite forces. As shown in Fig. 1.1(a) , when a bar is stretched, its length increases. This induced stress is known as tensile stress . The relative increase in length compared to the original length is termed as tensile strain . Tensile stress acts perpendicular to the area and tends to pull the section apart. Let: P = Applied pull (force) A = Cross-sectional area L = Initial length dL = Increment in length σ = Tensile stress ε = Tensile strain When a section x–x divides the bar, the internal resisting force balances the applied...

Simple Stresses and Strains In strength of materials we study how solid bodies behave when different kinds of forces act on them. Whenever you load a bar, a rod or any machine part, its shape or size changes a little. If the load becomes too large, the member may finally fail. To understand and design safe components, we use the basic ideas of stress and strain . 1.1 Introduction Imagine pulling a steel rod. Your hands apply a force, so the rod tries to stretch. The material of the rod opposes this stretching because of the internal bonding between its particles. This internal opposition is what makes the material “strong”. Up to a certain level of load the rod comes back to its original length as soon as the force is removed. This region is called the elastic region. Within this limit the internal resistance developed in the material is in step with (proportional to) the amount of deformation. If the load is increased beyond the elastic limit, the mater...