Summary

This document provides an introduction to stress, defining it as force per unit area. It discusses three main types of stress (tensile, compressive, and shear) and their effects on materials. The document also includes examples of stress in various structures like bridges and buildings and explains characteristics of each type of stress.

Full Transcript

Part one : stress I. Introduction Stress is de ned as force per unit area within materials that arises from externally applied forces, uneven heating, or permanent deformation and that permits an accurate description and prediction of elastic, plastic, and uid behaviour....

Part one : stress I. Introduction Stress is de ned as force per unit area within materials that arises from externally applied forces, uneven heating, or permanent deformation and that permits an accurate description and prediction of elastic, plastic, and uid behaviour. stress refers to the internal resistance offered by a material to an applied force or load. It is a fundamental concept used to analyze the behavior of structures under various forces, ensuring their safety and stability. Stress is typically measured as force per unit area and is expressed in units such as Pascals (Pa) or Newtons per square meter (N/m²). Stress is given by the following formula: σ = F / A II. Types of Stress Stress can be classi ed into three main types based on the nature of the applied force: 1. Tensile Stress caused by forces that pull or stretch the material. 2. Compressive Stress caused by forces that compress or squeeze the material. 3. Shear Stress caused by forces acting parallel to the material’s surface. Understanding stress is critical in designing structures like buildings, bridges, and roads, as it helps engineers determine the material’s strength and ability to withstand external loads without failure. Proper analysis ensures that structures are both ef cient and durable, capable of withstanding forces like gravity, wind, and seismic activity. fi fi fl fi A. Tensile Stress Tensile stress is a quantity associated with stretching or tensile forces. It is responsible for the elongation of the material along the axis of the applied load. Tensile stress is de ned as: The magnitude F of the force applied along an elastic rod divided by the cross-sectional area A of the rod in a direction that is perpendicular to the applied force. Tensile Stress Tensile stress occurs when a material is subjected to forces that pull it apart or elongate it. This type of stress causes the material to stretch along the direction of the applied force. It is de ned mathematically as the force acting per unit cross- Here is an illustration of a steel cable under tensile sectional area of the material: stress, as used in a suspension bridge. It shows how the cable stretches under pulling forces while.supporting the structure Characteristics of Tensile Stress 1. Direction: Acts along the longitudinal axis, pulling the material outward. 2. Effect on Material: Increases the length of the material while reducing its cross- sectional area. 3. Material Behavior: The material will stretch up to its elastic limit and may eventually break if the stress exceeds the ultimate tensile strength (UTS). fi fi Examples 1. The stress in a steel cable used in suspension bridges. 2. The stretching of a rubber band when pulled. 3. Tension in ropes or wires used for lifting loads. Applications Tensile stress is essential in the design of structural components like bridges, tension members, and cables, where materials are expected to bear pulling forces without failure B. Compressive Stress Compressive stress is the force that is responsible for the deformation of the material such that the volume of the material reduces. It is the stress experienced by a material which leads to a smaller volume. High compressive stress leads to failure of the material due to tension. Compressive stress is the stress on materials that leads to a smaller volume. Compressive Stress Compressive stress occurs when a material is subjected to forces that push it together, causing it to shorten or compress. It is the opposite of tensile stress and typically results in a decrease in the length of the material while increasing its cross-sectional area. Compressive stress is mathematically de ned as: σ = F / A is the compressive force applied (in Newtons, N). is the cross-sectional area of the material (in square meters, m²). Characteristics of Compressive Stress 1. Direction: Acts inward, perpendicular to the cross-sectional area. 2. Effect on Material: Tends to shorten or compact the material. fi 3. Material Behavior: The material resists the applied force up to its compressive strength; beyond that, it may buckle or fracture. Examples 1. Columns in buildings: Supporting vertical loads from the structure above. 2. Concrete foundations: Withstanding the weight of the structure and external loads. 3. Books on a shelf: Exerting compressive stress on the shelf. Applications Compressive stress is a critical factor in the design of load-bearing components like columns, beams, and walls, ensuring the structure remains stable under compressive forces. Materials like concrete and stone are often chosen for such applications because of their high compressive strength. Here is an illustration demonstrating compressive stress on a concrete column. It highlights the forces acting downward and the material’s ability to withstand compression. C. Shear stress Shear Stress Shear stress occurs when a material is subjected to forces that act parallel or tangential to its surface, causing adjacent layers of the material to slide relative to one another. It is de ned mathematically as: Shear stresses = V / A Where: is the shear force applied (in Newtons, N). is the cross-sectional area where the force acts (in square meters, m²). Characteristics of Shear Stress 1. Direction: Acts parallel to the material’s surface. 2. Effect on Material: Tends to distort the material’s shape without changing its volume. 3. Material Behavior: Failure under shear stress often occurs along a plane where the material cannot resist the sliding forces. Examples 1. Rivets and bolts: Subjected to shear stress in mechanical joints. 2. Scissors: Cutting materials by applying shear forces. 3. Beams: Experiencing shear stress from forces applied at an angle to their length. Applications Shear stress is crucial in designing components like bolts, rivets, beams, and shafts, ensuring they can handle forces acting tangentially without failing. Materials like steel are often used for their high shear strength in structural and mechanical applications. fi Part two: Torsion Introduction Torque is a moment that twists a structure. Unlike axial loads which produce a uniform, or average, stress over the cross section of the object, a torque creates a distribution of stress over the cross section. To keep things simple, we're going to focus on structures with a circular cross section, often called rods or shafts. When a torque is applied to the structure, it will twist along the long axis of the rod, and its cross section remains circular. To visualize what I'm talking about, imagine that the cross section of the rod is a clock with just an hour hand. When no torque is applied, the hour hand sits at 12 o'clock. As a torque is applied to the rod, it will twist, and the hour hand will rotate clockwise to a new position (say, 2 o'clock). The angle between 2 o'clock and 12 o'clock is referred to as the angle of twist, and is commonly denoted by the Greek symbol phi. This angle lets us determine the shear strain at any point along the cross section. Torsion in civil engineering Torsion in Civil Engineering refers to the twisting of structural elements due to applied torque or moments that cause rotational deformation about the longitudinal axis. This phenomenon occurs when forces act eccentrically or asymmetrically on structural components like beams, shafts, or bridges. Torsion leads to shear stresses within the material and can result in cracking, twisting deformation, or even structural failure if not properly accounted for. The shear stress due to torsion is determined using the formula , where is the shear stress, is the applied torque, is the distance from the center to the point of interest, and is the polar moment of inertia of the cross-section. To resist torsion, structural design often includes reinforcement, particularly in concrete beams, and uses shapes like closed sections (e.g., circular or rectangular tubes) that are more effective at resisting twisting forces. Torsion commonly appears in structures like bridge girders, spiral staircases, and transmission shafts. Proper analysis and design are crucial to address combined loading conditions such as bending, shear, and torsion, ensuring the stability and safety of the structure. Torsional Behavior of High Strength Concrete Members Strengthened by Mixed Steel Fibers Reference :onlinelibrary.wiley Equations torsion Max = T × r/J T: Torque applied to the shaft. r: Distance from the center of the shaft to the point where the shear stress is being measured (outer radius for maximum stress). J: Polar moment of inertia of the cross-section. This equation calculates the maximum shear stress at the outer surface of a shaft subjected to torsion. = T × L /J × G T: Torque applied to the shaft. L: Length of the shaft. J: Polar moment of inertia of the cross-section. G: Shear modulus (modulus of rigidity) of the material. This equation determines the angular deformation (angle of twist) in a shaft due to the applied torque. These equations are essential in the design and analysis of shafts and other components subjected to torsion. They ensure that the material can withstand applied stresses without excessive deformation or failure. 𝜽 𝝉  Simple example : A hollow cylindrical steel shaft is 1.5 m long and has inner and outer diameters respectively equal to 40 and 60 mm (a) What is the largest torque that can be applied to the shaft if the shearing stress is not to exceed 120 MPa? (b) What is the corresponding minimum value of the shearing stress in the shaft? Example: A shaft composed of segments AC, CD, and DB is fastened to rigid supports and loaded as shown in Fig. P-323. For bronze, G = 35 GPa; aluminum, G = 28 GPa, and for steel, G = 83 GPa. Determine the maximum shearing stress developed in each segment. References I. Hibbeler – Mechanics of Materials II. Singer III. onlinelibrary.wiley IV. MECHANICS OF MATERIALS Ferdinand P. Beer

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