Ms chapter 5

Ms chapter 5

• 1.
  1. Essential thatengineer understands the behavior of various materials with respect to strengths and deformations when subject to various types of loads. 3. Strength of materials basically is a study of behavior of machine components and structure components when exposed to external loads. 5. The knowledge of internal reaction that the materials offer to the external loads, the change that are produced in a body become the basis of selection of materials for a particular application.
• 2.
  Definition of strainand stress On the application of a load (force), the material of a body undergoes a change. This change in size or shape is called strain. Strain, e defined as how much the length of the object has been changed (ΔL) compared to its original length (Lo). e = ΔL/Lo Strain : Percentage that something has deformed
• 3.
  The internal resistance(counterforce) offered by the material of the body to this change is called stress. •Counterforce tends to return the atoms to their normal position •The total resistance developed is equal to the external force Stress (σ) can be equated to the load per unit area or the force (F) applied per cross-sectional area (A) perpendicular to the force. The SI unit for stress is the Pascal (symbol Pa), which is a shorthand name for one Newton (Force) per square metre (Unit Area). stress on an object defined as the force per unit area
• 4.
  Types of Stress 1.Tensile stress When a section is subjected to 2 equals and opposite pull, as a result of which the body tends to lengthen, the stress induced is called tensile stress. P P The corresponding strain is called tensile strain • ΔL is increase in length Tensile stress, σt = F/A • Tensile stress by Tensile strain, et = ΔL/Lo convention is considered as positive stress
• 5.
  2. Compressive stress When a section is subjected to 2 equal and opposite pushes, as a result of which the body tends to shorten its length, the stress induced is called compressive stress. P P The corresponding strain is called compressive strain • ΔL is decrease in length Compressive stress, σc = F/A • Compressive stress by Compressive strain, ec = ΔL/Lo convention is considered as negative stress
• 6.
  3. Shear stress If the external forces acting on a piece of material tend to slide one layer of the material over the next layer then stress and strain set up are called shearing stress and strain. P D’ C’ D C F Shearing stress, τ = θ θ A B A Shearing strain is measured by the angular displacement of the block. If ΔCBC’ or DAD’ = θ, then γ = CC’/BC
• 7.
  Results of Stress Whenexert forces on an object, there are three things that can happen: (3) The object can go back to its original form. If the object does that then the object was strained within its elastic limit. Elastic Limit: The relations below hold, and Hooke’s law is valid. (2) The object can retain its new shape. If this happens, the object was strained within its plastic limit. Plastic Limit: The relations below do not hold and the material is permanently deformed. (3) It can break. The object has been strained too much and fractures. Fracture: Forces exceed maximum limits derived from the relations below and material breaks.
• 8.
  When (1) occurs,the material obeys Hooke’s law F = -kΔL. The material will retain its original shape when the forces are removed. How much it moves from its original position when the forces are on it depend on the size and shape of the body and how the force is applied. The force can be applied in three ways: tension, compression, and shear: The change in shape for each of these three ways is determined from the: 1. elastic modulus for tension and compression, and 2. from the shear modulus for shear 3. If the pressure is distributed on all parts of the object, the volume of the material changes and the bulk modulus is used Finally, if the stress exceeds some maximum the object will break. The maximum stress depends on whether there is a tensile, compressive, or shear force.
• 9.
  Elasticity is aproperty of an object or material which will restore it to its original shape after distortion. A spring is an example of an elastic object - when stretched, it exerts a restoring force which tends to bring it back to its original length. This restoring force is in general proportional to the stretch described by Hooke's Law. Hooke's Law: One of the properties of elasticity is that it takes about twice as much force to stretch a spring twice as far. That linear dependence of displacement upon stretching force is called Hooke's law which can be expressed as F = -kΔL where F = force in the spring (N) k = spring constant (N/m) ΔL = elongation of the spring (m)
• 10.
  1. Tensile Stressor Compressive Stress Constant: Young’s Modulus, E = FLo /AΔL = Stress/Strain 2. Shear Stress Constant: Shear Modulus, G = FLo /AΔL = Stress/Strain 3. All directions Constant: Bulk Modulus, B = -ΔPVo /ΔV
• 11.
  Example: • Find thearea of a wire of diameter 0.75 mm in m2. What is the strain of a 1.5 m wire that stretches by 2 mm if a load is applied? • A wire made of a particular material is loaded with a load of 500 N. The diameter of the wire is 1.0 mm. The length of the wire is 2.5 m, and it stretches 8 mm when under load. What is the Young Modulus of this material? • A load of 200 N is applied to a steel wire 2 m long and diameter 0.5 mm. If the Young Modulus for steel is 2 x 10^11 Pa, the extension is:
• 12.
  Hooke’s Law essentiallystates that stress is proportional to strain Yield point Ultimate strength Elastic limit Elastic Plastic Rupture region region
• 13.
  Elastic Region •In thecontext of material behavior, a structural component is said to behave elastically if during loading/unloading the deformation is reversible. •In other words, when the loads are released the specimen will return to its original, undeformed configuration. As loads are increased and the stress in the specimen continues to rise, the material eventually reaches the elastic limit. Beyond this limit, any additional loading will result in some permanent change to the specimen geometry upon unloading. Any increase in stress beyond the yield point will cause the material to be deformed permanently. Also in this so-called yielding region, the deformation will be relatively large for small, almost negligible increases in the stress. This process, characterized by a near-zero slope to the stress-strain curve, is often referred to as perfect plasticity.
• 14.
  Plastic Region •When loadingis carried beyond the yielding region, the load needs to increase for additional strain to occur. •This effect is called strain hardening, and it is associated with an increased resistance to slip deformation at the microscale (for polycrystalline materials). Eventually, the stress-strain curve reaches a maximum at the point of ultimate stress. For many materials, the decrease in the cross-sectional area of the specimen is not readily visible to the naked eye until this limit point is passed. When the loading is continued beyond the ultimate stress, the cross- sectional area decreases rapidly in a localized region of the test specimen. Since the cross-sectional area decreases, the load carrying capacity of this region also decreases rapidly. The load (and stress) keeps dropping until the specimen reaches the fracture point.
• 15.
  rst we needto work out the area: = πr2 = π × (0.5 × 10-3)2 = 7.85 × 10-7 m2 E = Fl/Ae. 2 x 10^11 = (200 x 2) / 1.96 ^ -3 x e = (200 x 2 ) / 1.96 ^ -3 x 2 x 10^1 ress = F/A = 500 N ÷ 7.85 × e = 0.01 m = 10 mm 10-7 m2 = 6.37 × 108 Pa rain = e/l = 0.008 ÷ 2.5 = 0.0032 oung’s Modulus = stress/strain = 6.37 × 108 Pa ÷ 0.0032