ELASTISITY AND STRENGTH OF MATERIALS M Theory Let us consider the effect of a stretching force F applied to a bar (Fig. 1). The applied force is transmitted to every part of the body, and it tends to pull the material apart. This force, however, is resisted by the cohesive force that holds the material together. The material breaks when the applied force exceeds the cohesive force. If the force in Fig. 5.1 is reversed, the bar is compressed, and its length is reduced. Formulas 1. Tensile Stress o (N/m² or Pa) is the internal force per unit area acting on the material; it is defined as here F is the internal force acting on the body (N) H; A is the area on which the force is applied (m²). 2. Longitudinal (tensile) strain ɛ. The force applied to the bar in Fig. 1 causes the bar to elongate by an amount AL. The fractional change in length is called the longitudinal strain; that is, Here l is the length of the bar (m), lo is initial length and Al is the change in the length due to the applied force. 3. Hooke's law. In 1676 Robert Hooke observed that while the body remains elastic, the ratio of stress to strain is constant (Hooke's law); that is, o = ɛY, here ɛ is longitudinal strength and the constant of proportionality Y is called Young's modulus (Pa). Young's modulus has been measured for many materials (Table 1). 4. A Spring. A useful analogy can be drawn between a spring and the elastic properties of a material. Consider the spring shown in Fig. 2. Equilibrium position The force Fgpring required to stretch (or compress) the spring is directly proportional to the amount of stretch; that is, Fspring = -kAL The constant of proportionality k is called the spring constant (spring stiffness). 5. Spring constant (N/m) Y· A k = here Y is Young's modulus (Pa); A is cross-sectional area (m²); l is length of the string before deformation. 6. Potential energy of the string (J) A stretched (or compressed) spring contains potential energy; that is, work can be done by the stretched spring when the stretching force is removed. The energy E stored in the spring is k(Al)² 2 here k is spring constant(N/m) and Al is longitudinal strain (m). 7. Shear Stress o (N/m2 or Pa) is the internal force per unit area acting on the material. Unlike tensile and compressive stresses, the area is parallel to the applied force. Ax here F is the force acting on the body (N) H; A is the area on which the force is applied (m²). 8. Shear Strain Lo Δx where Ax is the amount by which the top surface moves, and Lo is the distance from the top surface, which moved a distance Ax, to the bottom surface which didn’t move at all. 9. Sher Modulus Where o is shear stress and ɛ – shear strain. 3. bone force 400 kN is required. Find rupture strength of bone having diameter 30 mm and thickness 3 mm, if to break the