Understanding Static Stress & Mechanical Failure in Engineering Materials, Study notes of Designs and Groups

Download Understanding Static Stress & Mechanical Failure in Engineering Materials and more Study notes Designs and Groups in PDF only on Docsity! Prof. Sengupta MET 301:Theories of failure 1/9 Chapter 2 Working Stress and Failure Theories A Simplified Approach We are interested in learning how static mechanical stress can cause failure in machine parts. Static stress means that the stress has been applied slowly and is maintained at a steady level. Failure from cyclic (or dynamic) stress and impact stress will be treated later. Here, we should also keep in mind that, there are many other factors such as, surface wear damage from friction, overheating, chemical corrosion, metallurgical fault or a combination of these and other factors may also cause failure. 1. Molecular Concept of Mechanical Failure Engineering materials have a crystalline molecular structure, which means the atoms (or molecules) of the material are arranged in a fairly ordered fashion and the atoms are held in fixed position with respect to each other by strong inter-atomic bond. An external mechanical force tends to displace these atoms from their original positions in the direction of the force, which is resisted by the inter-atomic forces. Up to a certain limiting level of the external force, the atoms are displaced to some extent, but are pulled back to their original position when the external force is removed. This phenomenon gives rise to the elastic behavior of material (elastic deformation), that is up to a certain stress level generally the displacement is proportional to force. Hook’s law essentially is the same that is stress is proportional to strain. If a force is applied parallel to an atomic plane (shear force) and the force is high enough, a plane of atoms may slide over the adjacent plane of atoms overcoming the inter-atomic forces of the immediately neighboring atoms. When this sliding occurs, atoms in the sliding plane will slip under the influence of new set of atoms. Conceptual model of this slip deformation is shown in the diagram below. After the slip has occurred, the positions of the atoms have changed permanently, resulting in a permanent change in shape or size of the part. This type of permanent deformation is called plastic deformation. Plastic deformation is not acceptable in most mechanical design situations, because the permanently deformed part may no longer serve its intended purpose, and from the mechanical design stand point we may say that the part has failed. For example, a landing gear of an aircraft deforms elastically during landing from the ground reaction forces, but we certainly don’t want it to be permanently (or plastically) deformed, because then the actuators or other mechanisms may not work properly during the next landing. Prof. Sengupta MET 301:Theories of failure 2/9 If a force applied normally across the atomic plane (tensile force) and the force is high enough, two adjacent planes of atoms may separate out from each other producing a crack. The initial crack will reduce the cross-sectional area of the load bearing surface, thus helping the crack to propagate until the entire surfaces separates from the inter-atomic plane. This type of failure, which produces separation of atomic planes, is called fracture or rupture type failure. Obviously this type of failure will also not be acceptable in mechanical design. A conceptual model of fracture failure is shown below. In macro scale, the materials are aggregate of randomly oriented grains. This means that the atomic planes in different grains are randomly oriented (see figure below). The implication is that, the externally applied force may act as shear force or tensile force to different grains, depending on their orientation of atomic plane. As a result of this, the microstructure of a material can influence whether there will be slip or a crack from an externally applied force. Prof. Sengupta MET 301:Theories of failure 5/9 4. Theories of Failure We will study four important failure theories, namely (1) maximum shear stress theory, (2) maximum normal stress theory, (3) maximum strain energy theory, and (4) maximum distortion energy theory. Out of these four theories of failure, the maximum normal stress theory is only applicable for brittle materials, and the remaining three theories are applicable for ductile materials. The failure theories have been formulated in terms of three principal normal stresses (S1, S2, S3) at a point. For any given complex state of stress (x,y,z, xy,yz,zx), we can always find its equivalent principal normal stresses (S1, S2, S3). Thus the failure theories in terms of principal normal stresses can predict the failure due to any given state of stress. When the external loading is uniaxial, that is S1= a positive or negative real value, S2=S3=0, then all failure theories predict the same as that has been determined from regular tension/compression test. 4.1. Maximum shear stress theory: APPLICABLE FOR DUCTILE MATERIAL – Failure starts by yielding (plastic deformation). This theory postulates that failure will occur in a machine part if the magnitude of the maximum shear stress (max) in the part exceeds the shear strength (yp)of the material determined from uniaxial testing. max => yp 22 ; 2 ; 2 max 133221 ypSSSSSSS         Dividing both side by 2,   ypSSSSSSS  212121 ;;max Using a design factor of safety Nfs, the design equation is,   fs yp N S SSSSSS  212121 ;;max Prof. Sengupta MET 301:Theories of failure 6/9 Example 1: Use Maximum Shear Stress theory to determine the Factor of Safety Nfs , when the stress at a point is given by S1 = -10,000 psi, S2=20,000 psi, S3=0, and the yield strength of the part material Syp=51,000 psi.     7.1 000,30 000,51 )000,20000,10 000,51 ;;max ;;max 212121 212121       SSSSSS S N N S SSSSSS yp fs fs yp Example 2: Use Maximum Shear Stress theory to determine the Factor of Safety Nfs , when the stress at a point is given by S1 = 10,000 psi, S2=20,000 psi, S3=0, and the yield strength of the part material Syp=51,000 psi.   76.2 000,20 000,51 )200000 000,51 ;;max 212121      SSSSSS S N yp fs S1= 10,000 S2=20,000 S3=0 max=Max Radius = 10,000   Mohr Circle S1= -10,000 S2=20,000 S3=0 max=Max Radius = 15,000   Mohr Circle Prof. Sengupta MET 301:Theories of failure 7/9 4.2. Maximum normal stress theory APPLICABLE FOR BRITTLE MATERIAL – Failure starts from a crack (fracture). This theory postulates that failure will start in a machine part if the maximum tensile stress in the part exceeds the ultimate tensile strength of the part material (SUT), or if the maximum compressive stress in the part exceeds the ultimate compressive strength of the part material (SUC) as determined from uniaxial testing. For a given stress condition, if maxT is the max tensile stress, and maxC is the max compressive stress, then the design equation is  fsCfsTfs C UC fsC T UT fsT NNMinimumNand S Nand S N , maxmax    Example 3: The stresses on a Class 25 Gray Cast Iron part is shown below. Find the factor of safety. From table 14-16 in textbook for Class 25 Grey CI SUT = 25,000 psi & SUC =100,000 psi The Max Tensile stress = 10,000, NfsT= 25,000/10,000= 2.5 The Max Compressive stress = 20,000, NfsC = 100,000/20,000 = 5 The Nfs of the part is 2.5 (smaller of the two). 4.3. Maximum strain energy theory This theory postulates that failure will occur when the strain energy per unit volume due to the applied stresses in a part equals the strain energy per unit volume at the yield point in uniaxial testing. Strain energy is the energy stored in a material from elastic deformation, which is, work done during elastic deformation. Work done per unit volume = strain x average stress. During tensile test, stress increases from zero to Syp, that is average stress = Syp/2. Elastic strain at yield point = Syp/E, where E is the elastic modulus of elasticity. Strain energy per unit volume during uniaxial tension = average stress x strain = Syp 2/2E When the applied stress is (S1, S2, S3) then it can be shown (see textbook for derivation) that the strain energy stored in the part = [S1 2 + S2 2 + S3 2 - 2( S1S2 + S2S3 + S3S1 )]/2E, where  is Poisson’s ratio. 20 kpsi 20 kpsi 10 kpsi 10 kpsi