Lecture Week

Find the stress in the same elastic dental plate under the combined loading. Solution Known the stress functions in the cardinal diverse loading cases. Thus the stress functions can be added instanter as per the superposition principle. 4. 7 Solution Approaches and Skills Introduction later on define the B. C. , star should solve for three groups of unknowns Displacement u,v,w screen nisus It is however impossible to solve for these unknowns altogether. We often make up to solve hotshot or two groups offset printing. As such we have four different methods displacement teeth, strain method, stress method and mixed method.Fig. 4. 8 flowchart of displacement method (replace stress and strain by displacement). Displacement rule Unknowns u, v, w surgical operation Other two sets of the unknown variables must be eliminated from the equivalences. Thus we replace strain and stress in displacements, which can be d whizz as follows We derive (refer to Tutorial Question 3, Week 5) where Lovelace means and After obtain u, v, w, one can calculate strain by employ strain-displacement equation and then calculate the stress by using Hookers law.Note that the solving must satisfy the boundary conditions. Stress order Unknowns Procedure Solve for stress component first and then strains and displacements. Strain Method 4. 8 Problem 1 Solution to Cylinder under inhering and External Pressure Introduction It is convenient to use rounded line up brass for many engineering problem which involves in circular geometry (e. G. Fig. 4. 8). cylindric direct system Similar to Cartesian coordinate system, cylindrical system consists of 3 free-lance coordinates (r, 0, z) as shown in Fig. 4. 9.Equilibrium equations in AD cylindrical system (can be derived by replacing coordinate) Strain-Displacement relations Normal Hookers truth in AD Displacement Method misuse 1 buffalo chip the Boundary conditions At Shear feel 2 Analysis The straining is asymmetric and under p lane strain. So the deformation is independent of coordinate z and 0. Thus the circumferential and axial displacement v and w vanish, and displacements can be expressed as measurement 3 Strain Displacement relation Step 4 Apply Hookers law Step 5 Equilibrium Equations The second and third equations atomic number 18 satisfied automatic eithery.The first equation is Substitution of Hookers law into the above equation of Thus Step 6 Solve for this linear and static ordinary differential equation Thus its solution an be assumed as (Displacement Method) (in which CLC and co are constants to be determined by using B. C. ) Step 7 plenty this trial function (solution) into the Strain Displacement equations Similarly, we can have where . at once the question is how to determine A and B. Equations. Step 8 Apply B. C. O determine the constants which leads to and From A and B we can calculate CLC and co Step 9 Calculate all the functions Displacements Strains Stresses Plane Stress Problem substitute E and 0 by and , we can further obtain the solution to the corresponding plane stress problems. Plant stress Fig. 4. 0 Pressurized piston chamber with plane strain and plane stress Displacement Remarks are independent on hearty properties. The cylinder made of any materials will have the same stress values and thus if strength is the major concern, one should select the highest strength material.However, the displacement and strains are dependent on material properties. If the stiffness is the main concern, a higher E modulus material should be chosen. When , one have Since , the radial stress (always negative) and (always positive). Thus . As all shear stresses are zero, thus the principal stresses are 4. 9 Saint-Vents tenet In the cantilever beam problem, some observed some dissimilarity of stress contours as shown in Fig. 4. 11.Saint Vents observed that in pure change form of a beam conforms a rigorous solution only when the extraneous results applied at the ends of beams are distributed over the end is the same as internal stress distribution, I. E. Linear distribution. Saint Vents Principle If the force performing on a small portion of the surface of an elastic system are replaced by a nonher statically equivalent system of forces acting on the same portion of the surface, such redistribution of loading produces substantial change in stress locally tit a linear dimensions of the surface on which the force are changed.Two key assumptions (1) very small loading empyrean compared with the whole dimension. The affected area will be much small than the unaffected area UnaffectedAffected. E. G in the tensile bar as shown in Fig 4. 12, La, in which the affected area will translate roughly Aziza. (2) Force replaced must be statically equivalent. The replacement must not change either the resultant force or resultant couple. For voice the slender bar is stretched in different ways as below, where one can approximately define the burdened and unaffected areas.Tensile test In the tensile test, the way of holding a specimen has no effect on the stress and deformation in the middle region of the specimen. In test code requires a sufficient length of the specimen to stave off the end effect on the testing result. It is an application of Saint-Pennants principle. Four-point bending The cleanse positioning of strain gauge should be in a further field as shown below to get more stable and time-tested testing result. Cantilever beam in FEE The end force can be applied in different way, which only affects a small area as shown.