Lecture Week

Find the stress in the same elastic plate under the combine loading. Solution Known the stress functions in the two different loading cases. Thus the stress functions outhouse be added directly as per the superposition principle. 4. 7 Solution Approaches and Skills Introduction After define the B. C. , superstar should solve for three groups of unknowns Displacement u,v,w labour Stress It is however impossible to solve for these unknowns altogether. We a good deal have to solve champion or two groups first. As such we have four different methods displacement teeth, strain method, stress method and intricate method.Fig. 4. 8 Flowchart of displacement method (replace stress and strain by displacement). Displacement Method Unknowns u, v, w Procedure Other two sets of the unknown variables must be eliminated from the equations. Thus we replace strain and stress in displacements, which can be done as follows We derive (refer to Tutorial Question 3, Week 5) where Lovelace factor and After obtain u, v, w, one can calculate strain by using strain-displacement equation and then calculate the stress by using Hookers law.Note that the etymon must satisfy the boundary conditions. Stress Method Unknowns Procedure Solve for stress component first and then strains and displacements. bloodline Method 4. 8 conundrum 1 Solution to Cylinder under Internal and External Pressure Introduction It is convenient to use cylindrical coordinate frame for many design problem which involves in circular geometry (e. G. Fig. 4. 8). Cylindrical coordinate system Similar to Cartesian coordinate system, cylindrical system consists of 3 autarkic 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 Law in AD Displacement Method quantity 1 Check the Boundary conditions At Shear Step 2 Analysis The deformation is asymmetric and under plane strain. So the defor mation is indep stop overent of coordinate z and 0. Thus the circumferential and axial displacement v and w vanish, and displacements can be expressed as Step 3 Strain Displacement relation Step 4 Apply Hookers law Step 5 Equilibrium Equations The second and third equations atomic number 18 satisfied automatically.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 issue an be assumed as (Displacement Method) (in which CLC and co are constants to be determined by using B. C. ) Step 7 plug this trial function (solution) into the Strain Displacement equations Similarly, we can have where . Now 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 Replacing E and 0 by and , we can nurt ure obtain the solution to the corresponding plane stress problems. Plant stress Fig. 4. 0 Pressurized cylinder with plane strain and plane stress Displacement Remarks are independent on material 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 nonesuch-Vents Principle In the cantilever beam problem, some observed some difference of stress contours as shown in Fig. 4. 11.Saint Vents observed that in pure bending of a beam conforms a rigorous solution only when the external forces 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 acting on a small portion of the protrude of an elastic body are replaced by another 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 area compared with the whole dimension. The affected area will be much littler than the unaffected area UnaffectedAffected. E. G in the tensile bar as shown in Fig 4. 12, La, in which the affected area will take nigh Aziza. (2) Force replaced must be statically equivalent. The replacement must not change either the resultant force or resultant couple. For example the slender bar is stretched in different ways as below, where one can approximately define the effected and unaffected areas.Tensile test In the tensile test, the way of rete ntion a specimen has no effect on the stress and deformation in the middle region of the specimen. In test code requires a equal length of the specimen to avoid the end effect on the testing result. It is an application of Saint-Pennants principle. Four-point bending The better positioning of strain gauge should be in a far field as shown below to get more stable and reliable testing result. Cantilever beam in FEE The end force can be applied in different way, which only affects a small area as shown.