Stress Tensor Components - Special Stress States - Hydrostatic Pressure - Statistics Assignment Help

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The default units used in numerical examples are m, kN, kNm, kP a. We thus do not indi cate units in examples as a rule. This is valid for the whole course. In design practice the units must be given at least for the results! 

The exercise complements section 2 of the lecture notes ’Stress, strain material failure, work, constitutive eqs” on the course web page, just ’Notes’ forthwith. They together should substi tute classes of the present 7th week of the semester. Questions concerning the stuff, address, please, to 

The purpose of the seventh week’s exercise is 1) extend the knowledge on stress and strength as taught in the elementary elasticity course 2) exercise the basic operations with the stress tensor and material failure criteria. 

It is recommended to try to solve each of the examples step by step on your own first and only when at loss at some point of the solution, resort to the text. This is how I tend to teach in contact classes. In this contactless form it requires more selfcontrol on your part, of course.  

The cube in the sketch is of differential dimension, wedge length dl. Analogous conclusion can be readily obtained for the second and third columns of the stress tensor matrix. The sketch can be used to testify to the assertion adopted without proof in the Notes, that the stress tensor and its matrix are symmetric. The equilibrium condition of moments with respect to the vertical axis through the cube center reads 

We can see two planes going through point P where the traction is known. They are the planes of the two sides of the beam and the tractions are zero. y and z are the normals of these planes. This entails zero columns 2 and 3 of the stress tensor matrix in this coordinate system:

Symmetry of the matrix allows to fill elements 2,1 and 3,1 with zeros and a single unknown el ement σxx remains. This is the reason why the stress state in frame beams can be simplified and approximate solutions of beams is used which assumes a uniaxial stress state. The assumption is acceptable if the beam is slender so that the uniaxial stress state on the edges can be assumed approximately inside the beam body. 

The second example is similar to the previous one just this time point P is on the curved surface of a general body. Axes x and y of the coordinate system lie in the tangent plane of the surface at point P. Assume that traction vector at plane x − y is {f} = {3, 4, 2}Tand the body is sub merged in a liquid with hydrostatic pressure 5 kP a. Determine as many elements of the stress matrix as possible. 

Normal stress has the direction of the normal vector and shear stress vanishes totally with this orientation of the cut plane. In this direction the material is subject to ’pure tension’ 2 kP a. The traction vector is shown in the central sketch in the above figure.

This is a very important plane stress state which can be assumed in metal sheets, concrete panels and other important structural elements. 

A bit different special stress state occurs in sufficiently long (best infinite) prismatic bodies whose load and supports does not vary in the longitudinal direction. In their cross-sections vanishes the shear stress but the normal stress remains nonzero. When axes x and y lie in the cross-section plane, the stress matrix differs from that of the plane stress by the nonzero σzz entry. This is the plane strain state (there are no extensions/contractions in the longitudinal direction). The stress tensor in example 3 meets the conditions of the plain strain state with longitudinal axis of the body in the x direction. 

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