File:Isotropic stress noavg.svg

From Infogalactic: the planetary knowledge core
Jump to: navigation, search
Original file(SVG file, nominally 480 × 480 pixels, file size: 74 KB)

Summary

Illustration of isotropic normal tensile stress. Top left: Each face of a cube of homogeneous material is under traction by a force withmagnitude F, applied evenly over the entire face. Top right: the force applied by the material across an imaginary section S parallel to one of the faces is also F, so the stress is normal and equal to F/A, where A is the area of S (and of the face). Bottom left: if S is a diagonal cut parallel to one edge, the force across S will have to balance the two forces shown. The magnitude of that force is F*sqrt(2), but the area is also A*sqrt(2), so the stress is again perpendicular to S and equal to F/A. Top right: if S is a cut across three corners of the cube, as shown, the force across S would have to balance the three forces shown, each equat to F/2. The magnitude of the force across S is Fsqrt(3)/2, but the area of S is A*sqrt(3)/2, so the stress is again normal to S and equal to F/A. Drawn by J.Stolfi with Inkscape, according to the style of <a href="//commons.wikimedia.org/wiki/File:Axial_stress.svg" title="File:Axial stress.svg">File:Axial_stress.svg</a>.

Licensing

Lua error in package.lua at line 80: module 'strict' not found.

File history

Click on a date/time to view the file as it appeared at that time.

Date/TimeThumbnailDimensionsUserComment
current12:34, 9 January 2017Thumbnail for version as of 12:34, 9 January 2017480 × 480 (74 KB)127.0.0.1 (talk)Illustration of isotropic normal tensile stress. Top left: Each face of a cube of homogeneous material is under traction by a force withmagnitude F, applied evenly over the entire face. Top right: the force applied by the material across an imaginary section S parallel to one of the faces is also F, so the stress is normal and equal to F/A, where A is the area of S (and of the face). Bottom left: if S is a diagonal cut parallel to one edge, the force across S will have to balance the two forces shown. The magnitude of that force is F*sqrt(2), but the area is also A*sqrt(2), so the stress is again perpendicular to S and equal to F/A. Top right: if S is a cut across three corners of the cube, as shown, the force across S would have to balance the three forces shown, each equat to F/2. The magnitude of the force across S is Fsqrt(3)/2, but the area of S is A*sqrt(3)/2, so the stress is again normal to S and equal to F/A. Drawn by J.Stolfi with Inkscape, according to the style of <a href="//commons.wikimedia.org/wiki/File:Axial_stress.svg" title="File:Axial stress.svg">File:Axial_stress.svg</a>.
  • You cannot overwrite this file.

The following page links to this file: