{"id":1357,"date":"2024-09-23T14:58:54","date_gmt":"2024-09-23T12:58:54","guid":{"rendered":"https:\/\/solveredu.com\/?p=1357"},"modified":"2026-01-27T23:52:36","modified_gmt":"2026-01-27T22:52:36","slug":"rozci%c4%85ganie-%c5%9bciskanie-statycznie-niewyznaczalne-zadania","status":"publish","type":"post","link":"https:\/\/solveredu.com\/en\/post\/rozci%c4%85ganie-%c5%9bciskanie-statycznie-niewyznaczalne-zadania\/","title":{"rendered":"Tension compression statically indeterminate - tasks"},"content":{"rendered":"

In this entry:<\/p>\n\n\n\n

    \n
  1. How to solve statically non-determinable tasks<\/u><\/a><\/li>\n\n\n\n
  2. Example of a solution to a concentrated force task<\/u><\/a><\/li>\n\n\n\n
  3. Example solution of a continuous load task<\/a><\/li>\n<\/ol>\n\n\n\n

    The normal stress induced by tensile and compressive forces in a prismatic bar was discussed in the entry Tension and compression<\/a>. In this post, we dealt with statically determinable examples, that is, examples where we had only one support reaction and were able to determine it from the equilibrium condition.<\/p>\n\n\n\n

    Statically indeterminate tasks<\/h1>\n\n\n\n

    Now we will deal with examples that are a little more difficult, that is, statically non-determinant. These are tasks where the bar is fixed at both its ends (you can say that it is placed between two non-sliding walls). In this case, we have two unknown reaction forces of these restraints and only one equilibrium equation, so that's why we talk about statically non-equivocal examples.<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    Equilibrium equations<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    Geometric condition<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    In this case, we use an additional geometric condition<\/strong>. This condition states that the total elongation of the rod must be equal to zero. Since both ends of the bar cannot move (they are restrained), so the total deformation of our system is equal to \"0\".<\/p>\n\n\n\n

    You may encounter tasks in which the bar is set vertically. It does not matter when it comes to the method of solution we proceed exactly the same.<\/p>\n\n\n\n

    When it comes to the number of fragments into which we divide our rod, two factors matter:<\/p>\n\n\n\n