# Intro to the Finite Element Method [lecture notes] by Y. Liu

By Y. Liu

Best nonfiction_6 books

Basic Elements of the Christian Life, Vol. 1

Booklet via Witness Lee, Watchman Nee

Additional resources for Intro to the Finite Element Method [lecture notes]

Example text

Bar and Beam Elements To calculate the support reaction forces, we apply the 1st and 3rd equations in the global FE equation. 0 × 10 4 N L u  L  3 and the 3rd equation gives, u1  EA   EA F3 = 0 − 1 1]u2  = ( − u2 + u3 ) [ L L u   3 = − 10 . ×10 4 N Check the results.! © 1998 Yijun Liu, University of Cincinnati 37 Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements Distributed Load q i j x qL/2 qL/2 i j Uniformly distributed axial load q (N/mm, N/m, lb/in) can be converted to two equivalent nodal forces of magnitude qL/2.

This expression can also be derived using other more rigorous approaches, such as the Principle of Minimum Potential Energy, or the Galerkin’s Method. Now, we evaluate (20) for the bar element by using (14) L k= − 1 / L  EA  1 − 1 E[− 1 / L 1 / L]Adx =   1 / L L  − 1 1    ∫ 0 which is the same as we derived using the direct method. Note that from (16) and (20), the strain energy in the element can be written as 1 U = u T ku 2 © 1998 Yijun Liu, University of Cincinnati (21) 31 Lecture Notes: Introduction to Finite Element Method Chapter 2.

Q L L qL/2 qL qL2/12 L © 1998 Yijun Liu, University of Cincinnati L 61 Lecture Notes: Introduction to Finite Element Method Chapter 2. 6 y p 1 E,I x 2 L Given: A cantilever beam with distributed lateral load p as shown above. Find: The deflection and rotation at the right end, the reaction force and moment at the left end. Solution: The work-equivalent nodal loads are shown below, y f m 1 E,I 2 x L where f = pL / 2, m = pL2 / 12 Applying the FE equation, we have © 1998 Yijun Liu, University of Cincinnati 62 Lecture Notes: Introduction to Finite Element Method Chapter 2.