Solution-Quiz-Fall2010-2_3

Solution-Quiz-Fall2010-2_3 - s K Q F = . 1 1 2 6 3 4 4 1 1...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
EML 4507 Fall 2010 Quiz 2-v3 Solution A stepped bar is clamped at both ends. A force of 10,000 N is applied as shown in the figure. Areas of cross sections of the two portions of the bar are 1×10 -4 m 2 and 2×10 -4 m 2 , respectively. Young’s modulus =10 10 Pa. Use three finite elements to solve the problem. a) Sketch the displacement field u ( x ) as a function of x on the graph below. b) Calculate the stress at a distance of 2.5 m from the left support. Solution Note that in uniaxial bar FE modeling there should be nodes at points of application of forces. Since the cross sectional area is assumed to be a constant within an element there should be nodes where area changes. Thus we require at least three elements for this problem as shown below. Write down the stiffness matrices of the three uni- axial bar elements: [ ] [ ] [ ] (1) 1 6 2 (2) 2 6 3 (3) 3 6 4 1 1 10 1 1 2 2 10 2 2 2 2 10 2 2 u k u u k u u k u - = - - = - - = - Assemble them to obtain the structural level equations: [ ]{ } { } s s
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: s K Q F = . 1 1 2 6 3 4 4 1 1 1 3 2 10 10,000 2 4 2 2 2 u F u u u F- -- = +-- - Delete the rows corresponding to u 1 and u 4 to obtain the global equations as [ ]{ } { } K Q F = 2 6 3 3 2 10 2 4 10,000 u u- = -+ Solving the above equation we obtain the displacements as: 3 3 2 3 2.5 10 m, 3.75 10 m u u--= + = + The cross section at 2.5 m from the left support is in Element 3. The force in an Element 3 is given by ( ) ( ) 3 (3) 4 3 AE P u u L =- Stress is given by P/A . Substituting the numbers we obtain ( ) ( ) ( ) 3 10 3 10 3.75 10 37.5MPa -=- = -1 m 1 m 1 m 1 m 1 1 m 1 m 2 3 1 2 3 4 1 2 3 0 u ( x ) mm x 2.5 3.75 (m) 10 kN...
View Full Document

Ask a homework question - tutors are online