Solution-Quiz-Fall2010-2_1

# Solution-Quiz-Fall2010-2_1 - 4 4 1 1 1000 1 2 1 10 1 3 2 2...

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EML 4507 Fall 2010 Quiz 2-v1 Solution A stepped bar is clamped at both ends. A force of 1000 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 9 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 5 2 (2) 2 5 3 (3) 3 5 4 1 1 10 1 1 1 1 10 1 1 2 2 10 2 2 u k u u k u u k u - = - - = - - = - Assemble them to obtain the structural level equations: [ ]{ } { } s s s K Q F = . 1 1 2 5 3
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Unformatted text preview: 4 4 1 1 1000 1 2 1 10 1 3 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 5 3 2 1 1000 10 1 3 u u-- = - Solving the above equation we obtain the displacements as: 3 3 2 3 6 10 m, 2 10 m u u--= - × = - × The cross section at 2.5 m from the left support is in Element 3. The force in 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 9 3 10 0 2 10 2MPa σ-=- - × = Variation of u ( x ) as a function of x : 1 m 1 m 1 m 1 m 1 1 m 1 m 2 3 1 2 3 4 1 2 3 0 -6 -2 1000 N x (m) u ( x ) mm...
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## This note was uploaded on 06/07/2011 for the course EML 4507 taught by Professor Kim during the Fall '11 term at University of Florida.

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