MIT2_094S11_hw_5_sol

# MIT2_094S11_hw_5_sol - 2.094 FINITE ELEMENT ANALYSIS OF...

This preview shows pages 1–3. Sign up to view the full content.

2.094 F INITE E LEMENT A NALYSIS OF S OLIDS AND F LUIDS S PRING 2008 Homework 5 - Solution Instructor: Prof. K. J. Bathe Assigned: 03/06/2008 Due: 03/13/2008 Problem 1 (10 points): ' 2 ij v ij ij G τκ ε δ =+ (a) ij ijrs rs C τ = (b) C = (c) Let’s start from equation (b). Using ( i ji ij j ) γ and () ijrs ij ir i rs js jr s C λδδ μδδ δδ = ++ , 11 1111 11 1122 1112 12 22 1133 33 23 1123 113 31 1 11 22 33 (2 ) CC CCC C εεεγγ λμ ελ λ =+++++ + + 22 22 33 11 ) λε μ ε + + 33 22 33 11 2) ( =++ + 12 1211 11 1222 1212 12 1223 1231 22 1233 33 23 3 1 12 C εεεγγγ =+++++= μ 23 23 μγ = 31 31 = Therefore, 000 0 0 0 0 2 2 2 C λλ μλ 0 0 ⎡⎤ ⎢⎥ = + + ⎣⎦ + Page 1 of 6

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Substituting (1 )(1 2 ) E ν λ = +− and 2(1 ) E μ = + , we obtain C in Table 4.3. Hence equation (c) is equivalent to equation (b). Now derive equation (a) from equation (b). () {} 33 3 rs ij rs ir js is jr rs ij rs ij rs rs ir js is jr vv ij ijrs rs ijrs rs rs rs rs rs v rr ij ji ir js rs is jr rs ij ij rr ij j CC εε τ ε εδλ δ εδ 3 v λδδε λδδ μδδ ε μδδ μδ ′′ ++ ⎛⎞ + + + == = + + ⎜⎟ ⎝⎠ =+ + 2 2 2 2 3 2 3 i ij i vj ij ij ij v vi j j G λε δ με
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 02/24/2012 for the course MECHANICAL 2.094 taught by Professor Klaus-jürgenbathe during the Spring '11 term at MIT.

### Page1 / 7

MIT2_094S11_hw_5_sol - 2.094 FINITE ELEMENT ANALYSIS OF...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online