{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

4020-F2011-HW3Solutions

# 4020-F2011-HW3Solutions - MSE 4020 HW3-SOLUTION Problem 1...

This preview shows pages 1–5. Sign up to view the full content.

1 MSE 4020 HW3-SOLUTION 09/27/2011 Problem 1 Fail if the following conditions are satisfied, ! 1 N + ! 2 N + ! 3 N 3 = ! 1 N + ! 2 N 3 " P yield # 3p yield 3p yield - 3p yield - 3p yield ! 2 N ! 1 N SAFE

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

View Full Document
2 Problem 2 Fail if the following conditions are satisfied, ! 1 N + ! 2 N + ! 3 N 3 = ! 1 N + ! 2 N 3 " P yield # 3p yield 3p yield ! 2 N ! 1 N SAFE
3 Problem 3 Fail if the following conditions are satisfied, ! 1 N + ! 2 N 3 " P yield ! max N #! min N 2 " P yield \$ % ' ( 2p yield 2p yield - 2p yield - 2p yield ! 2 N ! 1 N SAFE 3p yield 3p yield

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

View Full Document
4 Problem 4 (a) For homogeneous strain, in general we can write down following displacement field for 2D problem: u x = r ! x + s ! y + t u y = m ! x + n ! y + k where, r, s, t, m, n, k are constants to be determined from given geometry. For Point A(0,0), after deformation, its position is unchanged, thus, u x = r ! 0 + s ! 0 + t u y = m ! 0 + n ! 0 + k " # \$ % \$ t = 0 k = 0 ' ( \$ ) \$ For Point B(2,3), after deformation, its position becomes (1.9,3.05), thus,
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

4020-F2011-HW3Solutions - MSE 4020 HW3-SOLUTION Problem 1...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online