me211solution_13 - y x z 900 x = = 300 psi , y = z = 0 3(1)...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
x z y 900 300 3(1) x psi σ == , 0 yz σσ Generalized Hook s Law: [] 3 3 11 ( ) 300 (0 0) 0.375(10 ) 800(10 ) xx y z E εσ ν ⎡⎤ =− + = + = ⎣⎦ 3 3 () 0 ( 3 0 0 0 ) 0 . 3 7 5 ( 1 0 800(10 ) yy x z E + = −+ = ) n New length to x direction: 3 ' 6(1 ) 0.375(10 )) 6.00225 x ai ε =+=+ = New length to y direction: 3 ' 3(1 ) 0.375(10 ) ) y b εν =+ 1 3 tan 6 θ ⎛⎞ = ⎜⎟ ⎝⎠ New angle: 1 3 ' tan 0.01 6 θθ Δ= ° ' tan( ') ' b a = 3 1 3 0.375(10 ) ) tan tan 0.01 6 6.00225 −° = 0.164 =
Background image of page 1

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

View Full DocumentRight Arrow Icon
15-85. Plane stress: 3 0 σ = [] 11 2 1 () 3 E εσ ν =− + 6 12 6 1 350(10 ) (0.3)( 0) 200(10 ) σσ + LL (1) 22 1 1 3 E + 6 21 6 1 250(10 ) (0.3)( 0) 200(10 ) −= −+ (2) Solving Eqs. (1) and (2) yields: 1 60.4 MPa = 2 31.9 MPa
Background image of page 2
15-102 Since the aluminum is confined at its sides, 0 xy εε == and since it is not restrained in z direction, . 0 z σ = Applying the generalized Hooke s Law with the additional thermal strain, 1 () xx y z T E εσ ν α ⎡⎤ =− + + ⎣⎦ Δ 1 yy x z T E + + Δ 1 zz x y T E + + Δ 6 3 1 0 0.35( 0) 13.1(10 )(200) 10(10 ) σσ + + LL (1) 6 3 1 0 0.35( 0) 13.1(10 )(200) 10(10 ) yx + + (2) 6 3 1 0 0.35( ) 13.1(10 )(200) 10(10 ) zx y + + (3) Three unknowns can be found from three equations.
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 10

me211solution_13 - y x z 900 x = = 300 psi , y = z = 0 3(1)...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online