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CONTINUUM MECHANICS for ENGINEERS

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(b) Project each of the stress vectors obtained in (a) onto the primed<br />

axes to determine the nine components of [ σ .<br />

ij ′ ]<br />

(c) Verify the result obtained in (b) by a direct application of Eq 3.5-1<br />

of the text.<br />

⎡143<br />

36 114⎤<br />

1 ⎢<br />

⎥<br />

Answer: [ σ ′ ]=<br />

⎢<br />

⎥<br />

MPa<br />

ij 36 166 3<br />

7<br />

⎣<br />

⎢114<br />

3 −15⎦<br />

⎥<br />

3.13 At point P, the stress matrix is given in MPa with respect to axes<br />

Px 1x 2x 3 by<br />

⎡6<br />

4 0⎤<br />

⎡2<br />

1 1⎤<br />

⎢<br />

⎥<br />

⎢ ⎥<br />

Case 1: [ σ Case 2:<br />

ij]=<br />

⎢<br />

4 6 0<br />

⎥ [ σ ij]=<br />

⎢<br />

1 2 1<br />

⎥<br />

⎣<br />

⎢0<br />

0 −2⎦<br />

⎥<br />

⎣<br />

⎢1<br />

1 2⎦<br />

⎥<br />

Determine <strong>for</strong> each case<br />

(a) the principal stress values<br />

(b) the principal stress directions.<br />

Answer: (a) Case 1: σ (1) = 10 MPa, σ (2) = 2 MPa, σ (3) = –2 MPa<br />

Case 2: σ (1) = 4 MPa, σ (2) = σ (3) = 1 MPa<br />

( 1) eˆ eˆ<br />

(b) Case 1: nˆ<br />

1+ 2 ( 2) eˆ eˆ<br />

=± , nˆ<br />

1−2 ( 3)<br />

=± , nˆ = meˆ3<br />

2<br />

2<br />

( 1) eˆ ˆ ˆ<br />

Case 2: ˆ 1+ e2 + e3<br />

( 2) – eˆ ˆ<br />

n = , ˆ<br />

1+ e2<br />

( 3) – eˆ ˆ ˆ<br />

n = , ˆ<br />

1– e2 + 2e3<br />

n =<br />

3<br />

2<br />

6<br />

3.14 When referred to principal axes at P, the stress matrix in ksi units is<br />

* [ σ ij]=<br />

⎡2<br />

0 0⎤<br />

⎢ ⎥<br />

⎢<br />

0 7 0<br />

⎥<br />

⎣<br />

⎢0<br />

0 12⎦<br />

⎥<br />

If the trans<strong>for</strong>mation matrix between the principal axes and axes<br />

Px 1x 2x 3 is<br />

⎡ 3 1<br />

⎢ 5<br />

1<br />

[ aij]= ⎢a<br />

a a<br />

2<br />

⎢ 3 −1<br />

⎣ 5<br />

−<br />

4<br />

−<br />

5<br />

21 22 23<br />

−<br />

4<br />

−<br />

5<br />

where a21, a22, and a23 are to be determined, calculate . σ ij<br />

⎤<br />

⎥<br />

⎥<br />

⎥<br />

⎦<br />

[ ]

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