7. semester Continuum mechanics Solution Exercise 4

7. semester Continuum mechanics
Solution Exercise 4
Question 1
⎧
⎪⎨
σ ij =
⎪⎩
Question 2
σ 0 0
0 2σ 0
0 0 0
⎫
⎪⎬
⎪⎭
⎧
⎪⎨ σ 0 0
t i = σ ij n j = 0 2σ 0
⎪⎩
0 0 0
Question 3
σnormal = t i n i =
⎧
⎪⎨
⎪⎩
⎫ ⎧
⎪⎬ ⎪⎨
⎪⎭ ⎪⎩
σ cos θ
2σ sin θ
0
cos θ
sin θ
0
⎫T ⎧
⎪⎬ ⎪⎨ cos θ
sin θ
⎪⎭ ⎪ ⎩
0
⎫
⎪⎬
⎪⎭ = ⎧
⎪⎨
⎪ ⎩
σ cos θ
2σ sin θ
0
⎫
⎪⎬
⎫
⎪⎬
⎪⎭
(1)
(2)
⎪⎭ = σ cos2 θ + 2σ sin 2 θ (3)
τ 2 = |t i | 2 − σ 2 normal (4)
= σ 2 cos 2 θ + 4σ 2 sin 2 θ − (σ cos 2 θ + 2σ sin 2 θ) 2 (5)
= σ 2 ( cos 2 θ(1 − cos 2 θ) + 4 sin 2 θ(1 − sin 2 θ) − 4 cos 2 θ sin 2 θ ) (6)
= σ 2 cos 2 θ sin 2 θ = 1 4 σ2 sin 2 (2θ) (7)
which gives τ = ± 1 σ sin (2θ)
2
Alternatively τ can be determined by projection of the stress vector t i on a vector perpendicular
to n i and in the 1 − 2 plane.
⎧
⎪⎨ − sin θ
τ = t i cos θ
⎪ ⎩
0
⎫
⎪⎬
⎪⎭
= σ cos θ(− sin θ) + 2σ sin θ cos θ = σ sin θ cos θ = 1 sin (2θ) (9)
2
In this way the sign of τ is determined directly.
1
(8)

2
Question 4
The maximum value for τ is obtained for θ = π 4 + n π 2
where n takes the values 1, 2, 3, 4.
τmax = ± 1 2 σ (10)
The value of τmax corresponds to half the dierence between the 2 principal stresses.
The corresponding normal stress is given by:
σnormal = σ 1 2 + 2σ 1 2 = 1 (σ + 2σ) (11)
2
The value of the normal stress takes the average value of the two principal stresses.
Question 5
Fig. 1: Plot of normal- and shear stress
PS: Observe the similarity with Mohr's circle.