Solutions of Homework 7 y 1.
(A) Pressure at x=0, x=l resulting in uniform stress σxx:
E eff
xx
xx
Also: xx
p
h
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xx
l
1
x
o
1 xx xx E
Then: E eff E
t z
(B) The constraint in y direction at y=0 and y=h results in εyy=0 (more specifically u 2=0) and σ11= -p.
Then: yy
0
1
1 yy xx yy => yy xx
E
1 2 xx Also: xx 1 xx xx yy E E 1
Finally, E eff
xx xx
E
1 2
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(plane strain modulus)
(C) Faces y=0, y=h, z=0 and z=t are constrained so that v=0 and w=0 respectively on the four faces.
In this case, yy
zz 0 because v=w=0;
Also: yy
zz 0
yy 0
1
=> yy
1
1 yy xx yy zz = yy xx zz E E
xx zz
Similarly, zz
(1)
xx yy
From (1) and (2), we can write yy Then: xx
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1
xx yy E
(2)
zz
2
xx 1 zz => E eff xx xx
1
E 1
2 2 1
2.
We first check whether equilibrium is satisfied. When strains are calculated from stresses, we will also check for compatibility. It is easily shown that the following equations are satisfied:
xx , x xy, y 0
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xy , y yy, y 0
The strains in this problem are found by:
yy xx
1
xx yy E 1 yy xx E 1 xy E
xx
(1)
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(2) (3)
Substituting stress expressions in (1)-(3) gives:
P 3 l
y 3 y 2 y 1 3 y y 3 4 2 h 5 h h 2 2 h h
(4)
3 2 2 3 l 2 P 1 3 y y x y 3 y y yy 2 4 4 E h h h 5 h h 2 h 2 2 h
(5)
2
x xx 4 E h 2 h
1 3
xy 4 2 5 h
2
2
x h
(6)
We can verify that compatibility is satisfied for (4) –(6).
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For small strains:
xx ux , x yy u y , y xy
1 2
u
x, y
u y, x
Then we integrate (4) and (5) :
u x xx ( x, y )dx f ( y )
(7)
( x , y )dy g ( x )
(8)
u y
yy
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2
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In order to determine f(y) and g(x), we take into account that for x=0 => u x=0 due to symmetry.
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Thus, we can compute f(y) directly from (7). Also we calculate g(x) by substituting to
xy
1 2
u
x, y
From (7): u x
u y, x
P
..... x
E
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f ( y) f ( y) 0
Then substituting u y from (8) in xy
1 3
1
u 2
x, y
u y, x 1 4 2 2 h
2
x h
2 2 2 31 P y2 P 3 l y2 x x 1 3 y 1 4 4 3 6 8 g , ( x ) x h h 5 h h 2h 2 2 E 2 h E h3 2 h h
where g,x(x) is a function of x only.
The equality holds for all y and also for y=0: 2 3 2 3 x2 l 1 x 1 3 x g ( x ) C x 2 4 h E 4 h 3 h h 10 h 4 h
3 x 2
P 3
2
The constant C represents rigid body translation and thus can be omitted.
3
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3.
Determine: (a) The in-plane principal stresses and their directions if E=200 GPa and υ=0.3 (b) The absolutely maximum shear stress.
a xx We have:
Solving for
b
1
c
1
4
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xx
3 4
yy
3 4
2 xy
These are the 60º rosette equations
3 3 xx yy 2 xy 4 4 4
εxx, εyy, εxy,
we get εxx = 190µε, εyy = -130 µε, εxy = 577 µε
190 130 577 Then: 1,2 2 2 2 and, xy (max) ( 1 2 ) / 2 330 2
190 130
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1
2
2 1 360 , 2 300
The orientations of principal axes are:
2 p tan 1
577 320
p 30.5 (and θ p=120.5º)
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The principal stresses are also in the same coordinate frame as the principal strains.
1 POINT
Using Hooke’s law, we have:
1
2
200 109
1 2 0.3 11 0.3 11 22
1 0.3 1 2 0.3
200 109
Pa 0.4 0.3 11 0.3 22 62.3 1.3 0.4
200 109
0.4 0.3 22 0.3 11 106 39.2 MPa 0.52
Then: max
1 2 2
50.75 Pa
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All componens of stress in x 3 direction is 0 MPa.
4
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The maximum shear stress is obtained by calculating the radius of the largest Mohr’s circles formed by the principal stresses MPa.
4.
σ1=62.3
MPa,
σ2=-39.2
MPa and
σ3=0.
Hence the maximum shear stress is 50.75
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One can treat this problem as a plane stress problem. The stiffness matrix is known for the x 1-x2 coordinates. Also, because this is a quasi-isotropic material, [C ij] is the same at every 60º rotation. If the stress and the strain are known at the x 1-x2 coordinate, then they are connected through [C ij]. The same is true for a rotation that is a multiple of 60º. 2 POINTS
Assume a strain
ε11 applied
on the x 1-x2 coordinate such that
ε22 =
0, ε12 = 0.
11 C11 C12 0 11 C11 C12 0 11 C1111 0 22 C12 C22 0 0 C12 11 (1) 22 C12 C22 12 0 0 C44 12 0 0 C 44 0 0
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If we apply the same strain at 60º in the x 1-x2 coordinate system we should have the same stress as in the x1-x2 coordinate system.
11 0 ij ' 0 0 0 C 11 11 ij ' 0 C 12 11 ij ' Q ij QT ij QT ij ' Q
5
Also:
1 2 Q 3 2
3
1 POINT
2 1
2
ij QT ij ' Q 11
1
3
4
4 3
3
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4
4
ij ' Q ij QT ij QT ij' Q 1 3 2 2 Similarly: Q 3 1 2 2 1 3 C11 C12 4 4 T ' ij Q Q ij 11 3 (C11 C12 ) 4
3 4
(C11 C12 )
3 1 C11 C12 4 4
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In the x 1-x2 frame, [εij] and [σij] should satisfy (1):
11 C11 C12 C 22 12 C22 12 0 0
0 11 0 22 11 C44 12
1 4 3
C11
3 4 1
C 12
C11 C12 4 4 3 (C11 C 12 ) 4
C11 C 12 0
C12 C22 0
0 0 C 44
1
4 3
11 4 3 4
By comparing LHS to RHS in the last equation, we have:
C12 C 22 C44 C11 C 12
6
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