AE 321 – Solution of Homework #8 Problem #1.
Surface #2
T
M
Surface #3
Surface #1
Formulate the problem: Traction BVP
First apply static equilibrium to find all forces applied to this beam. A reaction force T is required to be applied at surface #1 to maintain force equilibrium. At the same surface, a moment, M, is needed to counteract the moment due to the vertical force, T, at the beam free end. Here, we will only use the force T at surface #1 to compute the surface stresses since we have not talked about the use of moment boundary conditions yet. [As we just saw in class the moment, M, will result in an axial stress that varies linearly through the beam thickness.]
5 POINTS First we apply Cauchy’s formula on each beam surface to find the surface stresses from the given tractions: Surface #1 for Ti=0 and n={-1,0,0}: T 0 T x xxnx xyny xz nz xx 0 T T Ti ij nj Ty yx nx yy ny yz nz yx T T 0 T z zx nx zy ny zz nz zx 0 x
y
3 PO POIN INTS TS
z
Surface #2 for Ti=0 and n={0,cosθ,sinθ} where the unit vector is defined de fined in the schematic below (for simplicity we used a circular cross-section but this is not necessary.):
1
x3 n
2 POINTS θ
x2
T 0 T x xxnx xyny xz nz cos xy sin xz 0 T 0 Ti ij n j Ty yx nx yy ny yz nz cos yy sin yz 0 T 0 T z zxnx zyny zznz cos zy sin zz 0 xy xz 0 3 POINTS yy yz 0 zy zz 0 x
y
z
Surface #3 for Ti=-Tey and n={1,0,0}: T 0 T x xxnx xyny xznz xx 0 T T Ti ij n j Ty yxnx yy ny yz nz yx T T 0 T z zxnx zyny zznz zx 0 x
3 POINTS
y
z
Given the above BCs this is tractions BVP and we will start with equilibrium equation assuming that body forces are negligible:
ij , j f i 0 ij , j 0
2 POINTS
We then use the stress-strain relations to compute the strains for an isotopic and homogeneous beam:
2
ij
1 2
ij
2 POINTS
ij kk 2 (2 3 )
At this point we check for compatibility:
11,22 22,11 2 12,12 0
12,13 13,12 23,11 11,23 0
22,33 33,22 2 23,23 0 23,21 21,23 31,22 22,31 0 33,11 11,33 2 31,31 0
2 POINTS
31,32 32,31 12,33 33,12 0
Finally, we compute the beam displacements by integrating the 6 strain-displacement equations for infinitesimal strain: Infinitesimal strain: ij
1 2
u
i, j
u j ,i
2 POINTS
This completes the formulation of this problem. No need to solve the equations.
Problem #2.
This is a 2D problem by assuming that the beam has unit thickness in the z direction. a) First, formulate the problem. Identify all tractions and unit normals on all surfaces. Then,
apply Cauchy’s formula on each surface. Surface #1 for Ti=Tex and n={1,0,0}: T T T x xxnx xyn y xx T Ti ij n j T 0 T n n 0 yx x yy y yx y x
3 POINTS
y
Therefore at x=ℓ:
xx ( x l, y) T yx ( x l, y) 0
3
y
Surface 2 T
T
x
2c
Surface 3
Surface 1
Surface 4 2ℓ
Surface #2 for Ti=0 and n={0,1,0}: T 0 T x xxnx xyny xy 0 Ti ij n j T 0 T n n 0 yx x yy y yy y x
2 POINTS
y
Therefore at y=c
xy ( x, y c) 0 yy ( x, y c) 0 Surface #3 for Ti=-Tex and n={-1,0,0}: T T T x xxnx xyny xx T Ti ij n j T 0 T n n 0 yx x yy y yx y x
y
Therefore at x=-ℓ
xx ( l, y) T yx ( x l , y) 0 Surface #4 for Ti=0 and n={0,-1,0}:
4
2 POINTS
T 0 T x xxnx xyny xy 0 Ti ij n j T 0 T n n 0 yx x yy y yy y x
2 POINTS
y
Therefore at y = -c:
xy ( x, y c) 0 yy ( x, y c) 0 Now that we found the stresses at the beam’s boundaries, we can proceed to make an assumption for the likely stress field. Starting from equilibrium equations:
xx,x xy, y 0 xy,x yy, y 0
2 POINTS
The applied stress is uniform and there is no shear component of stress. So we can safely say: xy 0 xy , x xy , y 0
2 POINTS
Then:
xx,x 0 xx C 1 yy, y 0 yy C 2 Then apply the BCs to determine the constants. At x=ℓ:
xx ( l , y) T C1 T 3 POINTS
At y=c:
yy ( x, y c) 0 C2 0 Finally the stress tensor is:
T 0 ij ( x, y) 0 0
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b) The stresses have been defined in part (a). The strains can be calculated by the following
expression:
ij
1
(1 ) ij ij kk E
where E and ν are constants. Expanding the above expression results in:
xx xx xy
1
T
xx yy E E 1
T
yy xx E E
3 POINTS
1
(1 ) xy 0 E
The displacements are determined by:
ij
1
ui, j u j,i 2
which for i=j=x becomes:
xx
1
T
ux, x u x, x u x, x u x xxdx f ( y) u x x f ( y) 2 E
2 POINTS
Similarly,
T u y y g ( x) E
2 POINTS
Need to determine f(x) and g(y), so use the expression for shear strain:
xy
1
1
ux, y uy ,x 0 f ( y) g (x) f ( y ) , g (x) 2 2
2 POINTS
Integrate to find f(y) and g(x):
f ( y ) 0 f ( y) y 1 g( x) 0 g ( x) x 1 6
2 POINTS
The first terms in the above expressions correspond to Rigid Body Rotations and the second terms to Rigid Body Translations. Finally, the displacement field will have the following form:
u x ( x, y)
T
x y 1 E T u y ( x, y) y x 1 E
2 POINTS
c) The rigid body terms are eliminated by assuming that the center of the beam undergoes no
rotations about the z axis and no translations on x and y directions. Thus:
u x ( x, y)
T
x E T
u y ( x, y)
E
2 POINTS y
d) The choice of boundary conditions (any equivalent BCs obeying St. Venant’s principle)
affects the stress/strain distribution in the bar at a length that is ~5-10 times the characteristic length were the BC is applied, in this case the beam thickness that is equal to 2c. Thus if we assume a 5 times factor:
2l 2c
l
(2) 5 10 c
2 POINTS
Note: an answer where the transient region is 10 times the characteristic length will also be taken as correct, namely:
2l 2c
l
(2) 10 20 c
Such answer is more conservative.
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Problem #3.
w
Surface 4
wl
wl
x
2c
Surface 3
Surface 1
y
Surface 2 2ℓ
a) For x=±ℓ the resultant of the forces can be estimated by using the definition of traction:
F TdS dF F TdS Fi Ti dS Fi ij n j dS T lim S 0 S S S S
i
i
Assume unit thickness and also from the schematic we have Tx = 0, Ty = -wl. Thus: c
S1
S3
c
T x 0 x (1)dS x (1)dS x ( x l, y) dy 0
T y wl xy (1)dS
and Ty wl xy (1)dS ,
S1
S 3
c
Thus :
xy
2 POINTS
2 POINTS
( x l , y)dy wl
c
The moment for beam unit thickness is given by:
c
M T x ydy x ( x l , y) ydy 0 c
8
2 POINTS
b) Apply Cauchy’s formula on surfaces #2 and #4:
Surface 2 for Ti= and n={0,1,0}: T 0 T x xxnx xyny xy 0 Ti ij nj T 0 T n n 0 yx x yy y y y x
2 POINTS
y
Therefore at y=c:
yx ( x, y c) 0 yy ( x, y c) 0 Surface 4 for T = wey and n={0,-1,0}: T 0 T x xxnx xyny xy 0 Ti ij n j T w T n n w yx x yy y y y x
2 POINTS
y
Therefore at y=-c
xy ( x, y c) 0 yy ( x, y c) w Similarly to part a), we solve for surface 1 and surface 3: c
( x l, y)dy 0 x
c c
xy
( l, y)dy wl
c
2 POINTS
c
( x l, y) ydy 0 x
c
c) Impose the boundary conditions on the stress functions to find the unknown constants:
For y=-c
9
2
xy ( x, y c) 0 2 A21 x 6 A23 x c 0 A21 3 A23 c2 0 yy ( x, y c) w 2 A20 2 A21 (c) 2 A23 ( c)3 w A20 A21c A23 c3
w
2 POINTS
2
For y=c 2
xy ( x, y c) 0 2 A21x 6 A23x c 0 A21 3A23c 2 0 yy ( x, y c) 0 2 A20 2 A21(c) 2 A23(c)3 0 A20 A21c A23c3 0
2 POINTS
By solving the system of equations above:
A20
w 4
, A21
3w 8c
, A23
w 8c3
2 POINTS
By substituting the stress field in the expressions for the resultant force and the resultant shear force, the expressions are satisfied trivially. By substituting in the expression for the resultant moments, A03 can be determined:
A03
w l 2
2 8c c2 5
2 POINTS
In order to derive the displacement field, the strain tensor must be defined. The stress tensor as found in part c) is:
3w l 2 2 3w 2 2 3 3w 3w 2 2 y 3 x y y x 3 xy 4c c 5 4c 3 4c 4c ij ( x, y) 3w 3w 2 w 3w w 3 x 3 xy y 3 y 4c 4c 2 4c 4c To check the validity of the expression, the stress tensor must satisfy equilibrium and the boundary conditions.
2 POINTS
10
Then, use the constitutive equation to calculate the strains:
ij
1
(1 ) ij ij kk
where E and ν are constants. Specifically:
1 vw
3w l 2 2 3w 2 3 w 3 w w xx y 2 3 x y 3 3 E 2 4 c c 5 4 c 4 c 2c 4c 3w 3w l 2 2 3w 2 3 w w 1 w yy y 2 3 x y 3 3 E 2 4c 2c 4c 4c c 5 4c (1 ) 3w 2 3w xy xy x E 4c3 4c
3 POINTS
The displacements can be derived from:
ij
1
ui, j u j ,i 2
For i=j=x the above expression gives:
xx
1
ux, x ux,x ux, x ux xxdx f ( y) 2
2 POINTS
3 l 2 2 w 1 3 3 3 1 u x vx y 2 x 3 x x y x 3 3 f ( y) 2 E 2c c 2c 2c c 5 2c Similarly,
w
3 3 l 2 2 3 2 y 4 1 g( x) u y y y x 2 3 3 3 2 E 4c 4c c 5 4c 2 4c 2c 2
2 POINTS
In order to define f(y) and g(x), we substitute in the expression below:
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xy
(1 ) 3w 2 3w ux, y u y, x xy x 2 E 4c3 4c
1
2 POINTS
w 3 l 2
2 1 3 3 w 3 2 2 1 2 x 3 x x 3 y x 3 3 f ( y) 3 xy g ( x) 4 E 2c c 5 2c 8E c 2c c 2c
f ( y) 0 3w l 2 1 x3 g ( x) 4 Ec 2c2 5 2 x 6c2
Integrate the last to find f(y) and g(x):
f ( y) 1 g ( x)
3w l 2
1 x
2 POINTS
x4
2
1 2 2 4 Ec 2c 5 2 2 24c
Finally, the displacement field will be:
w
3 l 2 2 1 3 3 3 1 u x vx y x x x y x 3 3 1 2 3 2 E 2c c 2c 2c c 5 2c w
3 3 l 2 2 3 2 y4 1 u y y y x 2 3 4c3 2c3 2 E 4 c 4 c c 5 4 c 2 2
2 POINTS
3w l 2
1 2 x4 x 2 1 8 Ec 2c2 5 2 12c e) Assuming u(0,y) = v(±ℓ,0)=0 the rigid body terms are:
1 0 1
3w l 2
1
l 4
l 2 5 2 12c 3w 5l 2 1 2 l 8 Ec 12c2 5 2
8 Ec 2c2
12
2
2 POINTS
and the displacement field is reduced to:
w
3 l 2 2 1 3 3 3 1 u x vx y 2 x 3 x x y x 3 3 2 E 2c c 2c 2c c 5 2c 3 3 l 2 2 3 2 y 4 1 w 2 u y y y x 2 3 3 3 2 E 4c 4c c 5 4c 2 4c 2c 3w l 2 1 2 x4 3w 5l 2 1 2 x 2 l 8 Ec 2c2 5 2 12c 8Ec 6c2 5 2
2 POINTS
Now we can draw a qualitative schematic of the deformed beam over a schematic of the undeformed beam using the simplification in part (d). To do so think of the following: How do the four corners of the deformed beam compare to the originally right angles of the undeformed beam? Do you have shear strains at the top and bottom beam corners? (NOTE THIS PART WAS NOT REQUIRED IN YOUR ANSWERS) The deformed shape is shown below. The corner angles remain 90 as no shear strains are ˚
present at the beam corners. The sides will be curved though because the u displacement is not a linear function of y.
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f) Because of partial symmetry in the way the beam is loaded, the maximum vertical
displacement is along the vertical centerline. Then, the derivative of the displacement in the y direction wrt. x at y=0 is:
u y w 2 3 y x 2 POINTS x 2E 2c3 3w l 2 1 4 x3 2 2 x 2 0 x 0 8 Ec 2c 5 2 12c The point of maximum vertical displacement will be the (0,0). The maximum vertical displacement will be given by:
u y
max
3w 5l 2
1 2 l 8 Ec 6c2 5 2
2 POINTS
NOTE: Maximum vertical displacement does not mean that the point (0,0) will be the lowest
point in the deformed beam. It means it is the point that is displaced vertically the most among all other points in the beam. This makes sense as the vertical stress changes from the maximum compressive value at the top of the beam (namely all points are pushed down) to zero at the bottom of the beam, i.e. those points do not “feel” any stress in the y-direction.
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