1.050 Engineering Mechanics
Lecture 23: Example – detailed steps
1
Problem statement
p
x
r
ρg
H
rigid substrate
(no displacement)
Note: p is applied pressure at the top of the soil layer
Goal: Determine
r r
r
z
Isotropic solid
(soil) on a rigid
substrate (infinitely
large in x-ydirections)
K,G given
r
ξ (x ), ε (x ), σ (x )
On the next few slides we will go through steps 1, 2, 3 and 4 to solve this problem.
2
1
Reminder: 4-step procedure to solve
elasticity problems
• Step 1: Write down BCs (stress BCs and displacement BCs),
analyze the problem to be solved (read carefully!)
• Step 2: Write governing equations for stress tensor, strain tensor,
and constitutive equations that link stress and strain, simplify
expressions
• Step 3: Solve governing equations (e.g. by integration), typically
results in expression with unknown integration constants
• Step 4: Apply BCs (determine integration constants)
3
Step 1: Boundary conditions
Write out all BCs in mathematical equations
Displacement BCs:
At z=H: Displacement specified
r
ξ d (z = H ) = (0,0,0)
or
ξ xd = 0, ξ yd = 0, ξ zd = 0
(no displacement at the interface between
the soil layer and the rigid substrate)
Stress BCs:
At z=0: Stress vector provided
r r
r
r
T d (n = −ez , z = 0) = pez
Note: Orientation of
surface and C.S.
4
2
Step 2: Governing equations
Write out all governing equations and simplify
Due to the symmetry of the problem (infinite in x- and y-directions),
the solution will depend on z only, and there are no displacements
in the
r
r
x- and y-directions (anywhere in the solution domain): ξ = ξ z ez
Governing eqn. for strain tensor:
1 ⎛ ∂ξ ∂ξ ⎞
ε ij = ⎜⎜ i + j ⎟⎟
2 ⎝ ∂x j ∂xi ⎠
Calculation of strain tensor simplifies
(symmetry):
Governing eqn. for stress tensor:
(cont’d next slide)
ε zz =
∂ξ z
∂z
(*)
Note : only 1
nonzero coefficient
of strain tensor
r
div σ + ρg = 0
5
Step 2: Governing equations (cont’d)
Gravity only in zdirection
Governing eqn. for stress tensor:
σ xx
∂x
σ xy
∂x
σ xz
∂x
+
+
+
σ xy
∂y
σ yy
∂y
σ yz
∂y
+
+
+
σ xz
∂z
σ yz
∂z
σ zz
∂z
+ ρg x = 0
+ ρg y = 0
+ ρg z = 0
Due to symmetry, only dependence on z-direction
σ xz
∂z
(1)
=0
σ zz
∂z
σ yz
∂z
=0
Final set of governing eqns. for stress tensor
+ ρg = 0
(note: g z = g )
6
3
Step 2: Governing equations (cont’d)
Link between stress and strain
Linear isotropic elasticity (considering that there is only one nonzero
coefficient in the strain tensor, ε zz ):
2 ⎞
⎛
3 ⎠
⎝
2 ⎞
⎛
σ 22 = ⎜ K − G ⎟ε 33
3 ⎠
⎝
4 ⎞
⎛
σ 33 = ⎜ K + G ⎟ε 33
3 ⎠
⎝
σ 11 = ⎜ K − G ⎟ε 33
(2)
7
Step 2: Governing equations (cont’d)
Now combine eqns. (*), (1) and (2):
Substitute (2) in (1):
4 ⎞
∂ε zz ⎛
⎜ K + G ⎟ + ρg = 0
3 ⎠
∂z ⎝
Substitute (*) in (4):
∂ 2ξ z
∂z 2
ρg
∂ 2ξ z
=−
4
∂z 2
K+ G
3
(4)
4 ⎞
⎛
⎜ K + G ⎟ + ρg = 0
3 ⎠
⎝
(5)
σ xz
∂z
=0
σ yz
∂z
Step 2 results in a set of differential eqns.
= 0 (6)
8
4
Step 3: Solve governing eqns. by integration
From (5):
ρg
∂ξ z
=−
z + C1 = ε zz
4
∂z
K+ G
3
(first integration)
⎛
⎞
⎟
4 ⎞⎜
ρg
⎛
z + C1 ⎟
σ zz = ⎜ K + G ⎟⎜ −
3 ⎠⎜ K + 4 G
⎝
⎟⎟
⎜
3
⎝
⎠
ξz = −
1 ρg
z 2 + C1 z + C2
2 K + 4G
3
(knowledge of strain enables
to calculate stress via eq. (2))
(second integration)
From (6):
σ xz
∂z
=0
σ yz
∂z
σ xz = const . = C3 σ yz = const . = C4
=0
9
Solution expressed in terms of integration constants Ci
Step 4: Apply BCs
Stress boundary conditions: Integration provided that
σ xz = const . = C3 σ yz = const . = C4
Stress vector at the boundary of the domain:
r r
r
r !
r
r
r
T d (n = −ez , z = 0) = pez = −σ xz ex − σ yz e y − σ zz ez
BC
Left and right side
must be equal,
therefore:
Stress vector due to stress tensor at z = 0:
r r
r
r
T ( n = −ez , z = 0) = σ ( z = 0) ⋅ ( −ez )
σ xz = C3 = 0, σ yz = C4 = 0
σ zz = − p
Note: Orientation of
surface and C.S.
10
5
Step 4: Apply BCs (cont’d)
Further,
⎛
⎞
⎟
ρg
4 ⎜
σ zz = K + G ⎜ −
z + C1 ⎟
3 ⎜ K + 4G
⎟⎟
⎜
3
⎝
⎠
⎛
⎝
4
3
⎞!
⎠
σ zz (z = 0) = C1 ⎜ K + G ⎟ = − p
(general solution)
(at z=0, see previous slide)
This enables us to determine the constant C1
C1 = −
p
4
K + G
3
11
Step 4: Apply BCs (cont’d)
Displacement boundary conditions:
ξz = −
1 ρg
p
z2 −
z + C2
4
2 K + 4G
K+ G
3
3
(general solution, with C1
included)
Displacement is known at z = H:
ξz ( z = H ) = −
1 ρg
p
!
H2 −
H + C2
= 0
4
2 K + 4G
K+ G
3
3
This enables us to determine the constant C2
C2 =
1
⎛ ρg 2
⎞
H + pH ⎟
⎜
4 ⎝ 2
⎠
K+ G
3
12
6
Final solution (summary): Displacement field, strain field, stress field
1
⎛ ρg 2
(H − z 2 ) − p(z − H ) ⎞⎟
⎜
4 ⎝ 2
⎠
K+ G
3
− ρgz + p
ε zz (z) =
4
K+ G
3
ξ z (z) =
σ zz (z) = − ρgz + p
13
Solution sketch
Displacement profile
−
ρg
H2
4 ⎞
⎛
2⎜ K + G ⎟
3 ⎠
⎝
−
p
H
4
K + G
3
Stress profile:
−p
ξz z = 0
z
σ zz z = 0
z
z=H
z=H
− ( p + ρgH )
14
7