Silva
Kevin Silva
MANE 7100 – HOMEWORK ASSIGNMENT 2
Problem 1
Question:
In two-dimensional elasticity theory, the stress function ∅(𝑥, 𝑦), defined by the relationships
𝜕2∅
𝜕𝑦 2
𝜕2∅
= 2
𝜕𝑥
𝜎𝑥𝑥 =
𝜎𝑦𝑦
𝜎𝑥𝑦 = 𝜎𝑦𝑥 = −
𝜕2∅
𝜕𝑥𝜕𝑦
Substitute the above expressions into the equilibrium equations and obtain the stress function equation.
Take into account also the equations of compatibility.
Solution:
Equilibrium equation
𝜕𝜎𝑖𝑗
+ 𝑋𝑖 = 𝜎𝑖𝑗 , 𝑗 + 𝑋𝑖 = 0
𝜕𝑥𝑗
𝑒𝑞 (𝑃1 − 1)
For this case, eq (P1-1) simplifies to
𝜕𝜎𝑥𝑥 𝜕𝜎𝑦𝑦 𝜕𝜎𝑥𝑧
+
+
+ 𝑋𝑥 = 0
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝜎𝑦𝑥 𝜕𝜎𝑦𝑦 𝜕𝜎𝑦𝑧
+
+
+ 𝑋𝑦 = 0
𝜕𝑥
𝜕𝑦
𝜕𝑧
Equilibrium equations
𝜕𝜎𝑥𝑥 𝜕𝜎𝑦𝑦
+
=0
𝜕𝑥
𝜕𝑦
𝑒𝑞 (𝑃1 − 2)
𝜕𝜎𝑦𝑥 𝜕𝜎𝑦𝑦
+
=0
𝜕𝑥
𝜕𝑦
𝑒𝑞 (𝑃1 − 3)
Then equilibrium equations (eq (P1-2) and (P1-3)) are automatically satisfied
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Constitutive equations (Strains from stress-strain relations, Hooke’s Law)
𝜖𝑥𝑥 =
1
[𝜎 − 𝜈(𝜎𝑦𝑦 + 𝜎𝑧𝑧 )]
𝐸 𝑥𝑥
𝑒𝑞 (𝑃1 − 4)
𝜖𝑦𝑦 =
1
[𝜎 − 𝜈(𝜎𝑥𝑥 + 𝜎𝑧𝑧 )]
𝐸 𝑦𝑦
𝑒𝑞 (𝑃1 − 5)
𝜖𝑥𝑦 =
1+𝜐
𝜎𝑥𝑦
𝐸
𝑒𝑞 (𝑃1 − 6)
For this case, eq (P1-4) and (P1-5) simplify to
𝜖𝑥𝑥 =
1
[𝜎 − 𝜈(𝜎𝑦𝑦 + 𝜎𝑧𝑧 )]
𝐸 𝑥𝑥
𝜖𝑦𝑦 =
1
[𝜎 − 𝜈(𝜎𝑥𝑥 + 𝜎𝑧𝑧 )]
𝐸 𝑦𝑦
Constitutive equations
𝜖𝑥𝑥 =
1
[𝜎 − 𝜈(𝜎𝑦𝑦 )]
𝐸 𝑥𝑥
𝑒𝑞 (𝑃1 − 6)
𝜖𝑦𝑦 =
1
[𝜎 − 𝜈(𝜎𝑥𝑥 )]
𝐸 𝑦𝑦
𝑒𝑞 (𝑃1 − 7)
𝜖𝑥𝑦 =
1+𝜐
𝜎𝑥𝑦
𝐸
𝑒𝑞 (𝑃1 − 8)
Plug stress function relationships into constitutive equations (eq (P1-6), (P1-7) and (P1-8))
𝜖𝑥𝑥
1 𝜕 2𝜙
𝜕 2𝜙
= [ 2 − 𝜈 ( 2 )]
𝐸 𝜕𝑦
𝜕𝑥
𝜖𝑦𝑦 =
1 𝜕 2𝜙
𝜕 2𝜙
[ 2 − 𝜈 ( 2 )]
𝐸 𝜕𝑥
𝜕𝑦
𝜖𝑥𝑦 =
1+𝜐
𝜕 2𝜙
(−
)
𝐸
𝜕𝑥𝜕𝑦
𝑒𝑞 (𝑃1 − 9)
𝑒𝑞 (𝑃1 − 10)
𝑒𝑞 (𝑃1 − 11)
Equation of compatibility
𝜕 2 𝜀𝑥𝑦 𝜕 2 𝜀𝑥𝑥 𝜕 2 𝜀𝑦𝑦
2
=
+
𝜕𝑥𝜕𝑦
𝜕𝑦 2
𝜕𝑥 2
𝑒𝑞 (𝑃1 − 12)
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Plug eq (P1-9), (P1-10) and (P1-11) into eq (P1-12)
2
𝜕2 1 + 𝜐
𝜕 2𝜙
𝜕 2 1 𝜕 2𝜙
𝜕 2𝜙
𝜕 2 1 𝜕 2𝜙
𝜕 2𝜙
[
(−
)] = 2 { [ 2 − 𝜈 ( 2 )]} + 2 { [ 2 − 𝜈 ( 2 )]}
𝜕𝑥𝜕𝑦 𝐸
𝜕𝑥𝜕𝑦
𝜕𝑦 𝐸 𝜕𝑦
𝜕𝑥
𝜕𝑥 𝐸 𝜕𝑥
𝜕𝑦
𝜕2 1 + 𝜐
𝜕 2𝜙
𝜕 2 1 𝜕 2𝜙
𝜕 2𝜙
𝜕 2 1 𝜕 2𝜙
𝜕 2𝜙
2
[
(−
)] = 2 { [ 2 − 𝜈 ( 2 )]} + 2 { [ 2 − 𝜈 ( 2 )]}
𝜕𝑥𝜕𝑦 𝐸
𝜕𝑥𝜕𝑦
𝜕𝑦 𝐸 𝜕𝑦
𝜕𝑥
𝜕𝑥 𝐸 𝜕𝑥
𝜕𝑦
𝜕 4𝜙
𝜕 4𝜙
𝜕 4𝜙
𝜕 4𝜙
𝜕 4𝜙
𝜕 4𝜙
−2 ( 2 2 ) − 2𝜐 ( 2 2 ) =
− 𝜈 ( 2 2) + 4 − 𝜈 ( 2 2)
𝜕𝑥 𝜕𝑦
𝜕𝑥 𝜕𝑦
𝜕𝑦 4
𝜕𝑥 𝜕𝑦
𝜕𝑥
𝜕𝑥 𝜕𝑦
𝜕 4𝜙
𝜕 4𝜙
𝜕 4𝜙
+
2
(
)
+
=0
𝜕𝑥 4
𝜕𝑥 2 𝜕𝑦 2
𝜕𝑦 4
Problem 2
Question:
A simply supported beam has length L = 1, breadth b = 1 and height h = 0.1. The elastic modulus of the
material is E = 1011 and its Poisson's ratio is 𝜐= 0.3. A downward load P = 105 N is applied at the
midpoint along the upper surface of the beam. Use the finite element method to determine approximate
solutions to this problem. Repeat the calculation assuming instead a distributed load Q = 105 N/m.
Compare your results against those obtained from elementary beam theory.
Solution:
See COMSOL and Maple files
Problem 3
Question:
The state of stress at a point inside an isotropic linear elastic body is given by the following stress tensor
components in Cartesian coordinates (all in MPa)
𝜎𝑥𝑥
𝜎 = [𝜎𝑦𝑥
𝜎𝑧𝑥
𝜎𝑥𝑦
𝜎𝑦𝑦
𝜎𝑧𝑦
𝜎𝑥𝑧
2 −3 1
𝜎𝑦𝑧 ] = [−3 4
5]
𝜎𝑧𝑧
1
5 −1
Assume E = 210GPa and v = 0.3 then determine the strain tensor at the point.
Solution:
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𝜖𝑥𝑥 =
1
1
[2𝑥106 𝑃𝑎 − 0.3(4𝑥106 𝑃𝑎 + (−1𝑥106 𝑃𝑎))]
[𝜎𝑥𝑥 − 𝜐(𝜎𝑦𝑦 + 𝜎𝑧𝑧 )] =
𝐸
210𝑥109 𝑃𝑎
𝜖𝑥𝑥 = 5.24𝑥10−6
𝜖𝑦𝑦 =
1
1
[4𝑥106 𝑃𝑎 − 0.3(2𝑥106 𝑃𝑎 + (−1𝑥106 𝑃𝑎))]
[𝜎𝑦𝑦 − 𝜐(𝜎𝑥𝑥 + 𝜎𝑧𝑧 )] =
9
𝐸
210𝑥10 𝑃𝑎
𝜖𝑦𝑦 = 1.76𝑥10−5
𝜖𝑧𝑧 =
1
1
[(−1𝑥106 𝑃𝑎) − 0.3(2𝑥106 𝑃𝑎 + 4𝑥106 𝑃𝑎)]
[𝜎𝑧𝑧 − 𝜐(𝜎𝑥𝑥 + 𝜎𝑦𝑦 )] =
𝐸
210𝑥109 𝑃𝑎
𝜖𝑧𝑧 = −1.33𝑥10−5
𝜖𝑥𝑦 =
1+𝜐
1
1 + 0.3
𝜎𝑥𝑦 =
𝜎𝑥𝑦 =
∗ (−3𝑥106 𝑃𝑎)
𝐸
2𝐺
210𝑥109 𝑃𝑎
𝜖𝑥𝑦 = −1.86𝑥10−5
𝜖𝑦𝑧 =
1+𝜐
1
1 + 0.3
𝜎𝑦𝑧 =
𝜎𝑦𝑧 =
∗ 5𝑥106 𝑃𝑎
𝐸
2𝐺
210𝑥109 𝑃𝑎
𝜖𝑦𝑧 = 3.10𝑥10−5
𝜖𝑧𝑥 =
1+𝜐
1
1 + 0.3
𝜎𝑧𝑥 =
𝜎𝑧𝑥 =
∗ 1𝑥106 𝑃𝑎
𝐸
2𝐺
210𝑥109 𝑃𝑎
𝜖𝑧𝑥 = 6.19𝑥10−6
𝜖𝑥𝑥
𝜖
𝜖 = [ 𝑦𝑥
𝜖𝑧𝑥
𝜖𝑥𝑦
𝜖𝑦𝑦
𝜖𝑧𝑦
𝜖𝑥𝑧
5.24𝑥10−6
𝜖𝑦𝑧 ] = [−1.86𝑥10−5
𝜖𝑧𝑧
6.19𝑥10−6
−1.86𝑥10−5
1.76𝑥10−5
3.10𝑥10−5
6.19𝑥10−6
3.10𝑥10−5 ]
−1.33𝑥10−5
Problem 4
Question:
Consider a long, thick-walled hollow metal cylinder (inner radius a, outer radius b). The cylinder is
subjected to inner pressure p and outer pressure q. A force balance on a small differential element of
area 𝑑𝑟 𝑥 𝑑𝜙 yield the equilibrium equation
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𝜎𝑟 + 𝑟
𝑑𝜎𝑟
− 𝜎𝜙 = 0
𝑑𝑟
Where 𝜎𝑟 and 𝜎𝜙 , are the radial and azimuthal (hoop) stresses. The radial and azimuthal (hoop) strains
in this case are given by
𝜖𝑟 =
𝑑𝑢
𝑑𝑟
𝜖𝜙 =
𝑢
𝑟
For linear elastic behavior, these strains are related to the stresses through Hooke’s law, i.e.
𝜖𝑟 =
1
(𝜎 − 𝜐𝜎𝜙 )
𝐸 𝑟
𝜖𝜙 =
1
(𝜎 − 𝜐𝜎𝑟 )
𝐸 𝜙
a. Combine all the above expressions to obtain a single differential equation for 𝜎𝑟
b. Solve the differential equation and obtain expressions for the radial (𝜎𝑟 ) and azimuthal/hoop
(𝜎𝜙 ) stresses as functions of radius r through the cylinder wall thickness.
c. Assume linear elastic behavior according to Hooke's law. Let the radial displacement of any
point on the cylinder wall be given by u(r) and obtain expressions for as a function of radius r
through the cylinder wall thickness.
d. If the ends of the cylinder are assumed to be stress free (i.e. plane stress conditions; 𝜎𝑧 = 0),
obtain an expression for the axial strain 𝜖𝑧 .
e. Obtain an expression for the radial displacement u.
f. Let a =1 m, b = 1.05 m, E = 1011 Pa, 𝜈 = 0.3, p = 105 Pa, and q = 106 Pa. Calculate and plot the
stresses, strains and displacement in this case.
g. Solve the problem in (f) above, using finite element method software (COMSOL and/or ANSYS).
Solution:
Part a. Combine all the above expressions to obtain a single differential equation for 𝜎𝑟
Beginning with eq (P4-3)
𝜖𝜙 =
𝑢
𝑟
𝑒𝑞 (𝑃4 − 3)
Take the derivative of eq (P4-3)
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𝜖𝜙 =
𝑢
𝑟
𝜖𝜙 𝑟 = 𝑢
𝑑𝜖𝜙 𝑟 + 𝜖𝜙 𝑑𝑟 = 𝑑𝑢
𝑒𝑞 (𝑃4 − 4)
Divide eq (P4-4) by dr
𝑑𝜖𝜙 𝑟
𝑑𝑢
+ 𝜖𝜙 =
𝑑𝑟
𝑑𝑟
𝑑𝜖𝜙 𝑟
𝑑𝑢
+ 𝜖𝜙 −
=0
𝑑𝑟
𝑑𝑟
𝑒𝑞 (𝑃4 − 5)
Substitute eq (P4-2) into eq (P4-5)
𝜖𝑟 =
𝑑𝑢
𝑑𝑟
𝑒𝑞 (𝑃4 − 2)
𝑑𝜖𝜙 𝑟
𝑑𝑢
+ 𝜖𝜙 −
=0
𝑑𝑟
𝑑𝑟
𝑒𝑞 (𝑃4 − 5)
𝑑𝜖𝜙 𝑟
𝑑𝑢
+ 𝜖𝜙 −
=0
𝑑𝑟
𝑑𝑟
𝑑𝜖𝜙 𝑟
+ 𝜖𝜙 − 𝜖𝑟 = 0
𝑑𝑟
𝑒𝑞 (𝑃4 − 6)
Substitute eq (P4-4) and eq (P4-5) into eq (P4-6)
𝜖𝑟 =
1
(𝜎 − 𝜐𝜎𝜙 )
𝐸 𝑟
𝑒𝑞 (𝑃4 − 4)
𝜖𝜙 =
1
(𝜎 − 𝜐𝜎𝑟 )
𝐸 𝜙
𝑒𝑞 (𝑃4 − 5)
𝑑𝜖𝜙 𝑟
+ 𝜖𝜙 − 𝜖𝑟 = 0
𝑑𝑟
𝑒𝑞 (𝑃4 − 6)
𝑟
𝑑 1
1
1
(𝜎𝜙 − 𝜐𝜎𝑟 ) + (𝜎𝜙 − 𝜐𝜎𝑟 ) − (𝜎𝑟 − 𝜐𝜎𝜙 ) = 0
𝑑𝑟 𝐸
𝐸
𝐸
𝑟
𝑑
(𝜎 − 𝜐𝜎𝑟 ) + (𝜎𝜙 − 𝜐𝜎𝑟 ) − (𝜎𝑟 − 𝜐𝜎𝜙 ) = 0
𝑑𝑟 𝜙
𝑒𝑞 (𝑃4 − 7)
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Solve eq (P4-1) for 𝜎𝜙
𝜎𝑟 + 𝑟
𝑑𝜎𝑟
− 𝜎𝜙 = 0
𝑑𝑟
𝜎𝜙 = 𝜎𝑟 + 𝑟
𝑑𝜎𝑟
𝑑𝑟
𝑒𝑞 (𝑃4 − 1)
𝑒𝑞 (𝑃4 − 8)
Plug eq (P4-8) into eq (P4-7)
𝜎𝜙 = 𝜎𝑟 + 𝑟
𝑑𝜎𝑟
𝑑𝑟
𝑒𝑞 (𝑃4 − 8)
𝑟
𝑑
(𝜎 − 𝜐𝜎𝑟 ) + (𝜎𝜙 − 𝜐𝜎𝑟 ) − (𝜎𝑟 − 𝜐𝜎𝜙 ) = 0
𝑑𝑟 𝜙
𝑟
𝑑
𝑑𝜎𝑟
𝑑𝜎𝑟
𝑑𝜎𝑟
((𝜎𝑟 + 𝑟
) − 𝜐𝜎𝑟 ) + ((𝜎𝑟 + 𝑟
) − 𝜐𝜎𝑟 ) − (𝜎𝑟 − 𝜐 (𝜎𝑟 + 𝑟
)) = 0
𝑑𝑟
𝑑𝑟
𝑑𝑟
𝑑𝑟
𝑟
𝑑𝜎𝑟
𝑑𝜎𝑟
𝑑2 𝜎𝑟
𝑑𝜎𝑟
𝑑𝜎𝑟
𝑑𝜎𝑟
+𝑟
+ 𝑟2
− 𝜐𝑟
+ 𝜎𝑟 + 𝑟
− 𝜐𝜎𝑟 + −𝜎𝑟 + 𝜐𝜎𝑟 + 𝜐𝑟
=0
2
𝑑𝑟
𝑑𝑟
𝑑𝑟
𝑑𝑟
𝑑𝑟
𝑑𝑟
3𝑟
𝑒𝑞 (𝑃4 − 7)
𝑑𝜎𝑟
𝑑2 𝜎𝑟
+ 𝑟2
= 0 𝑒𝑞 (𝑃5 − 7)
𝑑𝑟
𝑑𝑟 2
Refer to Maple file for parts a thru f. Refer to COMSOL file for part g.
Problem 5
Question:
Consider again a long, thick-walled hollow metal cylinder (inner radius a, outer radius b). The cylinder
is not pressurized but subjected instead to a steady state temperature gradient such that T(a) = Ta and
T(b) = Tb. For linear thermo-elastic behavior, the stress-strain relationships are given by,
𝜎𝑟 =
𝐸
(𝜖 + 𝜈𝜀𝜙 − (1 + 𝜐)𝛼𝑇)
1 − 𝜐2 𝑟
𝜎𝜙 =
𝐸
(𝜖 + 𝜈𝜖𝑟 − (1 + 𝜐)𝛼𝑇)
1 − 𝜐2 𝜙
Where 𝜖𝑟 = 𝑑𝑢/𝑑𝑟, 𝜖𝜙 = 𝑢/𝑟; 𝛼 is the thermal expansion coefficient and T(r) is the temperature.
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a. Combine all the above expressions to obtain a single differential equation for u(r).
b. Solve the differential equation, then substitute into the combines strain-displacement relationship
and into Hooke’s law to incorporate the boundary conditions 𝜎𝑟 (𝑎) = 𝜎𝑟 (𝑏) = 0 and obtain
expressions for the radial (𝜎𝑟 ) and azimuthal/hoop (𝜎𝜙 ) stresses and for u(r) as functions of
radius r through the cylinder wall thickness r and temperature T.
c. If the ends of the cylinder are assumed to be strain free (i.e. plane strain conditions; 𝜖𝑧 = 0),
obtain an expression or the axial stress 𝜎𝑧 .
d. The steady state temperature distribution across the cylinder is given by
ln(𝑏/𝑟)
ln(b/a)
Obtain expressions for 𝜎𝑟 , 𝜎𝜙 , 𝜎𝑧 , and u(r) in this case.
e. Let a = 1 m, b = 1.05 m, E = 1011Pa, 𝜈 = 0.3, 𝛼 = 10-5 (1/Celsius), Ta = 190 Celcius, Tb = 80
Celsius. Calculate and plot the stresses, strains and displacement in this case.
f. Solve the problem in (e) above, using finite element software (ANSYS)
𝑇(𝑟) = 𝑇𝑏 − (𝑇𝑏 − 𝑇𝑎 )
Solution:
Part a. Combine all the above expressions to obtain a single differential equation for u(r).
Since plane stress conditions are assumed, mechanical equilibrium requires that
𝑑𝜎𝑟 𝜎𝑟 − 𝜎𝜙
+
=0
𝑑𝑟
𝑟
𝑒𝑞 (𝑃5 − 1)
And you know that
𝜎𝑟 =
𝐸
(𝜖 + 𝜈𝜀𝜙 − (1 + 𝜐)𝛼𝑇)
1 − 𝜐2 𝑟
𝜎𝜙 =
𝐸
(𝜖 + 𝜈𝜖𝑟 − (1 + 𝜐)𝛼𝑇) 𝑒𝑞 (𝑃5 − 3)
1 − 𝜐2 𝜙
𝑒𝑞 (𝑃5 − 2)
Where
𝜖𝑟 =
𝑑𝑢
𝑑𝑟
𝜖𝜙 =
𝑢
𝑟
Plug eq (P5-2) and eq (P5-3) into eq (P5-1)
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𝑑
𝐸
𝑑𝑢
𝑢
( + 𝜈 − (1 + 𝜐)𝛼𝑇)]
[
2
𝑑𝑟 1 − 𝜐 𝑑𝑟
𝑟
𝐸
𝑑𝑢
𝑢
𝐸
𝑢
𝑑𝑢
[
2 ( 𝑑𝑟 + 𝜈 𝑟 − (1 + 𝜐)𝛼𝑇)] − [1 − 𝜐 2 ( 𝑟 + 𝜈 𝑑𝑟 − (1 + 𝜐)𝛼𝑇)]
1
−
𝜐
+
𝑟
= 0 𝑒𝑞 (𝑃5 − 4)
Simplify eq (P5-4)
𝑑𝑢
𝑢
𝑢
𝑑𝑢
− (1 + 𝜐)𝛼𝑇)]
[( + 𝜈 𝑟 − (1 + 𝜐)𝛼𝑇)] − [( 𝑟 + 𝜈
𝑑 𝑑𝑢
𝑢
𝑑𝑟
𝑑𝑟
=0
[( + 𝜈 − (1 + 𝜐)𝛼𝑇)] +
𝑑𝑟 𝑑𝑟
𝑟
𝑟
𝑑 𝑑𝑢
𝑢
1 𝑑𝑢
𝑢
𝑢
𝑑𝑢
+ (1 + 𝜐)𝛼𝑇] = 0
[( + 𝜈 − (1 + 𝜐)𝛼𝑇)] + [ + 𝜈 − (1 + 𝜐)𝛼𝑇 − − 𝜈
𝑑𝑟 𝑑𝑟
𝑟
𝑟 𝑑𝑟
𝑟
𝑟
𝑑𝑟
𝑑 𝑑𝑢
𝑢
1 𝑑𝑢
𝑢 𝑢
𝑑𝑢
[( + 𝜈 − (1 + 𝜐)𝛼𝑇)] + [ + 𝜈 − − 𝜈 ] = 0
𝑑𝑟 𝑑𝑟
𝑟
𝑟 𝑑𝑟
𝑟 𝑟
𝑑𝑟
𝑑 𝑑𝑢
𝑢
1 𝑑𝑢 𝜈𝑢 𝑢 𝜈 𝑑𝑢
+
− −
=0
[( + 𝜈 − (1 + 𝜐)𝛼𝑇)] +
𝑑𝑟 𝑑𝑟
𝑟
𝑟 𝑑𝑟 𝑟 2 𝑟 2 𝑟 𝑑𝑟
𝑑2 𝑢 𝜈 𝑑𝑢 𝜈𝑢
𝑑𝑇 1 𝑑𝑢 𝜈𝑢 𝑢 𝜈 𝑑𝑢
+
− 2 − (1 + 𝜐)𝛼
+
+
− −
=0
2
𝑑𝑟
𝑟 𝑑𝑟 𝑟
𝑑𝑟 𝑟 𝑑𝑟 𝑟 2 𝑟 2 𝑟 𝑑𝑟
𝑑2 𝑢 1 𝑑𝑢 𝑢
𝑑𝑇
+
− 2 = (1 + 𝜐)𝛼
2
𝑑𝑟
𝑟 𝑑𝑟 𝑟
𝑑𝑟
See Maple files for parts b thru e.
Problem 6
Question:
A Maxwell model (spring -modulus E - and dashpot - viscosity 𝜂 in series) is used to represent the stress
relaxation behavior of a certain polymer. The stress relaxation modulus can be approximately described
by
𝑡
𝐺(𝑡) = 𝐺𝑜 exp (− )
𝜏
Where G is in GPa, t is time in seconds, and 𝜏 = 𝜂/𝐸. Experimental measurements of G yielded the
values G(0) = 2GPa and G(104) = 1 GPa.
a. Compute the creep compliance.
b. Evaluate the strain at a time t = 103seconds following the sudden application of a stress 𝜎 =
102 𝑀𝑃𝑎
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Silva
Solution:
See Maple file for a and b.
Page 10 of 10