Department of Nuclear Engineering 22.314 : Structural Mechanics In Nuclear PowerTechnology Quiz No.1 Fall Term 2006 Open Book, 1.5 hours
Please state your assumptions and the definition of symbols appearing in your equations clearly. Note that there is an
22.314/1.56/2.084/13.14 Fall 2006
Problem Set VII
Due 11/02/06
This problem set illustrates applications of beam theory to Zircaloy Follower in a BWR for
calculation of curvature caused by Zircaloy growth ; Consult Notes X on Beam Theory.
ZIRCALOY FOLLOWE
22.314/1.56/2.084/13.14 Fall 2006 Problem Set VII Solution
Solution:
x z y Beam model Geometry and material properties: L=2.4m; W=200mm; T=7mm and E=75 GPa and =0.25 Boundary conditions: At z=0: u=v=w=0 M0=0
(a) According to the beam theory:
Cross section
22.314/1.56/2.084/13.14 Due 11/09/06
Fall 2006
Problem Set VIII
Consider the pre-stressed concrete containment example of Problem Set L.54. Evaluate the impact of changing the longitudinal tendon pitch from 165 mm to 200 mm on the required prestress level
22.314/1.56/2.084/13.14 Fall 2006
Problem Set VIII Solution
Solution:
Evaluate the impact of changing the longitudinal tendon pitch from 165 mm to 200 mm
on the required prestress level to prevent tensile in the concrete upon pressurization.
Material prop
22.314/1.56/2.084/13.14 Fall 2006 Problem Set IX Due 11/21/06
This problem set illustrates some features of seismic analysis. A response spectrum modal (lumped mass) method is to be used.
1) Geometry, Properties, Supports: The reactor component of intere
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Problem Set IX Solution
Solution:
The pipe can be modeled as in Figure 1.
R
Lumped mass
mB
PB
B
mA
PA
A
Pipe
Beam with the same support
Geometry and properties:
L=3m, R=0.105m and t=0.007m
E=200GPa, =8500kg/m3 and wa
22.31411.5612.084113.14Fall 2006
Problem Set V I
Due 10126106
his
p roblem i l l u s t r a t e s a mechanism by which c l a d stresses can be
reduced during up-power ramps by approaching f u l l power a t a slower
rate.
1.
Geometry (20C)
The c l a d outsi
22.314
Problem V Solution Fall 2006
Known
v =0.3
E =195000
A
= 150
oB :=260
sB =0.54-10 - 2
Derived properties
EmA =0
on
rsmB =sB - E
smY :=0.002
ry :
aY
=
smY - smA
mA .(oB - aA) . aAd Linear interpolation between A anb B to get yield stress arY
emB - sm
22.314/1.56/2.084/13.14 Fall 2006 Problem Set V Due 10/19/06
This problem illustrates some "plastic strain" concepts and some calculation methods for treating plasticity.
Uniaxial Stress-Strain Information: Consider a material that has the uniaxial stress
22.314/1.56/2.084/13.14
Fall 2006
Problem Set I
Due 09/19/06
1. A tensile test on 1020 steel gives the following results:
Load (kN) Diameter (mm)
0
12.8
22.2
28.5
50.2
12.2
51.2 (Max.)
10.4
43.6
Length (mm)
50.800
50.848
(yielding begins)
56.1
67.3
69.8 (
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Fall 2006
Problem Set I Solution
1. Original dimension:
2
D0 = 12.8 mm, A0 = D0 /4 = 128.68 mm2 ; and L0 = 50.800 mm.
(a) Stress at a load F = 22.2 kN:
= F /A0 = 172.5 MPa
Strain:
= (L L0 )/L0 = (50.848 - 50.8)/50.8 = 0.009448
Yo
22.314/1.56/2.084/13.14
Fall 2006
Problem Set II
Due 09/26/06
1. Consider a cylindrical vessel of inner radius R and wall thickness t with at ends. The
pressure inside the vessel is Pi and the surrounding pressure Po . What is the relative error in
estima
22.314/1.56/2.084/13.14
Fall 2006
Problem Set II Solution
1. Stress intensity = maxcfw_|r |, | z |, z r |
For thin wall approximation:
Pi + Po
2
Pi R2 Po (R + t)2
z =
(R + t)2 R2
Pi P o
t
=
(R + )
t
2
r =
(1)
(2)
(3)
Therefore:
Sthin = r =
Pi P o
t
Pi
22.314/1.56/2.084/13.14
Fall 2006
Problem Set III
Due 10/05/06
1. For the same vessel described in Problem II-2, nd out the maximum allowable pressure
of the coolant so as not to cause elastic failure under expected static loading and fatigue
conditions.
22.314/1.56/2.084/13.14
Fall 2006
Problem Set III Solution
We will consider the primary stress and secondary stress. Since the primary stresses are dened as
external stresses, the only primary stress for this problem is due to system pressure. The seconda
22.314/1.56/2.084/13.14
Fall 2006
Problem Set IV
Due 10/12/06
1. Consider a cantilever beam with one end xed into a wall. The dimensions are shown in the
gure below. A force F = bhy /2 is applied to the other end. y is the yield strength. Now a
bending mo
22.314/1.56/2.084/13.14
Fall 2006
Problem Set IV Solution
1. The stress due to force F is uniform with a value of y /2 as shown in Figure (a). The
stress due to pure bending in elastic regime is linearly symmetric as shown in Figure (b) In
elastic regime,