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Thermal Expansion
Related Topics
Wize University Physics Textbook (Master) > Heat and Temperature
Thermal Stress
3 Activities
How much taller does the Eiffel Tower become at the end of a day when the temperature has increased by 15
o
C. Its original height is 321m and you can assume it is made of steel.
5.2 m
5.2 cm
5.2 mm
I don't know
Check Submission
More Thermal Stress Questions:
Practice: Thermal Stress
An aluminum rod is attached to a steel bar as shown in the figure below so that they expand or stretch together. The length of these bars is
2
m
2\ m
2
m
at
10
°
C
10\degree C
10°
C
and the cross-sectional area of both bars are the same. How much is the stress in each bar at
100
°
C
100\degree C
100°
C
?
α
S
t
=
1.2
×
10
−
5
1
K
Y
S
t
=
20.0
×
10
10
N
m
2
α
A
l
=
2.4
×
10
−
5
1
K
Y
A
l
=
69
×
10
9
N
m
2
\begin{array}{ll} \alpha_{St}=1.2\times10^{-5} \frac{1}{K} & Y_{St}=20.0\times10^{10} \frac{N}{m^2}\\[10pt] \alpha_{Al}=2.4\times10^{-5} \frac{1}{K} & Y_{Al}=69\times10^9 \frac{N}{m^2} \end{array}
α
S
t
=
1.2
×
1
0
−
5
K
1
α
A
l
=
2.4
×
1
0
−
5
K
1
Y
S
t
=
20.0
×
1
0
10
m
2
N
Y
A
l
=
69
×
1
0
9
m
2
N
Practice: Thermal Stress
An aluminum rod is attached to a steel bar as shown in the figure below so that they expand or stretch together. The length of these bars is
2
m
2\ m
2
m
at
10
°
C
10\degree C
10°
C
and the cross-sectional area of both bars are the same. How much is the stress in each bar at
100
°
C
100\degree C
100°
C
?
α
S
t
=
1.2
×
10
−
5
1
K
Y
S
t
=
20.0
×
10
10
N
m
2
α
A
l
=
2.4
×
10
−
5
1
K
Y
A
l
=
69
×
10
9
N
m
2
\begin{array}{ll} \alpha_{St}=1.2\times10^{-5} \frac{1}{K} & Y_{St}=20.0\times10^{10} \frac{N}{m^2}\\[10pt] \alpha_{Al}=2.4\times10^{-5} \frac{1}{K} & Y_{Al}=69\times10^9 \frac{N}{m^2} \end{array}
α
S
t
=
1.2
×
1
0
−
5
K
1
α
A
l
=
2.4
×
1
0
−
5
K
1
Y
S
t
=
20.0
×
1
0
10
m
2
N
Y
A
l
=
69
×
1
0
9
m
2
N
When making cars companies cool aluminum bolts to
−
50
°
C
-50\degree\ C
−
50°
C
so that they perfectly fit into the car when inserted. After being inserted into the car the bolts warm up to room temperature which is
20
°
C
20\degree\ C
20°
C
.
Find the Stress caused from the bolt after it heats up to room temperature.
α
A
l
=
22
⋅
10
−
6
K
−
1
α_{Al}=22\cdot10^{-6}\ K^{-1}
α
A
l
=
22
⋅
1
0
−
6
K
−
1
A
3
m
3\ m
3
m
bronze bar with a cross-sectional area of
2.5
c
m
2
2.5\ cm^2
2.5
c
m
2
is placed between two rigid walls as shown below. At
T
=
−
15
°
C
T=-15\degree C
T
=
−
15°
C
, the gap is
Δ
=
2
m
m
\Delta =2\ mm
Δ
=
2
mm
What temperature will make the stress in the bar to be
40
M
P
a
40\ ΜPa
40
M
P
a
?
(
α
b
r
=
18.0
×
10
−
6
1
K
\alpha_{br}=18.0\times10^{-6} \frac{1}{K}
α
b
r
=
18.0
×
1
0
−
6
K
1
and
Y
b
r
=
80
×
10
9
P
a
Y_{br}=80\times10^9\ Pa
Y
b
r
=
80
×
1
0
9
P
a
)
A copper sphere with a radius of
5
c
m
5\ cm
5
c
m
at
T
=
80
°
C
T=80\degree C
T
=
80°
C
is placed into a
2
k
g
2\ kg
2
k
g
steel vessel which is full of water and ice.
What is the final temperature of the system if the final radius of the copper sphere is
4.996
c
m
4.996\ cm
4.996
c
m
?
If 1/3 of the initial ice-water mixture in the steel vessel is water, how much ice is melted in this process?(gr)
A compound bar made up of
15
c
m
15\ cm
15
c
m
of Cu with
A
c
u
=
25
c
m
2
A_{cu}=25\ cm^2
A
c
u
=
25
c
m
2
and
20
c
m
20\ cm
20
c
m
of steel with
A
S
t
=
10
c
m
2
A_{St}=10\ cm^2
A
S
t
=
10
c
m
2
and
T
=
293
K
T=293\ K
T
=
293
K
is fixed at both ends as shown in the diagram below. How much is the stress in each metal if the system is heated uniformly by a
700
w
a
t
t
700\ watt
700
w
a
tt
heater for 5 minutes?
C
C
u
=
390
J
k
g
⋅
K
C
S
t
=
470
J
k
g
⋅
K
Y
C
u
=
11
×
10
10
N
m
2
Y
S
t
=
20
×
10
10
N
m
2
α
C
u
=
1.7
×
10
−
5
1
K
α
S
t
=
1.2
×
10
−
5
1
K
ρ
C
u
=
9
k
g
/
c
m
3
ρ
S
t
=
7.85
k
g
/
c
m
3
\begin{aligned} C_{Cu}=390\ \frac{J}{kg\cdot K} && C_{St}=470\ \frac{J}{kg\cdot K}\\ Y_{Cu}=11\times10^{10}\ \frac{N}{m^2} && Y_{St}=20\times10^{10}\ \frac{N}{m^2}\\ \alpha_{Cu}=1.7\times10^{-5} \frac{1}{K} && \alpha_{St}=1.2 \times10^{-5} \frac{1}{K}\\ \rho_{Cu}=9\ kg/cm^3 && \rho_{St}=7.85\ kg/cm^3 \end{aligned}
C
C
u
=
390
k
g
⋅
K
J
Y
C
u
=
11
×
1
0
10
m
2
N
α
C
u
=
1.7
×
1
0
−
5
K
1
ρ
C
u
=
9
k
g
/
c
m
3
C
S
t
=
470
k
g
⋅
K
J
Y
S
t
=
20
×
1
0
10
m
2
N
α
S
t
=
1.2
×
1
0
−
5
K
1
ρ
S
t
=
7.85
k
g
/
c
m
3
New Practice Question
At 80 degrees of Celsius, a steel tire 10 mm thick and 80 mm wide that is to be shrunk onto a locomotive driving wheel 2 m in diameter just fits over the wheel which is at temperature of 25°C. How much is the stress on the tire when the combination cools down to 25°C? Neglect the deformation of the wheel.
(
α
S
t
=
11.7
×
10
−
6
1
K
a
n
d
Y
S
t
=
200
×
10
9
P
a
)
\left(\alpha_{St}=11.7\times10^{-6}\ \frac{1}{K}\ and\ Y_{St}=200\times10^9\ Pa\right)
(
α
S
t
=
11.7
×
1
0
−
6
K
1
an
d
Y
S
t
=
200
×
1
0
9
P
a
)
Input your answer in Pa, to 4 sig. figs., do NOT include units.
New Practice Question
An aluminum rod is attached to a steel bar as shown in the figure below so that they expand or stretch together. The length of these bars is 2𝑚 at 10℃ and the cross-sectional area of both bars is the same. How much is the stress in each bar at 100℃?
α
S
t
=
1.2
×
10
−
5
1
K
Y
S
t
=
20.0
×
10
10
N
m
2
\alpha_{St}=1.2\times10^{-5}\ \frac{1}{K}\ \ \ \ \ \ \ \ \ \ Y_{St}=20.0\times10^{10}\ \frac{N}{m^2}
α
S
t
=
1.2
×
1
0
−
5
K
1
Y
S
t
=
20.0
×
1
0
10
m
2
N
α
A
l
=
2.4
×
10
−
5
1
K
Y
A
l
=
69.0
×
10
9
N
m
2
\alpha_{Al}=2.4\times10^{-5}\ \frac{1}{K}\ \ \ \ \ \ \ \ \ \ Y_{Al}=69.0\times10^9\ \frac{N}{m^2}
α
A
l
=
2.4
×
1
0
−
5
K
1
Y
A
l
=
69.0
×
1
0
9
m
2
N
Thermal Stress
An aluminum rod is attached to a steel bar as shown in the figure below so that they expand or stretch together. The length of these bars is
2
m
2\ m
2
m
at
10
°
C
10\degree C
10°
C
and the cross-sectional area of both bars are the same. How much is the stress in each bar at
100
°
C
100\degree C
100°
C
?
α
S
t
=
1.2
×
10
−
5
1
K
Y
S
t
=
20.0
×
10
10
N
m
2
α
A
l
=
2.4
×
10
−
5
1
K
Y
A
l
=
69
×
10
9
N
m
2
\begin{array}{ll} \alpha_{St}=1.2\times10^{-5} \frac{1}{K} & Y_{St}=20.0\times10^{10} \frac{N}{m^2}\\[10pt] \alpha_{Al}=2.4\times10^{-5} \frac{1}{K} & Y_{Al}=69\times10^9 \frac{N}{m^2} \end{array}
α
S
t
=
1.2
×
1
0
−
5
K
1
α
A
l
=
2.4
×
1
0
−
5
K
1
Y
S
t
=
20.0
×
1
0
10
m
2
N
Y
A
l
=
69
×
1
0
9
m
2
N
Thermal Stress
An Aluminium rod of length of 4 meters and a cross sectional area of
10
c
m
2
10\ cm^2
10
c
m
2
weighs 1 kg. It is secured between two stable supports. There is no initial stress in the beam. If the temperature is initially 60 degrees and increases by 40 degrees. What is the stress in the beam. How much water at 20 degrees C must the 100 degree beam be put in to come to 50 degrees C.
α
A
l
=
22
⋅
10
−
6
K
−
1
α_{Al}=22\cdot10^{-6}\ K^{-1}
α
A
l
=
22
⋅
1
0
−
6
K
−
1
Y
A
l
=
65
⋅
10
9
P
a
Y_{Al}=65\cdot10^9\ Pa
Y
A
l
=
65
⋅
1
0
9
P
a
An Aluminium rod of length of 4 meters and a cross sectional area of
10
c
m
2
10\ cm^2
10
c
m
2
. It is secured between two stable supports. There is an initial stress in the beam of
5
⋅
10
7
N
m
2
5\cdot10^7\ \frac{N}{m^2}
5
⋅
1
0
7
m
2
N
. If the temperature increases by 30 degrees. What is the new stress in the beam.
α
A
l
=
22
⋅
10
−
6
K
−
1
α_{Al}=22\cdot10^{-6}\ K^{-1}
α
A
l
=
22
⋅
1
0
−
6
K
−
1
Y
A
l
=
65
⋅
10
9
P
a
Y_{Al}=65\cdot10^9\ Pa
Y
A
l
=
65
⋅
1
0
9
P
a
Thermal Stress
We are planning to close a hole using an aluminum ball but the ball is too small to cover the entire hole. What could we do to make it fit the hole?