Homework 3: Indeterminate axial member & Torsion
Due date: Feb 10, 2023
1. An aluminum shell (Ea = 70 GPa, αa = 23.6 x 10-6 /ºC) fully
bonded to a steel core (Es = 200 GPa, αs = 11.7 x 10-6 /ºC)
and is unstressed at a temperature of 20ºC. Considering
only axial deformations, determine the stress in the
aluminum shell when the temperature reaches 180ºC.
2. The cylindrical member as shown in the figure has diameter of 10 mm. The member issubjected
to an axial load P = 2.5 kN and a torque T = 9 N·m. Determine the state of stress in the form
σ x τ xy τ xz 
τ yx σ y τ yz  at the following points:


τ zx τ zy σ z 
(a) point m located on x-axis
(b) point n located at 4 mm from point o at 45 measured from x-axis
(c) point o (centroid of the cross-section)
Hint: Note that since the axial load P and Torsion T act together, at each point of interest, there will be a
normal stress component from load P and shear stress components from torsion T.
3. Determine the required diameter d, for the segment AB of the solid brass shaft shown, if the
permissible shear stress is τall = 50 MPa and the total angle of twist between A and C is not larger than
1.5º. The torques applied at point B and end C are TB = 2 kN.m and TC = 0.5 kN.m, respectively. Shear
modulus for brass is Gbrass = 40 GPa.
d
TB
A
Dia. = 40 mm
TC
2m
B
C
1m
cove
Es:
As:
70x
20
Pa
C205:
m
114.15
x 18
PL
(6ply:
En Ab
Aluminum shell
copla:
for
a
A: (30- 202)
== 1.04 x 100
ml
6 : (6p3y + ( 61) a
↳12 - 23 ( CSTS:
Teen + tea : #
1
K:
1
+
En Aa
EsA S
I
=
<สอบถาม " 3 (14:15 x
K=
2.46 x
↑=103
=(23.8
ห " +ห ยก ง (1.64
-(
a) (
x 20
stress
b-
STS
11.5x156) ( 180 - 20 ) = 3.40
x
10
N
8
in aluminum
Su : :
ม่
x10 Y
1808
2.46 * 10-
มี
วิ
steel
23.40x10" =
1.64
x
153
4
MPa
1
·J
a)
+24
622 = 2.55 10 =
31.
+162
Tax =
25
24
:- 45.84 MPO
Officiences,
&22:31.83 MPa
#CA104
36.87
Mpa
·
·
·
&
·
21.83
45.84
TCOS 45 = 25.93 MPa
Th2 : TTh = - Thin 45 :
=924 x 10 ) =
45.04
·
i
442: Tea:
T:
มี
ประชา
I' )
is fit
*
as a
aC
=- 9
↳Hey sp
poster,
+27
622:
-12 =
2)
28.92
MPG
·
บ
·
·
-2593
2594
:
37.83
·
G
C
TA =
1.5 KN
= 2KN
·
=- 0.5 KN
·
:ao
=
5
↳
50 Mia =
(0.040)
40 20
MPa
1 ท
m
to MPa =
C
x
0
0.083
x
1
0.040( m
&: 30. 12
Gpa
TAG : Tela
a, t,
2 kN
40
X 2m
GPa+*
de =
on
GcTC
2
=0.5
4
0.003
15
m
min
KN
X
20.12 GPax
1 m
C0.040,
4