Tài liệu Bài tập sức bền vật liệu (Giáo trình tiếng Anh) pdf

EXERCISES OF CHAPTER 1 + 2 + 3
Exercise 1:
Two solid cylindrical rods AB and BC
are welded together at B and loaded as
shown (Fig. 1). Knowing that the
everage normal stress must not exceed
150 MPa in either rod, determine the
smallest allowable values of the the
diameters d
1
and d
2
.
Exercise 2:
The uniform beam is supported
by two rods AB and CD that
have cross-sectional areas of 10
mm
2
and 15 mm
2
, respectively
(Fig. 2). Determine the position
d of the distributed load so that
the average normal stress in
each rod is the same
Exercise 3:

Fig.3a Fig.3b
The stress-strain diagram for a polyester resin is given in the figure 3b. If the rigid
beam is supported by a strut AB and post CD, both made from this material, and

subjected to a load of P = 80 kN (Fig.3a), determine the angle of tiltof the beam
when the load is applied. The diameter of the strur is 40 mm and the diameter of the
post is 80 mm.
Exercise 4:
Fig. 1
Fig.2
125kN
125kN
60kN
0.9m
1.2m
Member AC is subjected to a
vertical force of 3 kN. Determine
the position x of this force so that
the compressive stress at C is
equal to the tensile stress in the
tie rod AB (Fig.4a). The rod has a
cross sectional area of 400 mm
2
and the contact area at C is 650
mm
2
.
Solution
Internal loading. The free body diagram for member AC is shown in Fig. (4b).
There are three unknowns, namely, F
AB
, F
C
, and x. The equilibrium of AC will give:

( ) ( )
(1)
+ (2)
y AB
C
A
C
F 0 F F 3000N 0
M 0 - 3000N (x) F 200mm 0
+ ↑ = ⇒ + − =
= ⇒ + =
∑
∑
N
Average Normal Stress. A necessary third equation can be written that requires the
tensile stress in the bar AB and the compressive stress at C to be equivalent: i.e.,
( )
3

CAB
AB
C
2 2
F
F
F 1.625F
400 mm 650 mm
σ = = → =
Substituting (3) into Eq. 1, solving for F
AB

then solving for F
C
, we obtain:

AB
C
F 1143 N; F 1857 N= =
The position of the applied load is determined from Eq. 2.: x = 124 mm
Note that 0 < x < 200 mm, as required.
Exercise 5:
The steel column is used to support the
symmetric loads from the two floors of a
building (Fig.5). Determine the loads P
1
and
P
2
if A moves downward 3 mm and B
moves downward 2 mm when the loads are
applied. The column has a cross-sectional
area of 645 mm
2
. E
st
= 200 Gpa.
Exercise 6:
Fig.4
Fig.5
3,6m
3,6m

Link BC is 6 mm thick, has a width w =
25 mm, and is made of a steel of 480-
MPa ultimate strength in tension (Fig.6).
What was the safety factor used if the
structure shown was designed to support
a 16-kN load P?
Exercise 7:
In the figure 6 of the precedent exercise, suppose that link BC is 6 mm thick
and is made of a steel with a 450-MPa ultimate strength in tension. What should be
its width w if the structure shown is beeing designed to support a 20-kN load P with a
factor of safety of 3?
Exercise 8:
Both portions of the rod ABC are made of an aluminum for
which E = 70 Gpa (Fig.7). Knowing that the magnitude of
P is 4 kN. Determine:
(a) the value of Q so that the deflection at A is zero;
(b) the corresponding deflection of B;
Fig.7
Exercise 9:
A rod consisting of two cylindrical portions AB and BC is
restrained at both ends. Portion AB is made of brass (E
b
= 105 GPa, α
b
= 20.9 x 10
-6
/
0
C) and portion BC is made
of aluminum (E

a
= 72 GPa, α
a
= 23.9 x 10
-6
/
0
C) (Fig.8).
Knowing that the rod is initially unstressed, determine
(a) the normal stresses induced in portions AB and BC by
a temperature rise of 42
0
C; (b) the corresponding
deflection of point B.
Fig.6
Fig.8
Exercise 10:
A axial centric force of magnitude P = 450 kN
is applied to the composite block shown by
means of a rigid end plate (Fig.9). Knowing that
h = 10 mm, determine the normal stress in (a)
the brass core, (b) the aluminum plate
Exercise 11:
The rigid bar is supported by the two short
white spruce wooden posts and a spring. If each
of the posts has an unloaded length of 1 m and a
cross-sectional area of 600 mm
2
, and the spring
has a stiffness of k = 2 MN/m and an

unstretched length of 1.02 m, determine the
vertical displacement of A and B after the load
is applied to the bar
Exercise 12
The rigid bar shown in the figure is fixed to
the top of the three posts made of steel and
aluminum. The posts each have a length of
250 mm when no load is applied to the bar
and the temperature is T
1
= 20
0
C.
Determine the force supported by each post
if the bar is subjected to a uniform
distributed load of 150 kN/m and the
temperature is raised to T
2
= 80
0
C. The
diameter of each post and its material
properties are listed in the figure.
Solution
Fig.9
Fig.10
40mm
40mm
60mm
150kN/m

300mm
300mm
Steel
Steel
E
st
= 200 Gpa
α
st
= 12(10
-6
)/
0
C
Al.
Aluminum
E
Al
= 70 Gpa
α
st
= 23(10
-6
)/
0
C
Fig. 11
Equilibrium:

( )

1) (
3
y st Al
F 0; 2N N - 90 10 0+ ↑ = → + =
∑
Compatibility: Due to load, geometry, and material symmetry, the top of each post
is displaced by an equal amount. Hence:
( )
( )

st Al
2+ ↓ δ = δ
The final position of the top of each post is equal to its displacement caused
by temperature, plus its displacement caused by the internal axial force.
( )
( ) ( )
(3)
st st st
T N
+ ↓ δ = − δ + δ
( )
( ) ( )
(4)
Al Al Al
T N
+ ↓ δ = − δ + δ
Introducing into (2) gives:
( ) ( )
( ) ( )
st st Al Al

T N T N
− δ + δ = − δ + δ
(5)
Using Eqs. (2) – (5) , we get:
( ) ( )
( )
( )
( )
( )
( ) ( )
( )
( )
( )
( )
 
 
 
 
 
 
 
 
st
-6 0 0 0
2
9 2
Al
-6 0 0 0
2
9 2

N 0.250m
- 12 10 / C 80 C - 20 C 0.250m +
p 0.02m 200 10 N/m
N 0.250m
= - 23 10 / C 80 C - 20 C 0.250m +
p 0.03m 70 10 N/m
or
( )
3
st Al
N = 1.270N -165.9 10
(6)
Solving (1) and (6) simultaneously yields:
N
st
= - 14.6 kNN
Al
= 119 kN
The negative value of N
st
indicate that this force acts opposite to that shown in
figure. In other words, the steel posts are in tension and the aluminum post is in
compression.
90 kN
N
st
N
st
N
Al

(δ
st
)
T
(δ
st
)
N
(δ
Al
)
T
(δ
Al
)
N
Final position
Initial position