MCEN30017 Mechanics and Materials

1. Given the material parameters (E
1
, α
1
, E
2
, α
2
) and thicknesses (h
1
, h
2
) of each of the
metals in the bimetallic strip shown in Figure 1. Consider the change in temperature,
ΔT. Discuss and show (with the help of sketches), how the thermal stress in each layer
is inversely proportional to their thicknesses?





Figure 1
Solution
Refer to the examples on Lecture 4.

2. An elastic panel X fits into the rigid socket Y of height h (Young’s modulus E, Poisson’s
ratio ν > 0). Determine the stress σx and the value of the displacement vR at the top edge
T for a constant pressure P. It is assumed that the elastic panel can move frictionless in
the socket mounting.


Figure 3
Solution:
In the panel a uniform plane stress state is present, where the stress component is
known: = −. Thus, according to Hook’s law
= − = +
= − = − −
As the panel cannot expand in x-direction, it holds = 0
Inserting this into above equations, the normal stress in x and y-directions:
= −
2
1
= −
1 − 2


Now, we can calculate displacement in vertical direction by integrating above equation:


=
() = ∫ = −
1 − 2

+
The lower edge of the panel does not experience a displacement, therefore, (0) =
0, and = 0. The vertical displacement at the top edge
= |(ℎ)| =
1 − 2

ℎ

3. On the thin-walled elastic shaft_1 a pipe_2 will be mounted by heat shrinking. Before
heat shrinking, both parts have identical geometrical dimensions but are made of
different materials. Which temperature difference is required to the mount pipe_2 on
the shaft_1? What is the pressure p between the shaft and the pipe after cooling, if it is
assumed that no stresses are present in axial direction?

Figure 4
Solution:
For the Pipe_2 to be mounted on the shaft_1 its radius has to increase by thermal expansion by
t. Thus in the heated state, the hoop strain has to have value
2 =
2( + ) − 2
2
=



Now Hooke’s law yields with 2 = 0 (the pipe is stress free in the heated state.)
2 = 2∆
∆ =
1
2



The pressure after cooling is obtained from the equilibrium equations

1 = −



2 = +



Hooke’s law,
11 = 1; 22 = 2
The strain
1 =
∆1

; 2 =
∆2


And the kinematic compatibility: ∆2 = ∆1 +
Combining the above equations yields
=
12
1 + 2
(


)
2


4. A homogeneous frustum of a pyramid (Young’s modulus E) with a square cross-section
is loaded on its top surface by a stress σ0. Determine the displacement field u(x) of a
cross-section at position x.

Figure 5
Solution:
The normal force = −0
2 is constant. From the kinematic relation = / and
Hooke’s law =


= / , The differential equation for the displacement u
()


= −0
2
The area A(x) follows
() = [ + ( − )

ℎ
]
2

Thus
[ + ( − )

ℎ
]
2

= −0
2
Solving above equation
= −
0
2


[ + ( − )

ℎ
]
2 ⟹ ∫
()
(0)
= −
0
2

∫

[ + ( − )

ℎ]
2

0

Substitute, = + ( − )

ℎ
, = ( − )

ℎ

() − (0) = −
0
2

ℎ
−
(−
1

)
−
ℎ +

= −
0
2

ℎ
−
(
1

−
1
−
ℎ +
)
The displacement u(0) of the top surface follows from the boundary condition that the
displacement has to vanish on the bottom edge x=h:
(ℎ) = 0 ⟹ (0) =
0
2

ℎ
−
(
1

−
1

) =
0ℎ


From this relation the displacement follows
() =
0
2

ℎ
−
(−
1

+
1
−
ℎ +
)

5. A cantilever beam which is curved in the shape of a quadrant of a circle is loaded as
shown in Figure 3. Consider the radius of curvature of the beam is R. Young’s modulus
E and second moment of inertia of the area about an axis perpendicular to the plane of
the paper through the centroid of the cross-section is I. Find the vertical displacement
of point A on the curved beam.

Figure 7
Solution


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