ME3 Mechanics 2012 Summer
INSTITIÚID TEICNEOLAÍOCHTA, SLIGEACH
INSTITUTE OF TECHNOLOGY, SLIGO
School of Engineering
Head of School of Engineering
Exam Series: Summer
Mr Shane Fanning
Academic year: 2011/2012
Course:
B Eng (Ord) in Mechanical Engineering
Stage:
3
Subject:
Mechanics
Internal
Examiner(s):
Sean Dalton
External
Examiner(s):
Dr. Charles McCartan
Mr. Michael Conlon
Instructions to Candidates
Time Allowed:
3 hours
Number of Questions on Paper:
6 (Six)
Number of Questions to be attempted:
5 (Five)
Compulsory Questions:
None
Any Other Special Instructions:
E = 200GN / m 2 ; υ = 0.3
Unless otherwise stated.
ME3 Mechanics 2012 Summer
Question 1
Part a:
Shown in Figure 1 is the cross section of a beam.
Determine:
1. The distance of the horizontal
neutral axis from the bottom edge.
2. The value of the second moment
of area about this axis
3. The stresses induced in the top
and bottom surfaces, if subjected
to a bending moment of 2500Nm.
[14]
Part b:
If a beam of identical cross section is simply supported over a span of 12m
calculate the maximum bending moment induced and the corresponding
maximum bending stress. The beam is made of steel ( ρ = 7800kg / m3 ).
[6]
Question 2
Part a
The shaft shown in Fig 2a has a diameter
of 75mm. Calculate the maximum shear
stress induced and the angle of twist per
unit length when subjected to a torque of
4520Nm. [ G = 80GN / m 2 ]
[8]
Part b
Determine the maximum internal diameter
allowable if the solid shaft is replaced by
the larger, hollow shaft shown in Fig 2b, if
the maximum shear stress is not to exceed
that in part a. (torque remains the same).
Determine the percentage saving in
material which results from using the
larger hollow shaft.
Determine also the corresponding percentage reduction in the angle of twist.
[12]
ME3 Mechanics 2012 Summer
Question 3
Part a
The element shown is part of a structural
member and is subjected to the following
stresses.
σ x = 40MN / m 2 tensile
σ y = 70MN / m 2 compressive
τxy = 30MN / m 2
Determine:
1. the principal stresses and their angles
2. the maximum shear stress in the plane of the element
3. the absolute maximum shear stress
4. the direct and shear stresses on the inclined plane shown
[16]
Part b
If the maximum tensile principal stress is not permitted to exceed 70 MN/m2
determine the largest value of applied shear stress (τxy) , which may be
applied if σx and σy remain unchanged.
[4]
Question 4
Part a:
Strain gauges mounted on the surface of a
pressure vessel register the following
strain values.
ε1 = 255x10−6
ε 2 = 60x10−6
Determine the values of the stresses
induced. ( E = 200GN / m 2 , ν = 0.3 ).
Determine the value of the internal pressure if the pressure vessel wall
thickness if 10mm.
[10]
Part b:
A round bar mounted as a cantilever, has a length of 500mm and a diameter
of 25mm and deflects 30mm when subjected to a vertical load of 1kN at its
free end.
Determine the value of the materials modulus of elasticity [E]
[6]
What will be the angle of twist if the same sample is subjected to a torque of
500Nm.
[ ν = 0.3 ]
[4]
ME3 Mechanics 2012 Summer
Question 5
Part a:
A simply supported by beam has a
length of 4.8m a cross section 100mm
wide by 250mm deep and carries a
concentrated load of 160kN at its mid
point.
Calculate the deflection resulting from
the application of this concentrated load.
( E = 200GN / m 2 ).
[6]
Part b:
If this load is subdivided into two loads of
60kN and 100kN, what will now be the
deflection at the points D and E?
[14]
Question 6
Part a
The 38mm diameter shaft AB shown carries a
tensile force of 200kN and torque of 500Nm
Calculate the factor of safety against failure
according to the:
• Maximum shear stress criteria (tresca)
• Maximum shear strain energy (von-mises)
( σ y = 250MN / m 2 )
[14]
Part b
Calculate the maximum value of T which may be applied before failure occurs
according to the Tresca failure criteria.
[6]
ME3 Mechanics 2012 Summer
Mechanics Formulas
General bending/torsion
General bending equation
bd 3
πd 4
M σ E
= =
I sq =
I rnd =
I
y R
12
64
2
Parallel axis theorem. I = I o + Ah
General torsion equation
T τ Gθ
= =
J r
L
Shafts in series
Composite shafts
Shafts in parallel:
Complex stresses/strains
σθ = 1 2 (σ x + σ y ) + 1 2 (σ x − σ y )Cos 2θ + τxySin 2θ
τθ = 1 2 (σ x − σ y )Sin 2θ − τ xyCos 2θ
σ1 , σ 2 =
1
2
(σ
x
+ σy ) ± 12
Strains from stresses
ε1 = E1 (σ1 − νσ 2 )
ε 2 = E1 (σ 2 − νσ1 )
(σ
− σ y ) + 4τ xy
2
x
2
σ2 =
)
E
1−ν 2
(
)(
T = Ta + Tb
Ta = Tb
θa = θ b
 σ − σx 
θp = Tan −1  p

 τxy 
Stresses from strains
σ 1 = 1−Eν 2 ( ε1 +νε 2 )
(
πd 4
32
θ = θa + θ b
J rnd =
ε 2 + νε1 )
Elastic constants
E = 3K (1 − 2υ)
E = 2G (1 + υ )
Slope and deflection of beams
Standard cases
Concentrated load
Distributed load
Cantilever
WL3
wL4
y max =
y max =
8EI
3EI
3
Simply supported
WL
5wL4
y max =
y max =
48EI
384EI
Pba 2 2
Deflection off centre load
δ=
( L − b − a2 )
6 EIL
Failure Theories
Max shear stress (tresca): σ 1 − σ 3 = σ y
Max shear strain energy (von mises):
2
2
2
σ1 + σ 2 − σ1σ 2 = σ y