PD4 Mechanics 2005 Summer
Question 1
Part a:
A compound bar consists of a brass
tube 60mm outside diameter and
48mm inside diameter and length
100mm surrounding a solid steel
bar of 45mm diameter of the same
length.
Calculate the stresses and strains
induced in both materials when
subjected to an externally applied
load or 150kN.
[14]
2
Esteel = 200GN/m
Ebrass = 100GN/m2
Part b: Determine also the change in length of the assembly.
[3]
Part c: In order to reduce the amount of material used. The diameter of the
steel section is to be reduced. Determine the minimum diameter of steel
section if the maximum stress is limited to 100MN/m2.
[3]
Question 2
numbers below beam represent distances in metres from left hand edge
The beam shown has a length of 8m and is simply supported at positions A
and D. The beam is loaded by concentrated loads of 4Kn, 6kN and 2kN
respectively and carries a distributed load of 2 kN/m, between B and C.





Determine the reactions at the supports.
Determine the shear force at each lettered location in the beam
(and intermediate locations if required) and draw a shear force
diagram.
Determine the bending moment at each lettered location in the
beam (and intermediate locations if required) and draw a bending
moment diagram.
Determine the absolute maximum bending moment
Determine the exact location of the point of contra-flexure
[2]
[6]
[6]
[2]
[4]
PD4 Mechanics 2005 Summer
Question 3
The diagram
opposite shows the
loading conditions
for a 40mm diameter
steel shaft
E = 200 GN/m2.
The shaft is supported in bearings at 0mm and 370mm respectively (and can
be considered as simply supported). It carries loads of 130, 140 and 150N.
Part a:
Using McCauley’s method, derive expressions for the slope and
deflection of the beam under the specified loading conditions.
N.B. Shaft stationary
[14]
Part b:
Calculate the magnitude and direction of the deflection under each of
the applied loads.
[6]
Question 4
Part a
A hollow steel drive shaft, 200mm OD and 100mm ID and 2 metres
long, is used to transmit 694KW while rotating at a speed of 90 RPM.
Determine:
1. the maximum shear stress in the material and
2. the angular twist in degrees.
[G = 80 GN/m2]
[12]
Part b
It is required to replace the drive shaft due to wear and hollow stock is
not available.
If a solid steel shaft is to be used instead to transmit the same power at
the same speed, calculate the diameter, which would be required if the
shear stress is not to exceed that calculated in part a.
[4]
Part c
Determine the percentage change in both the weight and then angle of
twist using the replacement shaft.
[4]
PD4 Mechanics 2005 Summer
Question 5
A material element is subjected to a horizontal tensile
stress of 60 MN/m2 and a vertical tensile stress of
80MN/m2, together with a shear stress of 50MN/m2,
those on the 60MN/m2 planes being counter clockwise
in direction.
Part a: Determine:
1. The normal and shear stresses on the 45 plane shown
2. The principal stresses and their angles to the X axis
3. The maximum shear stress in the material.
[6]
[6]
[4]
Part b
If the applied shear stress xy (currently 50) where to be increased, what
would be the maximum allowable value of xy if the maximum principal
[4]
stress is not to exceed 150MN/m2.
Question 6
Part a: Explain why strain gauges are sensitive to strain along the
direction in which they are placed, yet are not sensitive to strain across
the gauge.
[4]
Part b:
In order to determine the stresses in a pressure vessel a 0-45-90
degree strain gauge rosette is applied randomly to the pressure vessel.
The values registered by the strain gauges are as follows.
Gauge A = 400 x 10-6
Gauge B = 350 x 10-6
Gauge C = 125 x 10-6
Determine graphically or analytically the principal strains in the material.
Part c:
As the object is a pressure vessel, comment on the expected results
from the stress calculation and then calculate the values of the
maximum and minimum principal stresses in the material.
E=200MN/m2 , v = 0.3
Part d:
Calculate the angles of the principal strains and in turn demonstrate the
help of a sketch the orientation of the ‘randomly placed’ strain gauge to
the axis of the pressure vessel.
[8]
[4]
[4]
PD4 Mechanics 2005 Summer
Mechanics Formulas
Bending in Beams
M  E
 
I
y R
Parallel axis theorem. I  Io  Ah 2
Composite shafts
Shafts in series
Shafts in parallel:
I =  d4
64
I = bd3
12
Torsion in shafts
T  G
 
J R
L
Figure 1: Second moment of Area
  a  b
Ta  Tb
T  Ta  Tb
a  b
J= d4
32
Figure 2: Polar Second
moment of Area
Slope and deflection of beams
d2y
General expression M xx  EI 2
dx
Standard cases
Concentrated load
Cantilever
WL3
y max 
3EI
Simply supported
WL3
y max 
48EI
Distributed load
wL4
y max 
8EI
5wL4
y max 
384EI
I

α=Tan -1  xx Tanθ 
I

 yy

Angle of Neutral Axis Oblique loading:
Complex stresses/strains
  1 2  x   y   1 2  x   y Cos2   xy Sin 2
 
1
2

1 ,  2 
x
1
  y Sin 2   xy Cos2
2

x
 y   12

  y   4 xy
2
x
2
   x 
p  Tan 1  p

  xy 
Strains from stresses
1  E1 1  2  2  E1 2  1 
Stresses from strains
E
1  2 
1 
1  2
E
2  1 
2 
1  2




Rectangular (45o) Rosette
 
1
1  2 2  2  3 2
1, 2  1 3 
2
2
      1  2  
1

1, 2  tan 1  2 3
2
1  3

