DE5302 Strength of Materials 1 Exam Question Book 2017
DE5302 Strength of Materials 1 Exam Question Book 2017
DE5302 Strength of Materials 1 Exam Question Book 2017
Please Note:
1. This examination will be held in accordance with institutional examination
procedures and in conjunction with the rules and regulations imposed by The
Board.
2. Any candidate who aids, attempts to aid, obtains aid or attempts to obtain aid
from another candidate will be disqualified and further dealt with under the
institutional disciplinary procedures for cheating. This could result in the
offender’s expulsion from the institution.
3. Candidates are permitted to use non-programmable calculators. This explicitly
excludes lap-top computers, cell phones, i-phones, i-pads or any other device
capable of connecting by wireless to the Internet or other electronic media, or
source of potential aid. Violations of this rule will also be dealt with as acts of
cheating.
4. This is a CLOSED BOOK examination. You may not be in possession of books,
notes memoranda or reference materials other than those supplied in the
examination.
5. You may make use of the information in the Data, Formulae and Tables booklet
handed out with this paper.
Instructions:
1. All answers must be written in the answer book supplied, and on extra sheets
(including graph paper) as you may request during the examination. Enter your
Student ID in the space provided on the front cover.
2. Write your student ID and the question number on each loose page, including
graph paper, and enclose such extra sheets in the book.
3. CROSS OUT any work that you do not wish to have marked.
4. Show ALL working in calculations to obtain full marks. In any question that does
not ask for numerical answers to a specified accuracy, answers should be
rounded appropriately.
5. Answer EIGHT (8) questions from the available ten (10). Each question is worth
10 marks.
A component made from mild steel is under load. The element below shows the stresses
that exist at a point on the surface of the component. There is a weld across this section,
which is oriented at an angle of 55o with respect to the x axis.
(a) Determine the principal stresses and the principal directions at this point. (4 marks)
(b) Sketch the principal stress element, indicating clearly the principal stresses and their
directions. (2 marks)
(c) Determine the maximum shear stress and the associated direct stress at the point.
(2 marks)
(d) Determine the shear stress parallel to the weld. (2 marks)
A steel cable of diameter 12 mm wraps around a drum and is used to raise and lower a
cage in a mine shaft. The mass of the cage is 450 kg. The cage is 50 m below the drum
and is descending at a constant rate of 0.75 ms-1 when the drum suddenly stops. The
Young’s modulus of steel is 200 GPa and the yield strength is 350 MPa.
(a) Identify the bolts with the maximum load acting on them. (1 mark)
(b) Determine the maximum load acting on the bolts. (4 marks)
(c) Calculate the size of bolts required if the allowable shear stress in a bolt is 190 MPa.
(2 marks)
(d) List and illustrate three modes of failure in bearing type joints. (3 marks)
A bar of length 250 mm and width 100 mm has a load of 15 kN applied as shown below. A
fillet weld group as shown connects the bar to a plate.
Calculate the size of the fillet weld required if the allowable shear stress in the weld is 135
MPa.
An I-beam is nailed together with a spacing of s as shown below. The cross-section of the
beam is also indicated. All dimensions are in mm. The nails can each support a shear
force of 4.5 kN. There is a point load of 15 kN acting at its mid-point.
The cantilever beam carries a uniform load of 1 kN/m and a point load of 5 kN as shown.
The maximum deflection of the beam is not to exceed 1/360 of the beam length. The
allowable bending stress, σallowable, is 175 MPa. Assume E=200 GPa for the beam.
Select the lightest universal beam from the table in the appendices that will satisfy all the
above requirements. Ignore the weight of the beam.
A steel bar is 700 mm long is mounted in a forging machine. It has a rectangular cross-
section measuring 30 mm x 10 mm. It is fixed at either end. The length of bar between the
fixtures is 500 mm. Esteel = 200 GPa and σyield = 220 MPa.
Determine the maximum load P the bar can carry without buckling. Use a factor of safety
with respect to buckling of 1.5.
A steel shaft is supported by bearings and rotating at a constant velocity. The shaft is 0.25
m long. It has a diameter of 40 mm. At its midpoint, there is a notch of radius 2.5 mm. The
diameter of the shaft at the notch is 22 mm. There is a tensile axial load of 10 kN acting on
the shaft. A vertical downward load of 5 kN acts at the location of the notch.
σyield = 350 MPa, σuts = 600 MPa, σe = 300 MPa.
(a) What is the stress concentration factor for the shaft with the 10 mm radius notch under
tension (Kt,tension)? (1 mark)
(b) What is the stress concentration factor for the shaft with the 10 mm radius notch under
bending (Kt,bending)? (1 mark)
(c) What is the fatigue strength reduction factor (Kf) for this shaft? (2 marks)
(d) What is the reduced endurance strength for this shaft? (2 marks)
(e) Determine the mean and alternating direct stress at the change in cross-section.
(2 marks)
(f) Using Soderberg theory, calculate the factor of safety for this load case. (2 marks)
Determine the largest load P that can be applied to the assembly. Ignore the mass of the
rigid bar and the steel rod.
A 60o strain rosette is mounted on the surface of a machine component under load as
shown below. Young’s modulus E=200 GPa; Poisson’s ratio ν=0.3; σyield=300 MPa;
σuts=500 MPa.
(a) Determine the principle strains and their direction relative to the x-axis. (6 marks)
(b) Determine the principle stresses at the point. (2 marks)
(c) Using an appropriate failure theory, determine the factor of safety at this point.
Assume the stress normal to the surface is zero. (2 marks)
Formulary
Stress and strain: 2𝜏𝜏𝑥𝑥𝑥𝑥
Or 𝑡𝑡𝑡𝑡𝑡𝑡 2𝜃𝜃 = 𝜎𝜎𝑥𝑥 − 𝜎𝜎𝑦𝑦
𝐹𝐹 𝜎𝜎 𝑥𝑥
𝜎𝜎 = ; 𝐸𝐸 = ; 𝜀𝜀 = 1
𝐴𝐴 𝜀𝜀 𝐿𝐿
𝜀𝜀𝑦𝑦 = �𝜎𝜎 − 𝜈𝜈𝜎𝜎𝑥𝑥 �
F τ 𝐸𝐸 𝑦𝑦
= = Gθ
A 1 E
σ1 = (ε 1 + νε 2 )
(1 − ν 2 )
M σ E
= =
I y R 𝐸𝐸
𝜎𝜎2 = (𝜀𝜀 + 𝜈𝜈𝜀𝜀1 )
(1 − 𝜈𝜈 2 ) 2
T τ θ
= =G
J r L 2
𝜀𝜀𝑥𝑥 + 𝜀𝜀𝑦𝑦 ��𝜀𝜀𝑥𝑥 − 𝜀𝜀𝑦𝑦 � + 𝛾𝛾𝑥𝑥𝑥𝑥
2
𝑭𝑭𝑭𝑭 �
𝑭𝑭𝑭𝑭𝒚𝒚 𝜀𝜀1,2 = ±
𝝉𝝉 = or 𝝉𝝉 = 2 2
𝑰𝑰𝑰𝑰 𝑰𝑰𝑰𝑰
2
𝑭𝑭𝑭𝑭 ��𝜀𝜀𝑥𝑥 − 𝜀𝜀𝑦𝑦 � + 𝛾𝛾𝑥𝑥𝑥𝑥
2
𝒒𝒒 = ; 𝑭𝑭𝒔𝒔 = 𝒒𝒒 × 𝑩𝑩𝑩𝑩𝑩𝑩𝑩𝑩 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 𝛾𝛾
𝑰𝑰 =±
2 2
d 2y Mx 𝛾𝛾𝑥𝑥𝑥𝑥
=
dx 2
EI tan 2𝜃𝜃 =
𝜀𝜀𝑥𝑥 − 𝜀𝜀𝑦𝑦
E = 2 G (1 + ν ) Mechanics:
F = ma
E = 3 K (1 − 2 ν )
T = Iα
p
K =
δV I = mk 2
V Properties of sections:
F2 𝑏𝑏𝑑𝑑3
U shear =∫ dx 𝐼𝐼 = + 𝐴𝐴𝑦𝑦� 2
2 AE 12
bd 3
M 2 I =
U bending =∫ dx 12
2 EI
𝑏𝑏𝑏𝑏3
1 𝐼𝐼 = 3
E strain = Fx x
2
πd 4
I =
σ x +σ y σ x −σ y
2 64
σ 1, 2 = ± + τ xy2
2 2 πd 4
J =
32
2
σ x −σ y 𝜋𝜋𝑑𝑑3 𝜋𝜋𝑑𝑑3
τ max = ± + τ xy2 𝑍𝑍𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = ; 𝑍𝑍𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 =
2 16 32
𝜎𝜎𝑝𝑝 − 𝜎𝜎𝑥𝑥
𝑡𝑡𝑡𝑡𝑡𝑡 𝜃𝜃 =
𝜏𝜏𝑥𝑥𝑥𝑥
𝑥𝑥 = 𝐿𝐿𝐿𝐿𝐿𝐿 𝜎𝜎
𝜀𝜀 = 𝛼𝛼𝛼𝛼 +
𝐸𝐸
𝜎𝜎 = 𝐸𝐸𝐸𝐸𝐸𝐸
Other:
−𝑏𝑏 ± √𝑏𝑏2 −4𝑎𝑎𝑎𝑎 4 0.5𝜎𝜎𝑦𝑦
𝑥𝑥 ; 𝑉𝑉 = 𝜋𝜋𝑟𝑟 3 ; 𝐹𝐹𝐹𝐹𝐹𝐹 =
2𝑎𝑎 3 𝜏𝜏
𝜋𝜋�𝐷𝐷 4 −𝑑𝑑4 � 𝜋𝜋�𝐷𝐷 2 −𝑑𝑑2 � 0.5𝜎𝜎𝑦𝑦
For pipe section:- 𝐼𝐼 = 𝑚𝑚4 𝐴𝐴 = 𝑚𝑚2 𝐹𝐹𝐹𝐹𝐹𝐹 =
64 4 𝜏𝜏
Strain Rosettes:
Principal Strains in a 450 Strain Gauge Rosette:
1 √2
𝜀𝜀1 ; 𝜀𝜀2 = [𝜀𝜀𝐴𝐴 + 𝜀𝜀𝐶𝐶 ] ± �[𝜀𝜀𝐴𝐴 − 𝜀𝜀𝐵𝐵 ]2 + [𝜀𝜀𝐵𝐵 − 𝜀𝜀𝐶𝐶 ]2
2 2
Principal Strain direction for a 450 Strain Gauge Rosette:
2𝜀𝜀𝐵𝐵 − 𝜀𝜀𝐴𝐴 − 𝜀𝜀𝐶𝐶
tan 2𝜃𝜃 =
𝜀𝜀𝐴𝐴 − 𝜀𝜀𝐶𝐶
Principal Strains in a 600 Strain Gauge Rosette:
Fatigue
𝐾𝐾𝑓𝑓 −1
𝑞𝑞 =
𝐾𝐾𝑡𝑡 −1
Weld Stresses:
𝑻𝑻𝑻𝑻
Twisting shear stress 𝜏𝜏 = ;
𝑱𝑱𝒘𝒘
For welds in torsion 𝜏𝜏𝑡𝑡𝑀𝑀𝑀𝑀𝑀𝑀 = �(𝜏𝜏1 𝑡𝑡 )2 + (𝜏𝜏2 𝑡𝑡)2 × 2 × 𝜏𝜏1 𝑡𝑡 × 𝜏𝜏2 𝑡𝑡 × 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
Buckling
𝑳𝑳𝒆𝒆 𝝅𝝅𝟐𝟐 𝑬𝑬𝑬𝑬 𝝅𝝅𝟐𝟐 𝑬𝑬𝑬𝑬
Slenderness ratio 𝑺𝑺𝑺𝑺 = 𝑭𝑭𝒄𝒄𝒄𝒄 = 𝟐𝟐 or 𝑭𝑭𝒄𝒄𝒄𝒄 =
𝒌𝒌 𝑳𝑳 [𝑺𝑺𝑺𝑺]𝟐𝟐
� 𝒆𝒆 �
𝒌𝒌
𝝅𝝅𝟐𝟐 𝑬𝑬𝑬𝑬 𝐼𝐼
𝑭𝑭𝒄𝒄𝒄𝒄 = 𝑘𝑘 = � (I is least moment of inertia). Or 𝐼𝐼 = 𝐴𝐴𝑘𝑘 2
𝑳𝑳𝟐𝟐
𝒆𝒆 𝐴𝐴
𝟐𝟐𝟐𝟐𝟐𝟐 𝑬𝑬
Euler’s validity equation for slenderness ratio limit 𝑺𝑺𝑺𝑺 = �
𝝈𝝈𝒚𝒚
Rankine-Gordon:
𝝅𝝅𝟐𝟐 𝑬𝑬𝑬𝑬
𝑭𝑭𝒄𝒄𝒄𝒄 =
𝟏𝟏+𝒂𝒂 (𝑺𝑺𝑺𝑺)𝟐𝟐
𝝈𝝈𝒚𝒚
Take constant “a” as 𝒂𝒂 =
𝝅𝝅𝟐𝟐 𝑬𝑬
Strain Energy:
Bolted joints
Forces in eccentrically loaded bolts:
𝐹𝐹𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝐹𝐹𝐹𝐹𝐿𝐿2
Primary force per Bolt 𝐹𝐹𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 = 𝐹𝐹𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 =
𝑁𝑁𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 2[(𝐿𝐿1 )2 +(𝐿𝐿2 )2 ]
𝐹𝐹𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝐹𝐹𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 2
𝐹𝐹𝑀𝑀𝑀𝑀𝑀𝑀 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = �
+ � � + (𝐹𝐹𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 )2
2 2
𝐹𝐹𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 2 𝑀𝑀𝑟𝑟𝑛𝑛
𝐹𝐹𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑚𝑚𝑚𝑚𝑚𝑚 = �� � + (𝐹𝐹𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 )2; 𝐹𝐹 =
2 2 2 2
𝑟𝑟1 + 𝑟𝑟2 +𝑟𝑟3 + −−−− + 𝑟𝑟𝑛𝑛2