UFMFSS-30-2 Structural Mechanics
Structural Mechanics
Part 1, Stress Analysis
Version 25/10/2021
Dr Arnaud Marmier
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Introduction: how to use this module
1. There is no secret, you must put the hours in, and you must work outside of the
lectorials and tutorials. Attending these is nowhere near enough. Consider that you
should work around 10 hours per week per (standard) module. This makes sense, and
gives you a ~40 hours week full-time, and also corresponds to roughly 150 hours per
15 credit. With 3 or 4 contact hours per week, that leaves you 6 to 7 hours of
independent study per module per week. A bit more or a bit less is fine of course,
but if you are doing less than 5 hours per week of independent study on each module,
you are cheating yourself. For “Stress analysis”, the first part of “Structural
Mechanics”, independent study mostly means solving the weekly problems. I use the
term “independent” a bit loosely, I mean “without a member of staff”, and you can
work with a group of other students of course!
2. Make sure you have attended (or watched a recording of) the lectorial first! This is
important: the lectorials provide context and motivation, cover the important
formulae, and solve examples, often using automatic methods (such as spreadsheets)
3. The Efficient Engineer Youtube channel is very well done, with nice animations. You
might want to look at the relevant videos (signposted on blackboard and on this
booklet) before solving the weekly tutorials questions
https://www.youtube.com/c/TheEfficientEngineer
4. You should also watch the lectures for deeper dives into the theory and origin of
formulae. Strictly speaking you don’t have to, the lectorials are enough to give you a
working, practical, knowledge of the stress analysis methods. But the lectures will
help you understand the topic better.
5. Make sure that you attempt a few (typically the first two, more if you can) tutorial
problems, BEFORE attending the tutorial session. Again, this is important, you want
to use the tutorial to ask questions, to get clarification, to hone your skills, actively,
NOT just passively watch someone solve a problem (that already has one or two
video solution anyway!).
6. Practice CHECKING your results before looking at solutions. I know this is not
natural, but this is a fabulously useful skill, well worth the effort. It will become
second nature after a while
7. You can use MDSolids, spreadsheets and/or reputable online calculators for checking
purposes.
8. If you get stuck, and your next tutorial session is too far away, you can consult the
video solutions. If that doesn’t help enough, you can also post a precise query on the
forum. But first check that it has not been covered already before opening a new
thread.
9. BOOKS! If you get stuck, and nothing from the extensive material provided on this
module is helping, consider borrowing a book from the library.
10. The DEWIS tests (every two weeks or so) are staggered so that you must revisit a
topic a couple of weeks after the initial topic was introduced and discussed. These
tests are really basic, and if you have followed the previous steps, you should easily
get 100% at your first attempt.
1
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
11. This booklet also contains longer, exam type problems, and detailed solutions. These
are great when revising for the exam (they have been taken from past exams), but feel
free to look at them any time.
2
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Week 0: Revision: Reaction forces and moments, bending moment,
and 2nd moment of Area
Summary
From past experience, we have found that many students of stress analysis have forgotten
much from the 1st year stress module.
PLEASE REVISE AND MAKE SURE YOU CAN PERFORM BASIC CALCULATIONS
QUICKLY AND RELIABLY.
Three points in particular lead to many marks being wasted at the exams.
1/ Reactions at support: many students forget that a fixed support carries reaction forces
along x and y, but also a reaction moment (indeed, without it, the system would not be in
static equilibrium, and would rotate!)
2/ Calculation of moments at any given points on beams: the MCQ style format of year1
stress sort of pushed students to use the shear force and bending moment diagrams. This is
fine, but in order to progress in year 2, the direct method for the bending moment is much
preferable (directly calculate the moments from all the loads to the left -or the right- of the
point of interest)
3/ Calculation of 2nd moment of area: these appear very often, for bending and shear
problems in particular. They are not difficult calculations and as it is possible to express a
cross section as different combinations of rectangles, checking is straightforward. But most
students do not check their calculations, and silly errors on 2nd moments of area will carry
over and ruin the rest of a problem. Methods marks are all and good, but getting the right
answers will yield more marks.
Questions 0.1-0.6
a) Calculate the reactions at the support for the beam depicted in figures 0.1 to 0.6.
b) Calculate the bending moment at any point (as functions of x) for the beams depicted
in figures 0.1 to 0.6. You might have to use different functions for different segments
of each beam.
c) Check your results by using MDSolids or an equivalent online calculator.
Figure 0.1: simply supported beam with point load
3
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Figure 0.2: simply supported beam with point moment
Figure 0.3: cantilever beam with point load
Figure 0.4: simply supported beam with UDL
Figure 0.5: cantilever beam with UDL
Figure 0.6: cantilever beam with UDL and point moment
Questions 0.7-0.9
a) Calculate the 2nd moment of area Ixx (assuming standard axes positions) for the cross
sections depicted in figures 0.6 to 0.8. Use two different combinations of rectangles
(including negative perhaps) in each case.(You might first have to calculate the
position of the centroid, for the neutral axis)
4
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
b) Check your results by building a spreadsheet.
c) Check your results by using MDSolids or an equivalent online calculator.
Figure 0.7: symmetric cross section
Figure 0.8: asymmetric box-shape cross section
Figure 0.9: asymmetric, T-shape cross section
5
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Week 1: Deformation under axial loading, principle of
superposition,
statically
indeterminate
problems,
stress
concentrations and elastoplastic materials
Summary
Deformation under axial loading
𝐹𝑖 𝐿𝑖
𝛿=∑
𝐸𝑖 𝐴𝑖
𝑖
𝐹𝑑𝑥
𝛿=∫
𝐸𝐴
Hypotheses necessary for Principle of superposition:
1. The loading must be linearly related to the stress or displacement that is to be
determined.
2. The loading must not significantly change the original geometry or configuration of
the member.
Statically indeterminate problems
1. Static equilibrium equations are not enough, too many unknowns.
2. consider “extra” reactions as redundant and equivalent to an applied load.
3. Remove corresponding support(s), but write a displacement relation(s) (usually equal
to 0, but not always).
4. Use superposition to solve.
Stress concentrations
σ𝑚𝑎𝑥 = K ∙ σ𝑎𝑣𝑒𝑟𝑎𝑔𝑒
Elastoplastic materials
ε<
σ𝑌
→ σ = E. ε,
𝐸
ε>
σ𝑌
→ σ = σ𝑌
𝐸
Question 1.1
The step bar shown in Figure 1.1 is subjected to two axial forces
of P and Q (P=10kN and Q=30kN)
1. Determine the internal force on each section.
2. Determine the amount of stress at the midpoint of each
section.
3. Determine the deformation of the whole steel rod (
E=200GPa).
Figure 1.1: vertical bar
6
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 1.2
Determine the reactions at A and B for the steel
bar and loading shown in Fig. 1.2, assuming a
close fit at both supports before the loads are
applied.
Figure 1.2: Statically indeterminate vertical bar
Question 1.3
The three A-36 steel bars shown in Figure 1.3 are
pin connected to a rigid member. If the applied
load on the member is 15 kN, determine the force
developed in each bar. Bars AB and EF each have
a cross-section area of 50 mm2 and bar CD has a
cross-section area of 30 mm2
Figure 1.3: Three bars (statically indeterminate)
Question 1.4
Determine the largest axial load P that can be safely
supported by a flat steel bar consisting of two portions,
both 10 mm thick, and respectively 40 and 60 mm wide,
connected by fillets of radius r = 8 mm. Assume an
allowable normal stress of 165 MPa. The geometry of
the bar is illustrated in Figure 1.4.
Figure 1.4: Flat steel bar, stress
concentration
Question 1.5
A simple rod of length L=500mm and cross sectional area A=60 mm2 is made of elastoplastic
material with E = 200GPa and σyield = 300MPa. The rod is subjected to an axial force until it
is stretched to 7 mm, and the load is then removed.
What is the resulting permanent set?
7
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Exam Question I.1 (2019-2020)
A brass stepped bar (Young’s modulus E = 100 GPa) is subjected to two loads, as depicted in
Figure I.1-a). Section [AB], [BC] and [BD] have cross-section areas of 500 mm2, 1000 mm2,
and 1200 mm2respectively.
Figure I.1 (not to scale)
a) Calculate the support reaction at D.
b) Calculate the internal forces and stresses in sections [AB], [BC] and [CD].
c) Calculate the displacement at point A.
The bar is now constrained by a support at A, as depicted in Figure Q1-b).
d) Calculate the support reactions at A and D.
e) Calculate the internal forces and stresses in sections [AB], [BC] and [CD].
Exam Question I.2 (2018-2019)
A 20 mm diameter steel bar (Young’s modulus E = 200 GPa) is subjected to two loads, as
depicted in Figure I.2-a
Figure I.2
a) Calculate the support reaction at A.
b) Calculate the internal forces and stresses in sections [AB], [BC] and [CD].
8
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
c) Calculate the displacement at point D
The bar is now constrained by a support at D, as depicted in Figure I.2-b.
d) Calculate the support reactions at A and D.
e) Calculate the internal forces and stresses in sections [AB], [BC] and [CD].
Exam Question I.3 (resit 2018-2019)
An aluminium bar (Young’s modulus E = 70 GPa) is subjected to two loads, as depicted in
Figure I.3.
Figure I.3
Questions a)-c) refer to Figure I.3-a).
a) Calculate the support reaction at A.
b) Calculate the internal forces and stresses in sections [AB] and [BC].
c) Calculate the displacement at point C.
The uppermost load is now replaced by a support at C, as depicted in Figure I.3-b).
d) Calculate the support reactions at A and C.
e) Calculate the internal forces and stresses in sections [AB] and [BC].
Exam Question I.4 (resit 2017-2018)
The rigid bar EFGH in Figure I.4 is suspended from four identical wires of length L, Young’s
modulus E and cross section area A. (Note that this is a statically indeterminate problem)
9
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Figure I.4
a) How many support reactions are unknown ?
b) How many independent equations can you write using static equilibrium only?
c) Write these “static equilibrium” equations
d) Write enough “compatibility” equations to have a solvable system (hint: the displacements
at points E, F, G, and H are related as the bar is rigid and remains straight, but not horizontal)
e) Solve the system of equation to calculate the reactions at the support as functions of the
load P
10
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Week 2: Eccentric loading, Unsymmetric bending, and bending of
composite beams
Summary
Eccentric loading
Point C is centroid (on neutral axes)
𝐹=𝑃
𝑀 = 𝑃∙𝑑
𝐹 𝑀𝑦
−
𝐴
𝐼
(careful with signs! Always check tension/compression)
𝜎𝑥 = (𝜎𝑥 )𝑐𝑒𝑛𝑡𝑟𝑖𝑐 + (𝜎𝑥 )𝑏𝑒𝑛𝑑𝑖𝑛𝑔 =
Unsymmetric bending
Point C is centroid (on neutral axes)
𝐌 = 𝐌𝑦 + 𝐌𝑧
𝑀𝑦 = 𝑀 ∙ sin(𝜃)
𝑀𝑧 = 𝑀 ∙ cos(𝜃)
𝜎𝑥 = (𝜎𝑥 )𝑏𝑒𝑛𝑑𝑖𝑛𝑔/𝑦 + (𝜎𝑥 )𝑏𝑒𝑛𝑑𝑖𝑛𝑔/𝑧 =
𝑀𝑦 𝑧 𝑀𝑧 𝑦
−
𝐼𝑦
𝐼𝑧
(careful with signs! Always check tension/compression)
tan(𝜙) =
𝐼𝑧
tan(𝜃)
𝐼𝑦
Bending of composite beams loading
Strain is continuous, linear
𝑦
𝜀𝑥 = − 𝜌
Stress is discontinuous
𝐸1 𝑦
𝜌
𝐸2 𝑦
= 𝐸2 ∙ 𝜀𝑥 = −
𝜌
𝜎𝑥1 = 𝐸1 ∙ 𝜀𝑥 = −
𝜎𝑥2
Can replace material1 by material2, different area
11
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
𝑛=
𝐸2
𝐸1
Can then find position of neutral axis (centroid of
transformed sections)
Can also find stress distribution of transformed
section (usual bending methods), and finally
𝑀𝑦
𝜎1𝑥 = 𝜎𝑥 = −
𝐼
𝜎2𝑥 = 𝑛 ∙ 𝜎𝑥
Question 2.1
A cast-iron machine part is acted upon by a 3 kN-m couple as shown in Figure 2.1. Knowing
that E = 190 GPa, v=0.3 and neglecting the effects of fillets, determine
1. The maximum tensile and compressive stresses
2. The radius of curvature
3. The radius of anticlastic curvature
Figure 2.1 Cantilevered beam
Question 2.2
The largest allowable stresses for the cast iron link are 30 MPa in tension and 120 MPa in
compression. Determine the largest force P which can be applied to the link.
Figure 2.2: Link, eccentric loading
12
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 2.3
The rectangular cross section shown in Figure 3.3 is
subjected to a bending moment of 12 kN.m.
1. Determine the normal stress developed at each
corner of the section, and
2. Specify the orientation of the neutral axis.
Figure 2.3: Beam in unsymmetric bending
Question 2.4
A bar is made from bonded pieces of steel (Es =
200GPa) and brass (Eb = 100GPa). Determine the
maximum stress in the steel and brass when a
moment Mx=5 kN.m is applied to the bar.
Figure 2.4: Composite beam
Question 2.5
A concrete floor slab is reinforced with 16 mm diameter steel rods, as shown in Figure 3.5.
The modulus of elasticity is 200 MPa for steel and 25 MPa for concrete. With an applied
bending moment of 4.5 kN.m for each 300 mm width of the slab, determine the maximum
stress in the concrete and steel.
Figure 2.5: Composite beam
13
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Exam Question II.1 (2018-2019)
A composite beam (cross section depicted in Figure Q2) is made of an alumium foam core
(grey, E = 10 GPa) and aluminium skin (white, E = 70 GPa).
Figure II.1
a) Choose a reference material and calculate a composite beam ratio 𝑛.
b) Calculate the second moments of area 𝐼𝑦𝑦 and 𝐼𝑧𝑧 .
c) The beam is subjected to a positive bending moment 𝑀𝑦 of 3 kN·m (top in tension, bottom
in compression). Calculate the minimum and maximum values of stress in the core and in the
skin (four values).
d) The beam is subjected to a positive bending moment 𝑀𝑧 of 5 kN·m (left in tension, right in
compression). Calculate the minimum and maximum values of stress in the core and in the
skin (four values).
Exam Question II.2 (2018-2019)
A beam of cross section as depicted in Figure II.2 is subjected to a bending moment of 5
kN·m, acting at 25° from the y axis, and passing through the centre of mass of the crosssection.
Figure II.2
14
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
a) Determine the position of the centre of mass and of the the neutral axes y-y and z-z
b) Calculate the second moments of area 𝐼𝑦𝑦 and 𝐼𝑧𝑧 .
c) Calculate the moments 𝑀𝑦 and 𝑀𝑧 .
d) Calculate the maximum tensile and compressive stresses. Indicate their locations on a
sketch.
e) Calculate the angle of the neutral axis for this moment. Draw the neutral axis on the sketch
started in d).
Exam Question II.3 (resit 2018-2019)
A composite beam (cross section depicted in Figure II.3-a) is made of brass (strip A) (E =
101 GPa) and mild steel (strip B) (E = 210 GPa).
Figure II.3
Questions a)-c) refer to Figure II.3-a).
a) Calculate the position of the neutral axis.
b) Calculate the second moment of area of the composite beam 𝐼𝑦𝑦 .
c) If the beam is subjected to a positive bending moment 𝑀𝑦 of 10 kN·m, calculate the
minimum and maximum values of stress in the steel and in the brass (four values).
The height of the steel strip is now a variable, ℎ, as shown in Figure II.3-b).
d) Determine the height ℎ of the steel strip so that the neutral axis of the beam is located at
the seam of the two metals.
e) For ℎ as calculated in d), calculate the maximum bending moment this beam can support if
the allowable bending stress is 250 MPa for the steel and 55 MPa for the brass.
15
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Week 3: Slope, elastic curve and deflection of beams
Summary
𝑑 2 𝑦 1 𝑀(𝑥)
≈ =
𝑑𝑥 2 𝜌
𝐸𝐼
𝜃 ≈ 𝑡𝑎𝑛(𝜃) =
𝑦≈
𝑑𝑦
1
≈ ∫ 𝑀(𝑥)𝑑𝑥 + 𝐶1
𝑑𝑥 𝐸𝐼
1
∫ 𝑑𝑥 ∫ 𝑀(𝑥)𝑑𝑥 + 𝐶1 𝑥 + 𝐶2
𝐸𝐼
Question 3.1
For the loading shown in Figure 3.1, determine
a) the equation of the elastic curve for the
cantilever beam AB,
b) the deflection at the free end
Figure 3.1: cantilevered beam
c) the slope at the free end.
Question 3.2
For the loading shown in Figure 3.2, determine
a) the equation of the elastic curve for the
cantilever beam AB,
b) the deflection at the free end
Figure 3.2: Another cantilevered beam
c) the slope at the free end.
Question 3.3
The simply supported prismatic beam AB carries a
uniformly distributed load w per unit length (Figure 3.3).
Determine
a) the equation of the elastic curve
b) the maximum deflection of the beam.
Figure 3.3: Another beam
Question 3.4
Determine the reactions at the support for the beam in Figure
3.4.
Figure 3.4: Yet another beam
Question 3.5
For the uniform beam, determine the reaction at A, derive
the equation for the elastic curve, and determine the slope
at A. (Note that the beam is statically indeterminate)
Figure 3.5: Annoying beam
16
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 3.6
For the uniform beam, derive the equation of
the elastic curve (deflection).
Figure 3.6: Beam with discontinuity
Question 3.7
For the uniform beam, derive the equation of
the elastic curve (deflection).
Figure 3.7: Statically indeterminate beam with
discontinuity. Fun!
Exam Question III.1 (2019-2020)
A simply supported beam (Young’s modulus 𝐸, second moment of area 𝐼) is supported and
loaded as shown in Figure III.1.
Figure III.1. Simply supported beam
a) Calculate the reactions at the supports in term of the uniformly distributed load 𝑤 and
beam length 𝐿.
b) Determine the bending moment between A and B as a function of 𝑤, 𝑥 and 𝐿.
c) Determine the bending moment between B and C as a function of 𝑤, 𝑥 and 𝐿.
d) Express the slope at any point of the beam as functions of 𝐸, 𝐼, 𝑤, 𝑥 and 𝐿 (no need to
solve the integration constants at this stage).
e) Express the deflection at any point of the beam as functions of 𝐸, 𝐼, 𝑤, 𝑥 and 𝐿 (no need
to solve the integration constants at this stage).
f) Calculate all the integration constants as functions of 𝐿 and and 𝑤, and write the functions
for the deflection.
Note: it is possible to use Macauley’s methods to solve this problem in a more compact way. In this
case , subquestions b) and c) are aggregated.
Exam Question III.2 (2018-2019)
A simply supported beam (Young’s modulus 𝐸, second moment of area 𝐼) is supported and
loaded as shown in Figure III.2.
17
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Figure III.2
a) Calculate the reactions at the supports in term of the load 𝑃 and beam length 𝐿.
b) Determine the bending moment between A and B as a function of 𝑃, 𝑥 and 𝐿 (do not forget
to include the effect of the reaction at A!).
c) Determine the bending moment between B and C as a function of 𝑃, 𝑥 and 𝐿.
d) Express the slope at any point of the beam as functions of 𝐸, 𝐼, 𝑃, 𝑥 and 𝐿 (no need to
solve the integration constants at this stage).
e) Express the deflection at any point of the beam as functions of 𝐸, 𝐼,𝑃, 𝑥 and 𝐿 (no need to
solve the integration constants at this stage).
f) Calculate all the integration constants as functions of 𝐿 and and 𝑃.
Exam Question III.3 (resit 2018-2019)
A cantilevered beam (Young’s modulus 𝐸, second moment of area 𝐼) is supported and loaded
as shown in Figure III.3.
Figure III.3
a) Calculate the reactions at the supports 𝑅𝐴 and 𝑀𝐴 in term of the load 𝑤 and the length 𝐿.
b) Determine the bending moment between A and B as a function of 𝑤, 𝑥 and 𝐿. (do not
forget to include the effect of the reactions at A!)
c) Determine the bending moment between B and C.
d) Express the slope at any point of the beam as functions of 𝐸, 𝐼, 𝑤, 𝑥 and 𝐿 (no need to
solve the integration constants at this stage).
e) Express the deflection at any point of the beam as functions of 𝐸, 𝐼,𝑤, 𝑥 and 𝐿 (no need to
solve the integration constants at this stage).
f) Calculate all the integration constants as functions of 𝐿 and and 𝑤.
18
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Exam Question III.4 (2017-2018)
A simply supported beam (Young’s modulus 𝐸, second moment of area 𝐼) is supported and
loaded as shown in Fig. III.4.
Figure III.4
a) Calculate the reactions at the supports in terms of the load.
b) Determine the bending moment between A and B as a function of 𝑃 and 𝑥.
c) Determine the bending moment between B and C as a function of 𝑃, 𝑥 and 𝐿.
d) Express the slope at any point of the beam as functions of 𝐸, 𝐼, 𝑃, 𝑥 and 𝐿 (no need to
solve the integration constants at this stage)
e) Express the deflection at any point of the beam as functions of 𝐸, 𝐼, 𝑃, 𝑥 and 𝐿 (no need to
solve the integration constants at this stage)
f) Calculate all the integration constants as functions of 𝐿 and 𝑃
Exam Question III.5 (Mock Exam)
A simply supported beam (Young’s modulus 𝐸, second moment of area 𝐼) is supported and
loaded as shown in Fig. III.5.
Figure III.5
a) Calculate the reactions at the supports in terms of the load.
b) Determine the bending moment between A and B as a function of 𝑀, 𝑥 and 𝐿.
c) Determine the bending moment between B and C as a function of 𝑀, 𝑥 and 𝐿.
d) Express the slope at any point of the beam as functions of 𝐸, 𝐼, 𝑀, 𝑥 and 𝐿 (no need to
solve the integration constants at this stage)
e) Express the deflection at any point of the beam as functions of 𝐸, 𝐼, 𝑀, 𝑥 and 𝐿 (no need
to solve the integration constants at this stage)
f) Calculate all the integration constants as functions of 𝐿 and 𝑀
19
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Appendix 3.1: Singularity functions and first two integrals
20
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Week 4: Buckling, critical load, secant formula for eccentric loading
Summary
𝑃𝑐𝑟 =
𝜋 2 𝐸𝐼
(𝐾𝐿)2
Pinned ends
Fixed and free ends
Fixed ends
Pinned and fixed ends
𝐾
𝐾
𝐾
𝐾
𝑣𝑚𝑎𝑥 = 𝑒 [sec (√
=1
=2
= 0.5
= 0.7
𝜎𝑚𝑎𝑥 =
𝑃 𝐾𝐿
𝜋 𝑃
) − 1] = 𝑒 [sec ( √ ) − 1]
𝐸𝐼 2
2 𝑃𝑐𝑟
𝑃 𝑃𝑒𝑐
𝑃 𝐾𝐿
𝑃
𝑒𝑐
𝜋 𝑃
+
sec (√
) = [1 + 2 sec ( √ )]
𝐴
𝐼
𝐸𝐼 2
𝐴
𝑟
2 𝑃𝑐𝑟
Question 4.1
Determine the critical load of a steel tube that is 5 m
long and has a 100-mm outer diameter and a 16-mm
wall thickness. Use E= 200GPa.
Figure 4.1: Steel tube column
Question 4.2
A compression member of 500mm effective length
consists of a solid 25 mm diameter aluminium rod. In
order to reduce the weight of the member by 25%, the
solid rod is replaced by a hollow rod of the cross section
shown below. E=70GPa. Determine
a) the percent reduction in the critical load,
b) the value of the critical load for the hollow rod
Figure 4.2: Aluminium column
Question 4.3
The aluminium column in Fig. 4.3 is fixed at its bottom and is
braced at its top by cables so as to prevent movement at the
top along the x axis.
If it is assumed to be fixed at its base, determine the largest
allowable load P that can be applied. Use a factor of safety for
buckling of 3.0. Take E=70 GPa, σy=215 MPa, A=7.5 10-3 m2,
Ix=61.3 10-6 m2, Iy=23.2 10-6 m2
Figure 4.3: Rectangular column
21
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 4.4
The uniform column AB consists of a 2.5 m section of structural tubing having the cross
section shown.
a) Using Euler’s formula and a factor of safety of two, determine the allowable centric load
for the column and the corresponding normal stress.
b) Assuming that the allowable load, found in part a), is applied as shown at a point 20 mm
from the geometric axis of the column, determine the horizontal deflection of the top of
the column and the maximum normal stress in the column. Use E=200 GPa.
Figure 4.4: Tubular columns
Question 4.5
An axial load P is applied to the 32-mm-diameter steel rod
AB shown in Fig 4.5. For P = 37 kN and e = 1.2 mm,
determine
a) the deflection at the midpoint C of the rod,
b) the maximum stress in the rod.
Use E = 200 GPa.
Figure 4.5: Eccentric loading
22
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Week 5: Shear force and shear flow
Summary
First moment of area: 𝑄 = 𝐴𝑦𝑐
Horizontal (longitudinal) shearing force: 𝐻 =
Horizontal (longitudinal) shear stress: 𝜏 =
𝑉𝑄∆𝑥
𝐼
𝑉𝑄
𝐼𝑡
The horizontal shear force per unit length of beam or Shear flow: 𝑞 =
𝑉𝑄
𝐼
Question 5.1
Beam AB is made of three planks glued together and
is subjected, in its plane of symmetry, to the loading
shown in Fig 5.1. Knowing that the width of each
glued joint is 20 mm, determine the average shearing
stress in each joint at section n-n of the beam. The
location of the centroid of the section is given in Fig.
6.1 and the centroidal moment of inertia is known to
be I=8.63. 10-6 m4.
Figure 5.1: Beam in shear
Question 5.2
A beam is made of three planks, nailed together.
Knowing that the spacing between nails is 25 mm and
that the vertical
shear in
the
beam
is
V = 500 N, determine the shear force in each nail.
Figure 5.2: Wooden beam
Question 5.3
The American Standard rolled-steel beam shown has been
reinforced by attaching to it two 16×200-mm plates, using 18mm diameter bolts spaced longitudinally every 120 mm.
Knowing that the average allowable shearing stress in the bolts
is 90 MPa, determine the largest permissible vertical shearing
force.
Figure 5.3: Reinforced I-Beam
23
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 5.4
A timber beam AB of span 3 m and nominal width
120 mm (actual width 100 mm) is to support the
three concentrated loads shown. Knowing that for
the grade of timber used 𝜎𝑎𝑙𝑙 = 12 MPa and 𝜏𝑎𝑙𝑙 =
0.8 MPa, determine the minimum required depth d
of the beam.
Figure 5.4: Wooden beam
Question 5.5
Figure 5.5: Cantilevered beam
For the beam and loading shown in Fig. 5.5, consider section n-n and determine
a) the largest shearing stress in that section,
b) the shearing stress at point a.
24
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Exam Question V.1 (2018-2019)
A wooden beam (cross-section depicted in Figure V.1) is made of four planks nailed together.
It is subjected to a downward vertical shearing force of 500 N.
Figure V.1
a) Calculate the second moments of area 𝐼𝑦𝑦 .
b) Calculate the first moment of area for points located on the neutral axis.
c) Calculate the average shearing stress for points located on the neutral axis.
d) Knowing that the maximum allowable shear force in each nail is 200 N, calculate the
minimum spacing between rows of nails.
25
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Week 6: Shearing stresses in thin walled members
Summary
First moment of area: 𝑄 = 𝐴𝑦𝑐
Horizontal (longitudinal) shearing force: 𝐻 =
Horizontal (longitudinal) shear stress: 𝜏 =
𝑉𝑄∆𝑥
𝐼
𝑉𝑄
𝐼𝑏
The horizontal shear force per unit length of beam or Shear flow: 𝑞 =
𝑉𝑄
𝐼
Question 6.1
An extruded aluminum beam has the cross section
shown. Knowing that the vertical shear in the beam
is 150 kN, determine the shearing stress at points a
and b.
Figure 6.1: Beam cross-section
Question 6.2
Three planks are connected as shown in Fig. 6.2 by bolts
of 14-mm diameter spaced every 150 mm along the
longitudinal axis of the beam. For a vertical shear of 10
kN, determine the average shearing stress in the bolts.
Figure 6.2: Wooden beam and bolts
Question 6.3
The built-up beam shown in Fig 6.3 is made by gluing together
five planks. Knowing that in the glued joints the average
allowable shearing stress is 350 kPa, determine the largest
permissible vertical shear in the beam.
Figure 6.3: Glued I-Beam
26
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 6.4
Knowing that a given vertical shear V causes a maximum
shearing stress of 75 MPa in the hat-shaped extrusion
shown in Fig 6.4, determine the corresponding shearing
stress at points a and b.
Figure 6.4: Extruded beam 1
Question 6.5
Knowing that a given vertical shear V causes a maximum
shearing stress of 75 MPa in an extruded beam having the
cross section shown in Fig 6.5, determine the shearing
stress at the three points indicated.
Figure 6.5: Extruded beam 2
Exam Question VI.1 (2019-2020)
An extruded aluminium beam has the cross section shown in Figure VI.1. It is subjected to a
downward vertical shearing force of 90 kN.
Figure VI.1. Beam cross section
27
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
a) Calculate the relevant second moment of area.
b) Calculate the shear stress at point A.
c) Calculate the shear stress at point B.
d) Calculate the shear stress at point C.
Exam Question VI.2 (resit 2018-2019)
An extruded aluminium beam has the cross section shown in Figure VI.2. It is subjected to a
downward vertical shearing force of 50 kN.
Figure VI.2
a) Calculate the relevant second moment of area.
b) Calculate the shear stress at point A (on the neutral axis).
c) Calculate the shear stress at point B.
d) Calculate the shear stress at point C.
28
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Week 7: Curved beam and torsion
Summary
Curved beams
𝑀𝑦
Stress in a curved beam: 𝜎 = 𝐴𝑒(𝑅−𝑦) =
𝑀(𝑟−𝑅)
, where 𝑒 is the distance between the neutral
𝐴𝑒𝑟
axis and the centroid of the section, 𝑒 = 𝑅 − 𝑟̅ . 𝑅 is the radius at the neutral axis
1
𝐴
1
𝑅
=
1
∫ 𝑟 𝑑𝐴, 𝑟 the position of the point of interest from the centre of curvature, 𝑦 the position of
the point of interest from the neutral axis, and 𝑟̅ the position of the centroid from the centre of
curvature.
Figure 7.01: Curved beam variables and radius of curvature formulae
Torsion
𝑇
𝜏
Torsion formula: 𝐽 = 𝑅 =
𝐺𝜃
; circular bar: 𝐽 =
𝐿
Rectangular bars (𝑎 > 𝑏): 𝜏𝑚𝑎𝑥 = 𝑐
𝑇
2
1 𝑎𝑏
𝜋𝑑4
32
, 𝜃 = 𝐽𝑐
=
𝜋𝑟 4
2
𝑇𝐿
2 𝑎𝑏
3𝐺
Table 7.0: Rectangular bars in torsion
Figure 7.02: Stress concentration factors
29
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 7.1
A machine component has a T-shaped cross
section and is loaded as shown in Fig. 7.1.
Knowing that the allowable compressive
stress is 50 MPa, determine the largest force P
that can be applied to the component.
Figure 7.1: Curved beam and cross-section
Question 7.2
The curved bar shown in Fig 7.2 has a cross
section of 40×60 mm and an inner radius r1=15
mm.
A) For the loading shown determine the largest
tensile and compressive stresses.
B) Determine the percent error introduced in the
computation of the maximum stress by assuming
that the bar is straight. Consider the case when (a)
r1= 20 mm, (b) r1= 200 mm, (c) r1= 2 m.
Figure 7.2: Curved bar
Question 7.3
A hollow cylindrical steel shaft is 1.5 m long and
has inner and outer diameters respectively equal to
40 and 60 mm (Fig. 8.3). (a) What is the largest
torque that can be applied to the shaft if the shearing
stress is not to exceed 120 MPa? (b) What is the
corresponding minimum value of the shearing stress
in the shaft?
Figure 7.3: Basic hollow shaft
Question 7.4
The stepped shaft shown in Fig 7.4 must transmit 40
kW at a speed of 720 rpm. Determine the minimum
radius r of the fillet if an allowable stress of 36 MPa
is not to be exceeded.
Figure 7.4: Stepped shaft
30
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 7.5
Shaft BC is hollow with inner and outer diameters
of 90 mm and 120 mm, respectively. Shafts AB and
CD are solid and of diameter d. For the loading
shown, determine (a) the maximum and minimum
shearing stress in shaft BC, (b) the required
diameter d of shafts AB and CD if the allowable
shearing stress in these shafts is 65 MPa.
Figure 7.5: combined shaft
Question 7.6
Using τall=40 MPa, determine the largest torque that
may be applied to each of the brass bars and to the
brass tube shown in Fig 7.6. Note that the two solid
bars have the same cross-sectional area, and that the
square bar and square tube have the same outside
dimensions..
Figure 7.6: Prismatic bars
31
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Week 8: Stress Transformation, direct method
Summary
tan2𝜃𝑝 =
𝜎𝑚𝑎𝑥,𝑚𝑖𝑛 =
2𝜏𝑥𝑦
𝜎𝑥 − 𝜎𝑦
𝜎𝑥 + 𝜎𝑦
𝜎𝑥 − 𝜎𝑦 2
2
± √(
) + 𝜏𝑥𝑦
2
2
tan2𝜃𝑠 = −
𝜏𝑚𝑎𝑥
= √(
𝜎𝑥 − 𝜎𝑦
2𝜏𝑥𝑦
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
+
cos2𝜃 + 𝜏𝑥𝑦 sin2𝜃
2
2
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
𝜎𝑦′ =
−
cos2𝜃 − 𝜏𝑥𝑦 sin2𝜃
2
2
𝜎𝑥 − 𝜎𝑦
𝜏𝑥′𝑦′ = −
sin2𝜃 + 𝜏𝑥𝑦 cos2𝜃
2
𝜎𝑥′ =
𝜎𝑥 − 𝜎𝑦 2
2
) + 𝜏𝑥𝑦
2
Question 8.1
For the state of plane stress shown in Fig. 8.1,
determine (a) the principal planes and the principal
stresses, (b) the stress components exerted on the
element obtained by rotating the given element
counterclockwise through 30°.
Figure 8.1: State of stress
Question 8.2
For the given state of stress shown in Fig 8.2,
determine (a) the principal planes, (b) the principal
stresses, (c) the orientation of the planes of
maximum in-plane shearing stress, (d) the
maximum in-plane shearing stress, (e) the
corresponding normal stress.
Figure 8.2: State of stress
Question 8.3
For the given state of stress shown in Fig 8.3,
determine (a) the principal planes, (b) the principal
stresses, (c) the orientation of the planes of
maximum in-plane shearing stress, (d) the
maximum in-plane shearing stress, (e) the
corresponding normal stress.
Figure 8.3: State of stress
32
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 8.4
A single horizontal force P of magnitude 750 N is
applied to end D of lever ABD. Knowing that
portion AB of the lever has a diameter of 30 mm,
determine:
(a) the normal and shearing stresses on an element
located at point H and having sides parallel to the x
and y axes,
(b) the principal planes and the principal stresses at
point H.
Figure 8.4: Lever
Question 8.5
The steel pipe AB in Fig 8.5 has a 102-mm outer
diameter and a 6-mm wall thickness. Knowing that
arm CD is rigidly attached to the pipe, determine
the principal stresses and the maximum shearing
stress at point K.
Figure 8.5: Steel pipe
Question 8.6
The axle of an automobile is acted upon by the
forces and couple shown. Knowing that the
diameter of the solid axle is 32 mm, determine (a)
the principal planes and principal stresses at point
H located on top of the axle, (b) the maximum
shearing stress at the same point.
Figure 8.6: car axle
33
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Exam Question VIII.1 (Mock exam)
The bent rod in Fig. VIII.1 has a diameter of 20 mm and is subjected to the force of 400 N.
Point A is located on the bottom surface.
Figure VIII.1
Determine:
a) The axial stress at A
b) The stress due to bending at A
c) The principal stresses at point A
d) The maximum in-plane shear stress that is developed at point A
34
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Week 9: Stress Transformation, Mohr’s circle and 3D stress
Summary
Question 9.1
For the state of plane stress shown in Fig. 9.1, (a)
construct Mohr’s circle, determine (b) the principal
planes, (c) the principal stresses, (d) the maximum
shearing stress and the corresponding normal stress.
Figure 9.1: State of stress
Question 9.2
For the given state of stress shown in Fig 9.2, (a)
construct Mohr’s circle, determine (b) the principal
planes, (c) the principal stresses, (d) the maximum
shearing stress and the corresponding normal stress.
Figure 9.2: State of stress
Question 9.3
For the given state of stress shown in Fig 9.3, (a)
construct Mohr’s circle, determine (b) the principal
planes, (c) the principal stresses, (d) the maximum
shearing stress and the corresponding normal stress.
Figure 9.3: State of stress
35
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 9.4
For the state of stress shown in Fig 9.4, determine
the maximum shearing stress when (a) σy=40 MPa,
(b) σy=120 MPa. (Hint: Consider both in-plane and
out-of-plane shearing stresses.)
Figure 9.4: State of stress
Question 9.5
For the state of stress shown in Fig 9.5, determine
the range of values of τxz for which the maximum
shearing stress is equal to or less than 60 MPa.
Figure 9.5: State of stress
36
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Week 10: Failure criteria
Summary
Maximum-shearing-stress/Tresca (Ductile)
For σa and σb with the same sign,
 max 
a
2
or
b
2

Y
2
For σa and σb with opposite signs,
 max 
 a b
2

Y
2
Maximum-distortion-energy/Von Mises (Ductile)
 2vonmisses   a2   a b   b2   Y2
Maximum-normal-stress/Coulomb (Brittle)
 a  U
 b  U
Simplified Mohr’s criterion (Brittle)
37
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 10.1
The state of plane stress shown in Fig 10.1 occurs in
a machine component made of a steel with 𝜎𝑌 = 325
MPa.
I/ Using the maximum-distortion-energy criterion,
determine whether yield will occur when (a) 𝜎𝑜 =
200 MPa, (b) 𝜎𝑜 = 240 MPa, (c) 𝜎𝑜 = 280 MPa. If
yield does not occur, determine the corresponding
factor of safety.
Figure 10.1: State of stress
II/ Repeat the calculations, but with the maximumshearing-stress criterion.
Question 10.2
The 36-mm-diameter shaft is made of a grade of
steel with a 250 MPa tensile yield stress. Using the
maximum-shearing-stress criterion, determine the
magnitude of the torque T for which yield occurs
when P = 200 kN.
Figure 10.2: Steel shaft
Question 10.3
The state of plane stress shown is expected to occur
in an aluminium casting. Knowing that for the
aluminium alloy used 𝜎𝑈𝑇 = 80 MPa and 𝜎𝑈𝐶 = 200
MPa and using Mohr’s criterion, determine whether
rupture of the casting will occur.
Figure 10.3: State of stress
Question 10.4
The cast-aluminium rod shown is made of an alloy
for which 𝜎𝑈𝑇 = 60 MPa and 𝜎𝑈𝐶 = 120 MPa.
Using Mohr’s criterion, determine the magnitude of
the torque T for which failure should be expected.
Figure 10.4: Aluminium shaft
38
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Exam Question X.1 (2019-2020)
The bent rod depicted in Figure X.1 has a diameter of 40 mm and is subjected to a force of
1.5 kN. Point H is located on the surface, with coordinates (0, 75, 20).
Figure X.1
a) Calculate the second moment of area and the polar second moment of area of the rod.
b) Calculate the stress due to bending at point H.
c) Calculate the stress due to torsion at point H.
d) What is the state of stress at point H (two axial stresses, one shear stress)? Draw a state of
stress diagram, indicating clearly the directions and magnitude of the three stresses.
e) Using Mohr’s circle or a direct method, calculate the principal stresses at H.
f) Determine whether the part yields at point H or not, using the maximum distortion energy
(Von-Mises) criterion (Yield strength σY = 350 MPa).
Exam Question X.2 (resit 2018-2019)
The bent rod in Figure X.2 has a diameter of 15 mm and is subjected to forces of 500 N.
Point A is located on the bottom surface.
Figure X.2
39
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
a) Calculate the cross section area of the rod and the second moment of area from the rod’s
neutral axis.
b) Calculate the axial stress at A.
c) Calculate the stress due to bending at A.
d) What is the state of stress at A (two axial stresses, one shear stress)? Draw a state of stress
diagram, indicating clearly the directions and magnitude of the three stresses.
e) Using Mohr’s circle or a direct method, calculate the principal stresses at A.
f) Using Mohr’s circle or a direct method, calculate the maximum in-plane shear stress that is
developed at A.
40
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Week 11: Strain Transformation
Summary
𝜀𝑎 =
𝜎𝑎 𝜈𝜎𝑏
𝜎𝑏 𝜈𝜎𝑎
−𝜈
−𝜈
𝜏𝑎𝑏
𝐸
(𝜎𝑎 + 𝜎𝑏 ) =
(𝜀𝑎 + 𝜀𝑏 ), 𝛾𝑎𝑏 =
−
, 𝜀𝑏 = −
, 𝜀𝑐 =
,𝐺 =
𝐸
𝐸
𝐸
𝐸
𝐸
1−𝜈
𝐺
2(1 + 𝜈)
𝛾𝑥𝑦
𝑡𝑎𝑛2𝜃𝑝 =
𝜀𝑥 − 𝜀𝑦
𝜀𝑚𝑎𝑥,𝑚𝑖𝑛
=
𝜀𝑥 + 𝜀𝑦
𝜀𝑥 − 𝜀𝑦 2
𝛾𝑥𝑦 2
± √(
) +( )
2
2
2
𝜀𝑥′ = 𝜀𝑥 cos 2 𝜃 + 𝜀𝑦 sin2 𝜃 + 𝛾𝑥𝑦 cos𝜃sin𝜃
𝜀𝑥 + 𝜀𝑦 𝜀𝑥 − 𝜀𝑦
𝛾𝑥𝑦
+
cos2𝜃 +
sin2𝜃
2
2
2
𝜀𝑥 + 𝜀𝑦 𝜀𝑥 − 𝜀𝑦
𝛾𝑥𝑦
𝜀𝑦′ =
−
cos2𝜃 −
sin2𝜃
2
2
2
𝛾𝑥′𝑦′
𝜀𝑥 − 𝜀𝑦
𝛾𝑥𝑦
=−
sin2𝜃 +
cos2𝜃
2
2
2
𝜀𝑥′ =
Question 11.1
The strains determined by the use of the rosette
shown during the test of a machine element are
ε1 = 600 μ, ε2 = 450 μ, and ε3 = -75 μ
Determine,
(a) the in-plane principal strains,
(c) the in-plane maximum shearing strain.
Figure 11.1: Another rosette
Question 11.2
Using a 60º rosette, the following strains have been
determined at point Q on the surface of a steel
machine base:
ε1 = 40 μ, ε2 = 980 μ, and ε3 = 330 μ
Using the coordinate axes shown, determine at point
Q,
(a) the strain components εx, εy, and γxy,
(b) the three principal strains,
Figure 11.2: Strain rosette
(c) the maximum shearing strain. (Use υ = 0.29)
41
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 11.3
A single strain gage forming an angle β = 18° with a
horizontal plane is used to determine the gauge
pressure in the cylindrical steel tank shown in Fig.
11.3. The cylindrical wall of the tank is 6 mm thick,
has a 600 mm inside diameter, and is made of a
steel with E = 200 GPa and ν = 0.30. Determine the
pressure in the tank indicated by a strain gage
reading of 280μ.
Figure 11.3: pressure vessel
Exam Question XI.1 (resit 2019-2020)
Using a rosette strain gauge, the following strains have been determined at point Q on the
surface of a steel machine base (Young’s modulus E = 210 GPa, Poisson’s ratio υ = 0.31,
Yield strength σY = 450 MPa), as shown in Figure XI.1.
𝜀𝐴 = 400 𝜇
𝜀𝐵 = −200 𝜇
𝜀𝐶 = 300 𝜇
Figure XI.1
a) Calculate the strain components εx, εy, and γxy.
b) Calculate the three principal strains.
c) Calculate the maximum shearing strain.
d) Calculate the three principal stresses.
e) Determine whether the part yields or not, using the maximum distortion energy criterion
(Von-Mises criterion)
Exam Question XI.2 (2018-2019)
Using a 60º rosette strain gauge, the following strains have been determined at point Q on the
surface of a steel machine base (Young’s modulus E = 210 GPa, Poisson’s ratio υ = 0.31,
Yield strength σY = 300 MPa), as shown in Figure XI.2.
42
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
𝜀𝐴 = −500 𝜇
𝜀𝐵 = 750 𝜇
𝜀𝐶 = 250 𝜇
Figure XI.2
a) Calculate the strain components εx, εy, and γxy.
b) Calculate the three principal strains.
c) Calculate the maximum shearing strain.
d) Calculate the three principal stresses.
e) Determine whether the part yields or not, using the maximum distortion energy criterion
(Von-Mises criterion)
Exam Question XI.3 (2017-2018)
The following readings have been obtained from the strain gauge rosette shown in Fig. XI.3,
which is attached to the outer surface of an aluminium pressure vessel. All gauges are in the
same plane XY. Assume Young’s Modulus, E=70GPa and Poisson’s ratio, ν = 0.3.
𝜀𝐴 = 150 𝜇
𝜀𝐵 = 250 𝜇
𝜀𝐶 = 350 𝜇
Figure XI.3
Determine:
a) The values of εx, εy and γxy
b) The principal strains
c) The reading of gauge D
d) The angle that the principal strain makes with the +X axis
Exam Question XI.4 (resit 2017-2018)
The following readings have been obtained from the strain gauge rosette shown in Fig. XI.4,
which is attached on the free surface of a chain on a suspension bridge. Assume all gauges
are in the same plane XY. Young’s modulus for the wrought iron chain is E=193GPa and
Poisson’s ratio is ν = 0.3. The Yield stress of the wrought iron is σy=600 MPa.
43
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
𝜀𝐴 = 1000 𝜇
𝜀𝐵 = 500 𝜇
𝜀𝐶 = 300 𝜇
Figure XI.4
Determine:
a) The strain components εx, εy and γxy
b) The stress components σx, σy and τxy
c) The principal stresses and their principal planes
d) The Factor of Safety (FOS) based on Maximum-Shearing-Stress Criterion
44
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Solutions (Weekly Questions)
Question 1.1
1. FAB = 10 kN and FBC = -20 kN
2. σAB = 31.83 MPa and σBC = -7.074 Mpa
3. 𝛿𝑑 = 4.6 10−3 mm
Question 1.2
𝑅𝐵 = 577 kN
𝑅𝐴 = 323 kN
Question 1.3
𝐹𝐴𝐵 = 9.52 kN
𝐹𝐶𝐷 = 3.46 kN
𝐹𝐸𝐹 = 2.02 kN
Question 1.4
P=36.3 kN
Question 1.5
𝛿𝑑 = 6.25 mm
Question 2.1
1) 76.0 MPa and -131.3 MPa
2) 55.0 m
3) -183.m
Question 2.2
77.0 kN
Question 2.3
𝜎𝐵 = 2.25 MPa 𝜎𝐶 = −4.95 MPa
𝜎𝐷 = −2.25 MPa
𝜎𝐸 = 4.95 MPa
𝛼 = −79.4°
Question 2.4
σmax in brass = 78.125 MN/m2
σmax in steel = 156.25 MN/m2
Question 2.5
σcomp in concrete = 9.29 MN/m2
σtens in steel = 127.9 MN/m2
Question 3.1
𝑀
a) 𝑦 = 2𝐸𝐼0 (𝐿 − 𝑥)2
45
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
b) 𝑦𝐴 =
𝑀0 𝐿2
2𝐸𝐼
𝑀0 𝐿
c) 𝜃𝐴 = −
𝐸𝐼
Question 3.2
𝑃
a) 𝑦 = 6𝐸𝐼 (−𝑥 3 + 3𝐿2 𝑥 − 2𝐿3 )
𝑃𝐿3
b) 𝑦𝐴 = − 3𝐸𝐼
c) 𝜃𝐴 =
𝑃𝐿2
2𝐸𝐼
Question 3.3
𝑤
a) 𝑦 = 24𝐸𝐼 (−𝑥 4 + 2𝐿𝑥 3 − 𝐿3 𝑥)
5𝑤𝐿4
b) 𝑦𝑚𝑖𝑛 = − 384𝐸𝐼
Question 3.4
5
1
3
𝐴𝑥 = 0, 𝐴𝑦 = 8 𝑤𝐿, 𝑀𝐴 = − 8 𝑤𝐿2 , 𝐵 = 8 𝑤𝐿
Question 3.5
1
a) 𝑅𝐴 = 10 𝑤0 𝐿
𝑤
0
(−𝑥 5 + 6𝐿2 𝑥 3 − 𝐿4 𝑥)
b) 𝑦 = 120𝐸𝐼𝐿
𝑤 𝐿3
0
c) 𝜃𝐴 = − 120𝐸𝐼
Question 4.1
305 kN
Question 4.2
a) Reduction by 1/16=6.25%
b) 49.7 kN
Question 4.3
141 kN
Question 4.4
a) Pcr=237 kN, Pall=118.5 kN, σall=55.5 MPa
b) use P/PCr=1/2 in formulae, and quite easily ymax=25 mm, σall=144.4 MPa
Question 4.5
a) 1.658 mm. b) 78.9 MPa
Question 5.1
At joint a, τave=725 kPa
At joint b, τave=608 kPa
46
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 5.2
92.6 N
Question 5.3
193.5 kN
Question 5.4
Mystery question
Question 5.5
a) 920 kPa. b) 765 kPa
Question 6.1
At point a, τa=101.6 MPa
At point b, τb=79.6 MPa
Question 6.2
20.6 MPa
Question 6.3
4.28 kN
Question 6.4
a) 41.4 MPa. b) 41.4 MPa
Question 6.5
a) 33.7 MPa. b) 75.0 MPa. b) 43.5 MPa.
Question 7.1
8.55 kN
Question 7.2
A) σt=5.22 MPa, σc=-12.49 MPa
B)
a) %error=-34.4%
b) %error=6.0%
c) %error=0.6%
Question 7.3
a) 4.08 kN·m
b) 80 MPa
Question 7.4
a) r~11.7 mm (K=1.18, r/d~0.26)
47
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 7.5
a) τmax=86.2 MPa, τmin=64.7 MPa
b) d=77.8 mm
Question 7.6
a) T1=532 N·m
b) T2=414 N·m
c) T3=555 N·m (note that the formula for a hollow non-cylindric cross section is not given in
the notes or the summary. You would need to do a bit of research for this one. This would of
course never appear in the exam, but provides an interesting challenge for those of you who
are into this sort of things, going a bit beyond the basics)
Results
Question 8.1
Sx
Sy
Txy
100
theta
60
-48
Principal
plane
Shear plane Smax
Smin
Tmax
Sx'
Sy'
Txy'
30
-33.7
11.3
132.00
28.00
52.00
48.43
111.57
-41.32
Question 8.2
Sx
Sy
-40
Txy
-60
Principal
plane (°)
theta
35
0
a)
Shear plane
(°)
Smax
Smin
Tmax
S_ave
37.0
-8.0
-13.60
-86.40
36.40
-50.00
c)
b)
b)
d)
e)
Question 8.3
Sx
Sy
Txy
10
50
Principal
plane (°)
theta
-15
0
a)
Shear plane
(°)
Smax
Smin
Tmax
S_ave
18.4
-26.6
55.00
5.00
25.00
30.00
c)
b)
b)
d)
e)
Question 8.4
F (N)
L (m)
M (N.m) d (m)
I (m4)
sigma_y (MPa)
0.25
187.5
0.03
3.98E-08
70.74 a)
750
Torsion
F (N)
L (m)
T (N.m) d (m)
J (m4)
tau_xy (MPa)
750
0.45
337.5
0.03
7.95E-08
63.66 a)
Sx
Sy
0.00
Txy
70.74
theta
63.66
Principal
Shear plane
plane (°)
(°)
Smax
Smin
Tmax
0
-30.5
14.5
108.19
-37.46
72.83
b)
b)
b)
48
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 8.5
Bending
F (N)
L (m)
M (N.m) d_ext (m) d_int (m)
I (m4)
sigma_y (MPa)
10000
0.15
1500
0.102
0.09
2.09E-06
-36.55
Torsion
F (N)
L (m)
T (N.m) d (m)
d_int (m)
J (m4)
tau_xy (MPa)
10000
0.2
2000
0.102
0.09
4.19E-06
24.37
Sx
Sy
0.00
Txy
-36.55
Principal
plane (°)
theta
24.37
0
Shear plane
(°)
Smax
Smin
Tmax
26.6
-18.4
12.18
-48.74
30.46
Question 9.1
Question 9.2
49
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 9.3
Question 9.4
a)
b)
Question 9.5
40 MPa
Question 10.1
I/
(a) 1.228
(b) 1.098
(c) Yielding occurs
II/
(a) 1.083
(b) Yielding occurs
(c) Yielding occurs
Question 10.2
T=708 kN.m
Question 10.3
Rupture will occur
50
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Question 10.4
T=196.9 N·m
Question 11.1
(εx=725μ, εy=-75μ, γxy=173.2μ);
a) εmin=-84.3μ, εmax=734μ
b) γmax=819μ
Question 11.2
a) εx=40μ, εy=860μ, γxy=750μ
b) εa=-106μ, εb=1006μ, εb=-368μ
c) γmax=1374μ
Question 11.3
P=1.421MPa
51
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Detailed Video Solutions to the Weekly Questions
0.1 https://youtu.be/8QIJqcMLsuU
0.2 https://youtu.be/5_6PekaW1Ek
0.3 https://youtu.be/RHOfIeG5QB8
0.4 https://youtu.be/Nqshf1O-dcY
0.5 https://youtu.be/_YdlKNDkSbw
0.6 https://youtu.be/R0eQQENczt4
0.7 https://youtu.be/z27mJdsutBg
1.1 https://youtu.be/dNoLdw8wQjY or https://youtu.be/CQNyz6pzj6o
1.2 https://youtu.be/wG4hxoEJPX4 or (https://youtu.be/l7iYPxKHWAc and
https://youtu.be/nIbHnRY2llg )
1.3 https://youtu.be/DQp85y5h5Gs or https://youtu.be/lU7om0ASeQc
1.4 https://youtu.be/MoSlm7KkGf8 or https://youtu.be/dwThzHAZC14
1.5 https://youtu.be/2QsCVl5b--k or https://youtu.be/xYWLuQNSICk
2.1 https://youtu.be/4qYS5iAMTNY or https://youtu.be/jpzWqk4s0j0
2.2 https://youtu.be/zRn5AFW9YfM or https://youtu.be/Rx2dz7GiTzk
2.3 https://youtu.be/s3wDx30WHAw or https://youtu.be/k9-YVhkTnC8
2.4 https://youtu.be/0Em0yQAsfgk or https://youtu.be/gTcNLmHqrKk
2.5 https://youtu.be/Adqf3D_0aM8 or https://youtu.be/z_4-Ez8tOL0
3.1 https://youtu.be/AY9EI5oAfgs or https://youtu.be/DR2F9bTaeRc
3.2 https://youtu.be/bw5Ia6KDGys or https://youtu.be/3sOdBlVg99U
3.3 https://youtu.be/OLMA1w6NIuQ or https://youtu.be/PNCGiH3JYGY
3.4 https://youtu.be/-y9MvYxLc4Q or https://youtu.be/8LWEOfals64
3.5 https://youtu.be/Vd6eZDQiWHY or (https://youtu.be/7MkU1NYCiGM and
https://youtu.be/9iZ81MSlipk )
3.6 https://youtu.be/gyg_ouy_PzQ
3.7 https://youtu.be/cZ9O8od92Z0
3.8 Extra problems done with Macauley's method (including MDSolid primer on Beams),
https://youtube.com/playlist?list=PLfFSXDirJ0XEzjwGfJ7dCnqiMzziLrSWi
4.1 https://youtu.be/zHgZGYb93KQ or https://youtu.be/Cdpk5gU9aOg
4.2 https://youtu.be/oZsISDWL6mE or https://youtu.be/0PCFkIVn57g
4.3 https://youtu.be/TG5dHv-LG3s or https://youtu.be/QZdrtR81zaw
4.4
or https://youtu.be/_XZPB7LGqUM
4.5
or https://youtu.be/QTutDHFFwWg
4.6 Bonus videos on buckling by Iain Barton,
https://www.youtube.com/playlist?list=PL4hAxbg-uFAi6Spc0e0xT243oo9GlOmGd
52
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
5.1 https://youtu.be/PaUC1eelxXk or https://youtu.be/dVSE-sUb-7s
5.2 https://youtu.be/eRNTwsU5cLs or https://youtu.be/8eTFRvQ6SSQ
5.3 https://youtu.be/9fpVl-TloWQ or https://youtu.be/PmAWTSPypfI
5.4 https://youtu.be/3Z4HveXzhwM or https://youtu.be/YmRYwUJ2nx4
5.5 https://youtu.be/jQEDldeDjXs or https://youtu.be/cfurY_9qUrc
5.6 Bonus videos on shear stress by Iain Barton,
https://www.youtube.com/playlist?list=PL4hAxbg-uFAiLdvyZ5eEMtHzNrG8craVh
6.1 https://youtu.be/LIbfFbzca4k or https://youtu.be/TB6JBhFitGY
6.2 https://youtu.be/WSYC305ME7c or https://youtu.be/QiFr00IgVBg
6.3 https://youtu.be/1cp8R-O3qQA or https://youtu.be/Jq6-ogp1gpc
6.4 https://youtu.be/J61td5-RoMw or https://youtu.be/i1s67Pusmvo
6.5 https://youtu.be/XCwoDb7JdMU or https://youtu.be/i8STll_drQQ
6.6 Bonus videos on shear stress by Iain Barton,
https://www.youtube.com/playlist?list=PL4hAxbg-uFAgh2yQnyRQ1FegpgC1nCi3u
7.1 https://youtu.be/xZ5fCNbOAhw or https://youtu.be/MLwsUjvR4hE
7.2 https://youtu.be/41HDVSu3VpA or (https://youtu.be/cGnSd4arR0w and
https://youtu.be/B_mkGr40vnE )
7.3 https://youtu.be/OkIZiMky8Dg or https://youtu.be/cICw1OPPaoo
7.4 https://youtu.be/_zKYs3He2k0 or https://youtu.be/T4T20o-C70s
7.5 https://youtu.be/kvXhfiU6IyA or (https://youtu.be/43KWhuhMVeA and
https://youtu.be/yYta84lirK8 )
7.6
Or https://youtu.be/PwgvX8EaCGg
7.7 Bonus video on curved beams by Iain Barton, https://youtu.be/3b-UnsYpTG8
8.1 https://youtu.be/Wt1FbrfjEsE or https://youtu.be/xicw1qA1S7s
8.2 https://youtu.be/TO4c5FmOaDs or https://youtu.be/QGpydfP6Aq0
8.3 https://youtu.be/04xZ7F_Y8dY or https://youtu.be/A4ScAMbFfbU
8.4 https://youtu.be/Lwhw-V6gGt8 or https://youtu.be/z7h1ypXKeKc
8.5
or https://youtu.be/VeXBke6P79w
8.6
or https://youtu.be/Z_-d0IoUa-Y
8.7 Bonus video on stress transformation by Iain Barton, https://youtu.be/W8AodzvlXmg
9.1 https://youtu.be/XlGmO6EX_x4
9.2 https://youtu.be/DQBwstcgcRI
9.3 https://youtu.be/oEhu8ZPFy14
9.4 https://youtu.be/GmD8PvnLGao
9.5 https://youtu.be/6W5Ne4zDbeI
10.1
10.2
https://youtu.be/JDsUEn5si_4
https://youtu.be/9C-brw5BaKI
53
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
10.3
10.4
https://youtu.be/Hd0m8dVmRRk
https://youtu.be/9MQa_wVMqAg
11.1
11.2
https://youtu.be/fh8Ijq0yOHQ
https://youtu.be/GvfYgn2l2Ks
54
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Detailed Solutions to Exam Questions)
Exam Question I.1 (2019-2020)
A brass stepped bar (Young’s modulus E = 100 GPa) is subjected to two loads, as depicted in
Figure I.1-a). Section [AB], [BC] and [BD] have cross-section areas of 500 mm2, 1000 mm2,
and 1200 mm2 respectively.
Figure I.1
a) Calculate the support reaction at D.
Static: RD = 50 kN
b) Calculate the internal forces and stresses in sections [AB], [BC].and [CD].
Forces (positive upward):[AB] 0 kN, [BC] +50 kN, [AB] -50 kN (BC obviously in
tension, check) (note that starting top to bottom give this directly. Starting bottom to
top would give opposite values, which must be corrected to account for sign
convention, -compression/+ tension). This is the only vaguely subtle point here.
Stresses:[AB] 0 MPa, [BC] 50.0 MPa, [CD] -41.7 MPa
c) Calculate the displacement at point D
𝐹𝐿
Standard problem with 𝛿 = ∑𝑖 𝐸𝑖𝐴𝑖 .
𝑖 𝑖
Finally, 𝛿𝐷 = −0.1167 mm
The bar is now supported at D, as depicted in Figure I.1-b.
d) Calculate the support reactions at A and D.
Classic statically indeterminate 1D problem.
From static equilibrium: 𝑅𝐴 + 𝑅𝐷 − 50 = 0
Not enough, we need a second equation. As the displacement is 0 at A (constrained)
𝐹𝐿
, we can write 𝛿𝐴 = ∑𝑖 𝐸 𝑖𝐴𝑖 = ∑𝑖 𝐹𝑖 𝑘𝑖 = 0, which simplifies to 𝑘1 (𝑅𝐴 ) + 𝑘2 (𝑅𝐴 + 50) +
𝑖 𝑖
𝑘3 (𝑅𝐴 − 50) = 0
This is a trivial equation, and finally,𝑅𝐴 = 18.4 kN and 𝑅𝐷 = 31.6 kN
e) Calculate the internal forces and stresses in sections [AB], [BC].and [CD].
Forces (positive upwards): [AB] 18.4 kN, [BC] +68.4 kN, [CD] –31.6 kN (BC
obviously in tension, check)
Stresses :[AB] 36.8 MPa, [BC] 68.4 MPa, [CD] -26.3 MPa
55
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Exam Question I.2 (2018-2019)
A 20 mm diameter steel bar (Young’s modulus E = 200 GPa) is subjected to two loads, as
depicted in Figure I.2-a.
Figure I.2
a) Calculate the support reaction at A.
Static: Ra = -250 kN
b) Calculate the internal forces and stresses in sections [AB], [BC].and [CD].
Forces (positive to the right):[CD] 0 kN, [BC] +400 kN, [AB] 250 kN (BC obviously in
tension, check) (note that starting right to left give this directly. Starting left to right
would give opposite values, which must be corrected to account for sign convention,
-compression/+ tension). This is the only vaguely subtle point here.
Do not forget the stresses:[CD] 0 MPa, [BC] 1273 MPa, [AB] 796 MPa
c) Calculate the displacement at point D
𝐹𝐿
Standard problem with 𝛿 = ∑𝑖 𝐸𝑖𝐴𝑖 .
𝑖 𝑖
Lengths: all 180 mm
Areas: all 314 mm2
Young’s moduli: all 200 GPa
Finally, 𝛿𝐷 = 1.86 mm
The bar is now supported at D, as depicted in Figure Q1-b.
d) Calculate the support reactions at A and D.
Classic statically indeterminate 1D problem.
From static equilibrium: 𝑅𝐴 + 𝑅𝐷 + 250 = 0
Not enough, we need a second equation. As the displacement is 0 at D (constrained)
𝐹𝐿
, we can write 𝛿𝐷 = ∑𝑖 𝐸 𝑖𝐴𝑖 = 0, which simplifies to (𝑅𝐷 ) + (𝑅𝐷 + 400) + (𝑅𝐷 + 250) =
𝑖 𝑖
0 (same E, L and A)
This is a trivial equation, and finally,𝑅𝐷 = −216.6 kN and 𝑅𝐴 = −33.3 kN
e) Calculate the internal forces and stresses in sections [AB], [BC].and [CD].
56
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Forces (positive to the right):[CD] -216.6 kN, [BC] +183.3 kN, [AB] 33.3 kN (CD
obviously in compression, check)
Stresses:[CD] -690 MPa, [BC] 584 MPa, [AB] 106.1 MPa
Exam Question I.3 (resit 2018-2019)
An aluminium prismatic bar (Young’s modulus E = 70 GPa) is subjected to two loads, as
depicted in Figure I.3-a.
Figure I.3
a) Calculate the support reaction at A.
Static: Ra = - 120 kN (sign‼!)
b) Calculate the internal forces and stresses in sections [AB], [BC]
Forces (positive to the top): [BC] +90 kN, [AB] 120 kN (AB obviously in tension,
check) (note that starting top to bottom give this directly. Starting bottom to top would
give opposite values, which must be corrected to account for sign convention, compression/+ tension). This is the only vaguely subtle point here.
Stresses [BC] 360 MPa, [AB] 343 MPa
c) Calculate the displacement at point C
𝐹𝐿
Standard problem with 𝛿 = ∑𝑖 𝐸𝑖𝐴𝑖 .
𝑖 𝑖
Finally, 𝛿𝐷 = 3.01 mm
The uppermost load is now replaced by a support at C, as depicted in Figure Q1-b.
d) Calculate the support reactions at A and C.
Classic statically indeterminate 1D problem.
From static equilibrium: 𝑅𝐴 + 𝑅𝐶 + 30 = 0
Not enough, we need a second equation. As the displacement is 0 at C
𝐹𝐿
(constrained), we can write 𝛿𝐶 = ∑𝑖 𝐸 𝑖𝐴𝑖 = 0, which simplifies to
𝑖 𝑖
E and L)
𝑅𝐶
+
250
𝑅𝐶 +30
450
= 0 (same
This is a trivial equation, and finally,𝑅𝑐 = −10.7 kN and 𝑅𝐴 = −19.3 kN
57
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
e) Calculate the internal forces and stresses in sections [AB] and [BC].
Forces (positive upwards): [BC] -10.7 kN, [AB] 19.3 kN
compression, check)
(BC obviously in
Stresses: [BC] -42.85 MPa, [AB] 42.85 MPa
Exam Question I.4
The rigid bar EFGH in Fig. I.4 is suspended from four identical wires of length L, Youngs
modulus E and cross section area A. (Note that this is a statically indeterminate problem)
Figure I.4
a) How many support reactions are unknown?
4 unknowns
b) How many independent equations can you write using static equilibrium only?
2D, but nothing along y, so only 2 equations
c) Write these “static equilibrium equations
For instance
𝑅𝐴 + 𝑅𝐵 + 𝑅𝐶 + 𝑅𝐷 = 𝑃 (sum forces F=0)
And
2𝑅𝐴 + 2𝑅𝐵 + 0 − 𝑅𝐷 = 0 (moments at C)
d) Write enough “compatibility” equations to have a solvable system (hint: the displacements
at points E, F, G, and H are related as the bar is rigid and remains straight, but not horizontal)
𝛿𝐷 − 𝛿𝐴 = 3(𝛿𝐵 − 𝛿𝐴 ) and 𝛿𝐶 − 𝛿𝐴 = 2(𝛿𝐵 − 𝛿𝐴 )
And
2𝑅𝐴 − 3𝑅𝐵 + 0 + 𝑅𝐷 = 0 and 𝑅𝐴 − 2𝑅𝐵 + 𝑅𝐶 + 0 = 0
e) Solve the system of equation to calculate the reactions at the support as functions of the
load P
Whichever methods of solution, Gauss pivot works well here.
𝑃
2𝑃
3𝑃
4𝑃
𝑅𝐴 =
, 𝑅𝐵 =
, 𝑅𝐶 =
, 𝑅𝐷 =
10
10
10
10
Exam Question II.1 (2019-2020)
A composite beam (cross section depicted in Figure II.1) is made of an aluminium foam core
(grey, E = 10 GPa) and aluminium skin (white, E = 70 GPa).
58
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Figure II.1
a) Choose a reference material and calculate a composite beam ratio 𝑛.
Easiest is to use foam as reference: 𝑛 = 70/10 = 7. (Inverse would work too)
b) Calculate the second moments of area 𝐼𝑦𝑦 and 𝐼𝑧𝑧 .
Standard methods for 2nd moment of area, replacing the skin by its equivalent “7 time
stretched”. And
7
1
𝐼𝑦𝑦 = 12 ∗ [704 − 58 ∗ 503 ] + 12 ∗ [58 ∗ 503 ] = 9.17 × 106 mm−4
[704 − 50 ∗ 583 ] +
1
∗ [50 ∗
12
583
and
7
𝐼𝑧𝑧 = 12 ∗
] = 7.50 × 106 mm−4
c) The beam is subjected to a positive bending moment 𝑀𝑦 of 3 kN·m (top in tension, bottom
in compression). Calculate the minimum and maximum values of stress in the core and in the
skin (four values).
Bending equation around yy: 𝜎 = 𝑛
𝑀𝑦 ×𝑧
𝐼𝑦𝑦
In the core, n=1, y=+/-25mm and we get 𝜎𝑚𝑎𝑥 = ±8.17 MPa
In the skin, n=7, y=+/-35mm and we get 𝜎𝑚𝑎𝑥 = ±80.1MPa
d) The beam is subjected to a positive bending moment 𝑀𝑧 of 5 kN·m (left in tension, right in
compression). Calculate the minimum and maximum values of stress in the core and in the
skin (four values).
Bending equation around zz: 𝜎 =
𝑀𝑧 ×𝑦
𝐼𝑧𝑧
In the core, n=1, z=+/-25mm and we get 𝜎𝑚𝑎𝑥 = ±19.32 MPa
In the skin, n=7, z=+/-35mm and we get 𝜎𝑚𝑎𝑥 = ±163.3MPa
Exam Question II.2 (2018-2019)
A beam of cross section as depicted in Figure II.2 is subjected to a bending moment of 30 kN,
acting at 25° from the y axis, and passing through the centre of mass of the cross-section.
59
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Figure II.2
a) Determine the position of the centre of mass and of the the neutral axes y-y and z-z
Reference at bottom (for z) and centre (for y, by symmetry) of beam: 𝑧̅ = ∑𝑖
𝑧𝑖 𝐴𝑖
𝐴𝑖
=
47.2 mm
b) Calculate the second moments of area 𝐼𝑦𝑦 and 𝐼𝑧𝑧 .
Standard methods for 2nd moment of area and 𝐼𝑧𝑧 = 0.601 × 106 mm−4 and 𝐼𝑦𝑦 =
2.96 × 106 mm−4
c) Calculate the moments 𝑀𝑦 and 𝑀𝑧 .
Project moment on axes y and z: 𝑀𝑦 = 4.53 kNm and 𝑀𝑧 = 2.11 kNm (both positive)
d) Calculate the maximum tensile and compressive stresses. Indicate their location on a
sketch.
Bending equation around y: 𝜎 =
𝑀𝑦 ×𝑧
𝐼𝑦𝑦
(using right hand rule the top should be in
tension for this moment, and this formula does indeed gives tension when z is
positive –top–)
Bending equation around z: 𝜎 = −
𝑀𝑧 ×𝑦
𝐼𝑧𝑧
(using right hand rule the left should be in
tension, the – sign must be added so that the formula gives tension is when y is
negative –left–)
For point NW (top-left) (-30, +42.8), combining the two stresses, we get 𝜎𝑚𝑎𝑥 =
170.9 MPa
For point SE (bottom right) (+25, -47.2), combining the two stresses, we get 𝜎𝑚𝑖𝑛 =
−160.0 MPa
e) Calculate the angle of the neutral axis for this moment. Place the neutral axis on the sketch
started in d).
𝑀𝑦 ×𝐼𝑧𝑧
𝑡𝑎𝑛𝜃 = − 𝐼
𝑦𝑦 ×𝑀𝑧
and 𝜃 = 23.54°
Exam Question II.3 (resit 2018-2019)
A composite beam (cross section depicted in Figure II.3-a) is made of brass (strip A) (E =
101 GPa) and mild steel (strip B) (E = 210 GPa).
60
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Figure II.3
Questions a)-c) refer to Figure II.3-a).
a) Calculate the position of the neutral axis.
Reference at bottom (for z) of beam. Replace top strip by a wider one made of brass
(n = 210/101 = 2.08): 𝑧̅ =
∑𝑖 𝑧𝑖 𝐴𝑖
∑𝑖 𝐴𝑖
= 40.89 mm
b) Calculate the second moment of area of the composite beam 𝐼𝑦𝑦 .
Standard methods for 2nd moment of area (still with modified top strip) and 𝐼𝑦𝑦 =
5.94 × 106 mm−4
c) If the beam is subjected to a positive bending moment 𝑀𝑦 of 10 kN·m, calculate the
minimum and maximum values of stress in the steel and in the brass (four values).
Right hand rule, so top in tension, bottom in compression. Just use bending equation
around y: 𝜎 =
𝑀𝑦 ×𝑧
𝐼𝑦𝑦
with × 𝑛 correction in steel strip, and 𝜎𝑚𝑖𝑛/𝑏𝑟𝑎𝑠𝑠 = −68.8 MPa
𝜎𝑚𝑎𝑥/𝑏𝑟𝑎𝑠𝑠 = 15.3 MPa 𝜎𝑚𝑖𝑛/𝑠𝑡𝑒𝑒𝑙 = 30.7 MPa 𝜎𝑚𝑎𝑥/𝑠𝑡𝑒𝑒𝑙 = 101.8 MPa
The height of the steel strip is now a variable, ℎ, as shown in Figure II.3-b).
d) Determine the height ℎ of the steel strip so that the neutral axis of the beam is located at
the seam of the two metals.
Reference at bottom (for z) of beam. Replace top strip by a wider one made of brass
(n = 210/101 = 2.08): force
𝑧̅ = 𝑧̅ =
∑𝑖 𝑧𝑖 𝐴𝑖
∑𝑖 𝐴𝑖
= 50, (or balance of forces) and ℎ =
34.7 mm
e) For ℎ as calculated in d), calculate the maximum bending moment this beam can support if
the allowable bending stress is 250 MPa for the steel and 55 MPa for the brass.
Standard methods for 2nd moment of area (with modified top strip) and 𝐼𝑦𝑦 = 10.58 ×
106 mm−4
Invert bending equation around y: 𝑀𝑦 =
𝜎×𝐼𝑦𝑦
𝑧
In brass (bottom), 𝑀𝑦 = 11.64 kN ∙ m n steel (top), 𝑀𝑦 = 76 kN ∙ m So : 𝑀𝑚𝑎𝑥 =
11.64 kN ∙ m
61
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Exam Question III.1 (2019-2020)
A simply supported beam (Young’s modulus 𝐸, second moment of area 𝐼) is supported and
loaded as shown in Figure III.1.
Figure III.1. Simply supported beam
a) Calculate the reactions at the supports in term of the uniformly distributed load 𝑤 and
beam length 𝐿.
1
3
Trivial statics, 𝑅𝐴 = 8 𝑤𝐿, 𝑅𝐶 = 8 𝑤𝐿
b) Determine the bending moment between A and B as a function of 𝑤, 𝑥 and 𝐿.
1
[A-B] 𝑀 = + 8 𝑤𝐿𝑥
c) Determine the bending moment between B and C as a function of 𝑤, 𝑥 and 𝐿.
𝐿 2
1
1
[B-C] 𝑀 = − 2 𝑤 (𝑥 − 2) + 8 𝑤𝐿𝑥
d) Express the slope at any point of the beam as functions of 𝐸, 𝐼, 𝑤, 𝑥 and 𝐿 (no need to
solve the integration constants at this stage).
𝑑𝑦
1
[A-B] 𝐸𝐼 𝑑𝑥 = 16 𝑤𝐿𝑥 2 + 𝑀1
𝑑𝑦
𝐿 3
1
1
[B-C] 𝐸𝐼 𝑑𝑥 = − 6 𝑤 (𝑥 − 2) + 16 𝑤𝐿𝑥 2 + 𝑀3
e) Express the deflection at any point of the beam as functions of 𝐸, 𝐼, 𝑤, 𝑥 and 𝐿 (no need
to solve the integration constants at this stage).
1
[A-B] 𝐸𝐼𝑦 = 48 𝑤𝐿𝑥 3 + 𝑀1 𝑥 + 𝑀2
𝐿 4
1
1
[B-C] 𝐸𝐼𝑦 = − 24 𝑤 (𝑥 − 2) + 48 𝑤𝐿𝑥 3 + 𝑀3 𝑥 + 𝑀4
f) Calculate all the integration constants as functions of 𝐿 and and 𝑤, and write the functions
for the deflection.
Important note: The exact values of the integration constant depend on the exact
choice of functions. Here for instance, we chose not to develop the power term for
the moment in segment [BC]
Fixed at A: 𝑀2 = 0
𝐿 4
1
1
Fixed at C:0 = − 24 𝑤 (𝐿 − 2) + 48 𝑤𝐿4 + 𝑀3 𝐿 + 𝑀4
7
𝑀3 𝐿 + 𝑀4 = − 24∗16 𝑤𝐿4
1
𝐿 3
𝐿
1
𝐿
𝐿 4 1
Continuous at B: 48 𝑤𝐿 (2) + 𝑀1 2 = − 24 𝑤 (2 − 2)
𝐿 3
𝐿
𝑤𝐿 (2) + 𝑀3 2 − 𝑀4 = 0
48
Continuous derivative at B: 𝑀1 −𝑀3 = 0
and:
7
𝑀4 = 𝑀2 = 0 𝑀1 = 𝑀3 = − 24∗16 𝑤𝐿3
62
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
1
7
[A-B] 𝐸𝐼𝑦 = 48 𝑤𝐿𝑥 3 − 24∗16 𝑤𝐿3 𝑥
𝐿 4
1
1
7
[B-C] 𝐸𝐼𝑦 = − 24 𝑤 (𝑥 − 2) + 48 𝑤𝐿𝑥 3 − 24∗16 𝑤𝐿3 𝑥
Using Macauley’s method instead as it’s so compact and efficient:
𝐿
𝑤(𝑥) = 𝑅𝐴 ⟨𝑥 − 0⟩−1 − 𝑤 ⟨𝑥 − ⟩0
2
𝐿
𝑉(𝑥) = 𝑅𝐴 ⟨𝑥 − 0⟩0 − 𝑤 ⟨𝑥 − ⟩1
2
𝑤
𝐿
𝑀(𝑥) = 𝑅𝐴 ⟨𝑥 − 0⟩1 − ⟨𝑥 − ⟩2
2
2
𝑑𝑦
𝑅𝐴
𝑤
𝐿
⟨𝑥 − 0⟩2 − ⟨𝑥 − ⟩3 + 𝐶1
𝐸𝐼
(𝑥) =
𝑑𝑥
2
6
2
𝑅𝐴
𝑤
𝐿
⟨𝑥 − 0⟩3 − ⟨𝑥 − ⟩4 + 𝐶1 𝑥 + 0
𝐸𝐼𝑦(𝑥) =
6
24
2
The boundary condition at A (fixed) is trivial and immediately leads to 𝐶2 = 0 (𝑥 = 0)
Boundary condition at C (fixed)
𝑅𝐴 3 𝑤𝐿
𝐿
⟨L⟩ −
⟨𝐿 − ⟩4 + 𝐶1 𝐿 + 0
6
24
2
𝑤𝐿 3
𝑤
𝐶1 𝐿 = −
𝐿 +
𝐿4
6×8
24 × 16
7
𝐶1 = −
𝑤𝐿3
384
0=
Exam Question III.2 (2018-2019)
A simply supported beam (Young’s modulus E, second moment of area I) is supported and
loaded as shown in Fig. III.2.
Figure III.2
a) Calculate the reactions at the support as functions of L and P
1
2
Trivial statics, 𝑅𝐴 = 3 𝑃, 𝑅𝐶 = 3 𝑃
b) Determine the bending moment between A and B
1
[A-B] 𝑀 = 3 𝑃𝑥
c) Determine the bending moment between B and C
63
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
1
[B-C] 𝑀 = 3 𝑃𝑥 − 𝑃 (𝑥 −
2𝐿
3
)
d) Express the slope at any point of the beam (no need to elucidate the integration constants
at this stage)
𝑑𝑦
1
𝑑𝑦
1
[A-B] 𝐸𝐼 𝑑𝑥 = 6 𝑃𝑥 2 + 𝑀1
1
[B-C] 𝐸𝐼 𝑑𝑥 = 6 𝑃𝑥 2 − 2 𝑃 (𝑥 −
2𝐿 2
) + 𝑀3′
3
e) Express the deflection at any point of the beam (no need to elucidate the integration
constants at this stage)
[A-B] 𝐸𝐼𝑦 =
1
18
𝑃𝑥 3 + 𝑀1 𝑥 + 𝑀2
1
1
[B-C] 𝐸𝐼𝑦 = 18 𝑃𝑥 3 − 6 𝑃 (𝑥 −
2𝐿 3
3
) + 𝑀3′ 𝑥 + 𝑀4′
f) Calculate all the integration constants as functions of L, P, E and I
Fixed at A:
𝑀2 = 0
Fixed at C:
𝑀3′ 𝐿 + 𝑀4′ = − 81 𝑃𝐿3
Continuous at B:
4
2𝐿
2𝐿
𝑀1 ( 3 ) = 𝑀3′ ( 3 ) + 𝑀4′
, or 𝑀4′ = 0 (using the next boundary condition)
Continuous derivative at B: 𝑀1 = 𝑀3′
4
Solves quite easily, with 𝑀1 = 𝑀3′ = − 81 𝑃𝐿2 and 𝑀2 = 𝑀4′ = 0
This approach is essentially the same as Macauley’s method. The trick is to not simplify
the moment after B in x, but too keep powers of (𝑥 −
2𝐿
3
).
Macauley’s is even more compact:
1
2
Trivial statics, 𝑅𝐴 = 3 𝑃, 𝑅𝐶 = 3 𝑃
2𝐿 −1
⟩
3
2𝐿
𝑉(𝑥) = 𝑅𝐴 ⟨𝑥 − 0⟩0 − 𝑃 ⟨𝑥 − ⟩0
3
2𝐿
𝑀(𝑥) = 𝑅𝐴 ⟨𝑥 − 0⟩1 − 𝑃 ⟨𝑥 − ⟩1
3
𝑑𝑦
𝑅𝐴
𝑃
2𝐿
⟨𝑥 − 0⟩2 − ⟨𝑥 − ⟩2 + 𝐶1
𝐸𝐼
(𝑥) =
𝑑𝑥
2
2
3
𝑅𝐴
𝑃
2𝐿
⟨𝑥 − 0⟩3 − ⟨𝑥 − ⟩3 + 𝐶1 𝑥 + 0
𝐸𝐼𝑦(𝑥) =
6
6
3
𝑤(𝑥) = 𝑅𝐴 ⟨𝑥 − 0⟩−1 − 𝑃 ⟨𝑥 −
Boundary condition at C (fixed)
𝑅𝐴 3 𝑃 𝐿 3
⟨L⟩ − ⟨ ⟩ + 𝐶1 𝐿 + 0
6
6 3
P 2
P
𝐶1 = −
𝐿 +
𝐿2
3×6
6 × 27
4
𝐶1 = − 𝑃𝐿2
81
0=
64
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
It is also possible to “simplify” the moment function, but this usually a terrible idea,
because the functions before and after the discontinuities will have different integration
constants. See how tedious it can become below.
b) Determine the bending moment between A and B
1
[A-B] 𝑀 = 3 𝑃𝑥
c) Determine the bending moment between B and C
2
2
[B-C] 𝑀 = − 3 𝑃𝑥 + 3 𝑃𝐿
d) Express the slope at any point of the beam (no need to elucidate the integration constants
at this stage)
𝑑𝑦
1
[A-B] 𝐸𝐼 𝑑𝑥 = 6 𝑃𝑥 2 + 𝑀1
𝑑𝑦
1
2
[B-C] 𝐸𝐼 𝑑𝑥 = − 3 𝑃𝑥 2 + 3 𝑃𝐿𝑥 + 𝑀3
e) Express the deflection at any point of the beam (no need to elucidate the integration
constants at this stage)
1
[A-B] 𝐸𝐼𝑦 = 18 𝑃𝑥 3 + 𝑀1 𝑥 + 𝑀2
1
1
[B-C] 𝐸𝐼𝑦 = − 9 𝑃𝑥 3 + 3 𝑃𝐿𝑥 2 + 𝑀3 𝑥 + 𝑀4
f) Calculate all the integration constants as functions of L, P, E and I
Fixed at A:
𝑀2 = 0
Fixed at C:
𝑀3 𝐿 + 𝑀4 = − 9 𝑃𝐿3
Continuous at B:
2
3
4
𝑀1 𝐿−𝑀3 𝐿 − 2 𝑀4 = 27 𝑃𝐿3
2
Continuous derivative at B: 𝑀1 −𝑀3 = 9 𝑃𝐿2
Solve system and:
4
𝑀4 = − 81 𝑃𝐿3
14
𝑀3 = − 81 𝑃𝐿2
4
𝑀1 = 81 𝑃𝐿2
Exam Question III.3 (resit 2018-2019)
A cantilevered beam (Young’s modulus 𝐸, second moment of area 𝐼) is supported and loaded
as shown in Figure III.3.
Figure III.3
a) Calculate the reactions at the supports 𝑅𝐴 and 𝑀𝐴 in term of the load 𝑤 and the length 𝐿.
65
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
1
1
Trivial statics, 𝑅𝐴 = 4 𝑤𝐿, 𝑀𝐴 = − 32 𝑤𝐿2
b) Determine the bending moment between A and B as a function of 𝑤, 𝑥 and 𝐿. (do not
forget to include the effect of the reactions at A!)
1
1
1
[A-B] 𝑀 = − 2 𝑤𝑥 2 + 4 𝑤𝐿𝑥 − 32 𝑤𝐿2
c) Determine the bending moment between B and C.
1
𝐿
1
1
[B-C] 𝑀 = − 4 𝑤𝐿 (𝑥 − 8) + 4 𝑤𝐿𝑥 − 32 𝑤𝐿2 = 0, yes 0! Quite obvious really, as on the
right of the last load.
A bit of a choice to be made here, by simplifying the moment function, we break the
link at point B, and the integration constants at the left and at the right will be
different… But it’s hard to resist a 0.
d) Express the slope at any point of the beam as functions of 𝐸, 𝐼, 𝑤, 𝑥 and 𝐿 (no need to
solve the integration constants at this stage).
𝑑𝑦
1
1
1
[A-B] 𝐸𝐼 𝑑𝑥 = − 6 𝑤𝑥 3 + 8 𝑤𝐿𝑥 2 − 32 𝑤𝐿2 𝑥 + 𝑀1
𝑑𝑦
[B-C] 𝐸𝐼 𝑑𝑥 = 𝑀3
e) Express the deflection at any point of the beam as functions of 𝐸, 𝐼,𝑤, 𝑥 and 𝐿 (no need to
solve the integration constants at this stage).
[A-B] 𝐸𝐼𝑦 = −
1
24
𝑤𝑥 4 +
1
24
𝑤𝐿𝑥 3 −
1
32
𝑤𝐿2 𝑥 2 + 𝑀1 𝑥 + 𝑀2
[B-C] 𝐸𝐼𝑦 = 𝑀3 𝑥 + 𝑀4
f) Calculate all the integration constants as functions of 𝐿 and and 𝑃.
Fixed at A: 𝑀2 = 0
Fixed at A: 𝑀1 = 0
Continuous derivative at B: 𝑀3 =
−𝑤𝐿3
384
13𝑤𝐿4
Continuous at B: 𝑀4 = 24∗256
Using Macauley’s method instead is still more efficient, even if it obscures that 0. There
is also a bit of a fudge to account for the UDL with TWO step functions.
𝐿
𝑤(𝑥) = 𝑀𝐴 ⟨𝑥 − 0⟩−2 + 𝑅𝐴 ⟨𝑥 − 0⟩−1 − 𝑤⟨𝑥 − 0⟩0 + 𝑤 ⟨𝑥 − ⟩0
4
𝐿
𝑉(𝑥) = 𝑀𝐴 ⟨𝑥 − 0⟩−1 + 𝑅𝐴 ⟨𝑥 − 0⟩0 − 𝑤⟨𝑥 − 0⟩1 + 𝑤 ⟨𝑥 − ⟩1
4
𝑤
𝑤
𝐿
𝑀(𝑥) = 𝑀𝐴 ⟨𝑥 − 0⟩0 + 𝑅𝐴 ⟨𝑥 − 0⟩1 − ⟨𝑥 − 0⟩2 + ⟨𝑥 − ⟩2
2
2
4
𝑑𝑦
𝑅
𝑤
𝑤
𝐿
𝐴
𝐸𝐼
(𝑥) = 𝑀𝐴 ⟨𝑥 − 0⟩1 + ⟨𝑥 − 0⟩2 − ⟨𝑥 − 0⟩3 + ⟨𝑥 − ⟩3 + 𝐶1
𝑑𝑥
2
6
6
4
𝑀𝐴
𝑅
𝑤
𝑤
𝐿
𝐴
⟨𝑥 − 0⟩2 + ⟨𝑥 − 0⟩3 −
⟨𝑥 − 0⟩4 + ⟨𝑥 − ⟩4 + 𝐶1 𝑥 + 𝐶2
𝐸𝐼𝑦(𝑥) =
2
6
24
24
4
The boundary condition at A (fixed) is trivial and immediately leads to C1 = 0 and
𝐶2 = 0 (𝑥 = 0)
66
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
At first, it seems hard to believe that these functions simplify to constants and affine
functions right of B, but they do!
1
1
𝑤
1
1
𝑤
[A-B] 𝑀(𝑥) = − 32 𝑤𝐿2 + 4 𝑤𝐿𝑥 − 2 x2
𝑤
𝐿 2
[B-C] 𝑀(𝑥) = − 32 𝑤𝐿2 + 4 𝑤𝐿𝑥 − 2 𝑥2 + 2 (𝑥 − 4) = 0 !!!
And so on for the slope and the deflection.
Powerful method, even if the notation might sometimes obscure simplifications, once
developed.
Exam Question III.4 (2017-2018)
A simply supported beam (Young’s modulus E, second moment of area I) is supported and
loaded as shown in Fig. III.4.
Figure III.4
a) Calculate the reactions at the support as functions of L and P
3
1
Trivial statics, 𝑅𝐴 = 4 𝑃, 𝑅𝐶 = 4 𝑃
b) Determine the bending moment between A and B
3
[A-B] 𝑀 = 4 𝑃𝑥
c) Determine the bending moment between B and C
3
1
1
1
[B-C] 𝑀 = 4 𝑃𝑥 − 𝑃 (𝑥 − 4 𝐿) = − 4 𝑃𝑥 + 4 𝑃𝐿
It’s actually more efficient in the long run to use the first form of the function! So we
will do this way first (purple)
d) Express the slope at any point of the beam (no need to elucidate the integration constants
at this stage)
𝑑𝑦
3
𝑑𝑦
3
[A-B] 𝐸𝐼 𝑑𝑥 = 8 𝑃𝑥 2 + 𝐶1
1
2
1
[B-C] 𝐸𝐼 𝑑𝑥 = 8 𝑃𝑥 2 − 2 𝑃 (𝑥 − 4 𝐿) + 𝐶3
e) Express the deflection at any point of the beam (no need to elucidate the integration
constants at this stage)
1
[A-B] 𝐸𝐼𝑦 = 8 𝑃𝑥 3 + 𝐶1 𝑥 + 𝐶2
1
1
1
3
[B-C] 𝐸𝐼𝑦 = 8 𝑃𝑥 3 − 6 𝑃 (𝑥 − 4 𝐿) + 𝐶3 𝑥 + 𝐶4
f) Calculate all the integration constants as functions of L, P, E and I
Fixed at A: 𝐶2 = 0
Continuous derivative at B: 𝐶3 = 𝐶1
Continuous at B: 𝐶4 = 𝐶2 = 0
67
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
7
Fixed at C: 128 𝑃𝐿3 + 𝐶3 𝐿 = 0
, so
7
𝐶1 = − 128 𝑃𝐿2
Macauley’s is even more compact:
3
1
Trivial statics, 𝑅𝐴 = 4 𝑃, 𝑅𝐶 = 4 𝑃
𝐿
𝑤(𝑥) = 𝑅𝐴 ⟨𝑥 − 0⟩−1 − 𝑃 ⟨𝑥 − ⟩−1
4
𝐿
𝑉(𝑥) = 𝑅𝐴 ⟨𝑥 − 0⟩0 − 𝑃 ⟨𝑥 − ⟩0
4
𝐿
𝑀(𝑥) = 𝑅𝐴 ⟨𝑥 − 0⟩1 − 𝑃 ⟨𝑥 − ⟩1
4
𝑑𝑦
𝑅𝐴
𝑃
𝐿
⟨𝑥 − 0⟩2 − ⟨𝑥 − ⟩2 + 𝐶1
𝐸𝐼
(𝑥) =
𝑑𝑥
2
2
4
𝑅𝐴
𝑃
𝐿
⟨𝑥 − 0⟩3 − ⟨𝑥 − ⟩3 + 𝐶1 𝑥 + 0
𝐸𝐼𝑦(𝑥) =
6
6
4
Boundary condition at C (fixed)
𝑅𝐴 3 𝑃 3𝐿 3
⟨L⟩ − ⟨ ⟩ + 𝐶1 𝐿 + 0
0=
6
6 4
3P 2
27P 2
𝐶1 = −
𝐿 +
𝐿
4×6
6 × 64
7
𝐶1 = −
𝑃𝐿2
128
Next, is another, a bit clumsy method, if you simplify the moment after the point load
d) Express the slope at any point of the beam (no need to elucidate the integration constants
at this stage)
𝑑𝑦
3
[A-B] 𝐸𝐼 𝑑𝑥 = 8 𝑃𝑥 2 + 𝐶1
𝑑𝑦
1
1
[B-C] 𝐸𝐼 𝑑𝑥 = − 8 𝑃𝑥 2 + 4 𝑃𝐿𝑥 + 𝐶3′
e) Express the deflection at any point of the beam (no need to elucidate the integration
constants at this stage)
1
[A-B] 𝐸𝐼𝑦 = 8 𝑃𝑥 3 + 𝐶1 𝑥 + 𝐶2
1
1
[B-C] 𝐸𝐼𝑦 = − 24 𝑃𝑥 3 + 8 𝑃𝐿𝑥 2 + 𝐶3′ 𝑥 + 𝐶4′
f) Calculate all the integration constants as functions of L, P, E and I
Fixed at A: 𝐶2 = 0
1
Fixed at C: 𝐶3′ 𝐿 + 𝐶4′ = − 12 𝑃𝐿3
1
Continuous at B: 𝐶1 𝐿 − 𝐶3′ 𝐿 − 4𝐶4′ = 48 𝑃𝐿3
1
Continuous derivative at B: 𝐶1 − 𝐶3 = 32 𝑃𝐿2
Solve system and:
1
𝐶4′ = 384 𝑃𝐿3
68
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
11
𝐶3′ = − 128 𝑃𝐿2
7
𝐶1 = − 128 𝑃𝐿2
Same answer for 𝐶1 and 𝐶2 , but a bit more tedious!
Exam Question III.5 (Mock exam)
A simply supported beam (Young’s modulus 𝐸, second moment of area 𝐼) is supported and
loaded as shown in Fig. III.5.
Figure III.5
a) Calculate the reactions at the supports in terms of the load.
𝑀
Trivial statics, 𝑅𝐴 = − 𝐿 , 𝑅𝐶 =
𝑀
𝐿
b) Determine the bending moment between A and B as a function of 𝑀, 𝑥 and 𝐿.
𝑀
[A-B] 𝑀(𝑥) = − 𝐿 𝑥
c) Determine the bending moment between B and C as a function of 𝑀, 𝑥 and 𝐿.
𝑀
[B-C] 𝑀(𝑥) = − 𝐿 𝑥 + 𝑀
d) Express the slope at any point of the beam as functions of 𝐸, 𝐼, 𝑀, 𝑥 and 𝐿 (no need to
solve the integration constants at this stage)
𝑑𝑦
𝑀
𝑑𝑦
𝑀
[A-B] 𝐸𝐼 𝑑𝑥 = − 2𝐿 𝑥 2 + 𝐴1
2𝐿
[B-C] 𝐸𝐼 𝑑𝑥 = − 2𝐿 𝑥 2 + 𝑀 (𝑥 −
3
) + 𝐴3 This is not an obvious choice, but it simplifies
calculations later. Check the “green” solution to see what happens if you write
𝑑𝑦
𝑀
𝐸𝐼 𝑑𝑥 = − 2𝐿 𝑥 2 + 𝑀𝑥 + 𝐴3′ (Different integration constants!)
e) Express the deflection at any point of the beam as functions of 𝐸, 𝐼, 𝑀, 𝑥 and 𝐿 (no need
to solve the integration constants at this stage)
𝑀
[A-B] 𝐸𝐼𝑦 = − 6𝐿 𝑥 3 + 𝐴1 𝑥 + 𝐴2
𝑀
[B-C] 𝐸𝐼𝑦 = − 6𝐿 𝑥 3 +
𝑀
(𝑥 −
2
2𝐿 2
3
) + 𝐴3 𝑥 + 𝐴4
f) Calculate all the integration constants as functions of 𝐿 and 𝑀
Fixed at A: 𝐴2 = 0
Continuous derivative at B: 𝐴3 = 𝐴1
Continuous at B: 𝐴4 = 𝐴2 = 0
𝑀
Fixed at C: 𝐴3 𝐿 + 𝐴4 = 6𝐿 𝐿3 −
𝑀 𝐿 2
2
(3)
69
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Solve and:
1
𝐴3 = 𝐴1 = 9 𝑀𝐿
Macaulay’s singularity functions would work even better here.
𝑀
Trivial statics, 𝑅𝐴 = − 𝐿 , 𝑅𝐶 =
𝑀
𝐿
2𝐿 −2
⟩
3
2𝐿
𝑉(𝑥) = 𝑅𝐴 ⟨𝑥 − 0⟩0 + 𝑀 ⟨𝑥 − ⟩−1
3
2𝐿
𝑀(𝑥) = 𝑅𝐴 ⟨𝑥 − 0⟩1 + 𝑀 ⟨𝑥 − ⟩0
3
𝑑𝑦
𝑅𝐴
2𝐿
(𝑥) =
⟨𝑥 − 0⟩2 + 𝑀 ⟨𝑥 − ⟩1 + 𝐶1
𝐸𝐼
𝑑𝑥
2
3
𝑅𝐴
𝑀
2𝐿
⟨𝑥 − 0⟩3 + ⟨𝑥 − ⟩2 + 𝐶1 𝑥 + 0
𝐸𝐼𝑦(𝑥) =
6
2
3
Boundary condition at C (fixed)
𝑅𝐴 3 𝑀 𝐿 2
⟨L⟩ + ⟨ ⟩ + 𝐶1 𝐿 + 0
0=
6
2 3
𝑀
𝑀
𝐶1 = 𝐿 −
𝐿
6
2×9
1
𝐶1 = 𝑃𝐿2
9
𝑤(𝑥) = 𝑅𝐴 ⟨𝑥 − 0⟩−1 + 𝑀 ⟨𝑥 −
What if at the function is written a bit more naturally? See below:
a) Calculate the reactions at the supports in terms of the load.
𝑀
Trivial statics, 𝑅𝐴 = − 𝐿 , 𝑅𝐶 =
𝑀
𝐿
b) Determine the bending moment between A and B as a function of 𝑀, 𝑥 and 𝐿.
𝑀
[A-B] 𝑀(𝑥) = − 𝐿 𝑥
c) Determine the bending moment between B and C as a function of 𝑀, 𝑥 and 𝐿.
𝑀
[B-C] 𝑀(𝑥) = − 𝐿 𝑥 + 𝑀
d) Express the slope at any point of the beam as functions of 𝐸, 𝐼, 𝑀, 𝑥 and 𝐿 (no need to
solve the integration constants at this stage)
𝑑𝑦
𝑀
𝑑𝑦
𝑀
[A-B] 𝐸𝐼 𝑑𝑥 = − 2𝐿 𝑥 2 + 𝐴1
[B-C] 𝐸𝐼 𝑑𝑥 = − 2𝐿 𝑥 2 + 𝑀𝑥 + 𝐴3′
e) Express the deflection at any point of the beam as functions of 𝐸, 𝐼, 𝑀, 𝑥 and 𝐿 (no need
to solve the integration constants at this stage)
[A-B] 𝐸𝐼𝑦 = −
𝑀
6𝐿
𝑀
𝑥 3 + 𝐴1 𝑥 + 𝐴2
[B-C] 𝐸𝐼𝑦 = − 6𝐿 𝑥 3 +
𝑀
2
𝑥 2 + 𝐴3′ 𝑥 + 𝐴4′
f) Calculate all the integration constants as functions of 𝐿 and 𝑀
Fixed at A: 𝐴2 = 0
70
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
1
Fixed at C: 𝐴3′ 𝐿 + 𝐴4′ = − 3 𝑀𝐿2
Continuous at B:
2𝐿
3
𝐴1 −
2𝐿
3
2
𝐴3′ − 𝐴4′ = 9 𝑀𝐿2
2
Continuous derivative at B: 𝐴1 −𝐴3′ = 3 𝑀𝐿
2
𝐴4′ = 9 𝑀𝐿2
Solve system and:
5
𝐴3′ = − 9 𝑀𝐿
1
𝐴1 = 9 𝑀𝐿
Exam Question V.1 (2018-2019)
A wooden beam (cross-section depicted in Figure V.1) is made of four planks nailed together.
It is subjected to a downward vertical shearing force of 500 N.
Figure V.1
a) Calculate the second moments of area 𝐼𝑦𝑦 .
Standard technique, and 𝐼𝑦𝑦 = 4.95 106 mm4
b) Calculate the first moment of area for points located on the neutral axis
Again, standard technique 𝑄 = 𝐴𝑦𝑐 for the area above the point of interest , and
𝑄0 = 75.5 103 mm3
c) Calculate the average shearing stress for points located on the neutral axis
Directly from 𝜏 =
𝑉𝑄
𝐼𝑡
with t =40 mm here, and 𝜏 = 191 kPA
d) Knowing that the maximum allowable shear force in each nail is 200 N, calculate the
minimum spacing between nail rows.
Directly from 𝐻 =
𝑉𝑄∆𝑥
𝐼
with 2 nails per row,
Need to recalculate 𝑄 = 𝐴𝑦𝑐 at the joint, so 𝑄0 = 63.0 103 mm3 and ∆𝑥 = 62.8 mm
Exam Question VI.1
An extruded aluminium beam has the cross section shown in Figure VI.1. It is subjected to a
downward vertical shearing force of 90 kN.
71
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Figure VI.1. Beam cross section
a) Calculate the relevant second moment of area.
coef
b
1
2
2
d
100
10
15
y
15
115
10
A
107.5
57.5
5
Ay
1500
2300
300
4100
161250
132250
1500
295000
Ybar
h=y-ybar bd3/12 ah2
I (mm4)
71.95122 35.54878
28125 1895574
71.95122 -14.4512 2534792 480326.8
71.95122 -66.9512
2500 1344740
71.95122
6286056.911
Standard technique, and 𝐼𝑦𝑦 = 6.28 106 mm4
b) Calculate the shear stress at point A.
point A
coef
b
1
-1
d
120
100
y
A
50
35
90
82.5
points of attachments
A(y-ybar) Q (mm3) t (mm)
n
tau (kN/mm2)
6000 108292.7
10
2
-3500 -36920.7
MPa
71371.95
0.051093
51.1 MPa
Again, standard technique to first get first moment of Area “above” point A, 𝑄 = 𝐴𝑦𝑐
for the area above the point of interest, with 𝑄0 = 71.372 103 mm3
Then directly from 𝜏 =
𝑉𝑄
𝐼𝑡
with t =2x10 mm here, and 𝜏 = 51.1 MPa
c) Calculate the shear stress at point B.
point B
coef
b
1
0
d
100
10
y
15
50
A
107.5
90
points of attachments
A(y-ybar) Q (mm3) t (mm)
n
tau (kN/mm2)
1500 53323.17
15
2
0
0
MPa
53323.17
0.025448
25.4 MPa
d) Calculate the shear stress at point C.
point B
coef
b
1
2
d
100
10
y
15
50
A
107.5
102.5
points of attachments
A(y-ybar) Q (mm3) t (mm)
n
tau (kN/mm2)
1500 53323.17
10
2
1000 30548.78
MPa
83871.95
0.060041
60.0 MPa
Exam Question VI.2 (resit 2018-2019)
An extruded aluminium beam has the cross section shown in Figure VI.2. It is subjected to a
downward vertical shearing force of 50 kN.
72
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Figure VI.2
a) Calculate the relevant second moment of area.
Standard technique, and 𝐼𝑦𝑦 = 1.396 106 mm4
b) Calculate the shear stress at point A (on the neutral axis).
First 𝑄 = ∑ 𝐴𝑦𝑐 = 24750 mm3 , then from 𝜏 =
𝑉𝑄
𝐼𝑡
with t =12 mm here, and 𝜏 =
73.8 MPa
c) Calculate the shear stress at point B.
First 𝑄 = ∑ 𝐴𝑦𝑐 = 17400 mm3 , then from 𝜏 =
𝑉𝑄
𝐼𝑡
with t =20 mm here, and 𝜏 =
31.1 MPa
d) Calculate the shear stress at point C.
First 𝑄 = ∑ 𝐴𝑦𝑐 = 23400 mm3 , then from 𝜏 =
𝑉𝑄
𝐼𝑡
with t =12 mm here, and 𝜏 =
69.8 MPa
Exam Question VIII.1 (Mock)
The bent rod in Fig. VIII.1 has a diameter of 20 mm and is subjected to the force of 400 N.
Point A is located on the bottom surface.
Figure VIII.1
Determine:
a) The axial stress at A
73
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
𝐹
400
By definition, 𝜎 = 𝐴 = 𝜋×0.012 = 1.273 MPa
b) The stress due to bending at A
Engineering bending formula, 𝜎 =
𝑀𝑦
𝐼
=
(400∗0.25)×−0.01
𝜋/4×0.014
= −127.3 MPa in compression
c) The principal stresses at point A
Total stress along the bar is sum as both axial and bending occur in the same
direction here, 𝜎𝑦 = −127.3 + 1.273 = −126 MPa still in compression
There is no component in the transverse direction, and no shear either. So the
principal stresses are directly 𝜎𝑥 and 𝜎𝑦 , at 0 MPa and −126 MPa
d) The maximum in-plane shear stress that is developed at point A
The maximum in plane shear stress is the radius of Mohr’s circle, and therefore
directly 63 MPa
Exam Question X.1 (2019-2020)
The bent rod depicted in Figure X.1 has a diameter of 40 mm and is subjected to a force of
1500 N. Point H is located on the surface, with coordinates (0, 75, 20).
Figure X.1
a) Calculate the second moment of area and the polar second moment of area of the rod.
𝜋
𝜋
𝐴 = 𝜋 × 202 = 1257 mm2 𝐼 = 4 × 204 = 125664 mm4 𝐽 = 2 × 204 = 251327 mm4
b) Calculate the stress due to bending at point H.
Engineering bending formula, 𝜎 =
𝑀𝑦
𝐼
=
(1500∗0.200)×20
125664
= 47.7 MPa in tension
c) Calculate the stress due to torsion at point H.
Engineering torsion formula, 𝜏 =
𝑇𝑟
𝐽
=
(1500∗0.350)×20
251327
= 41.8 MPa
d) What is the state of stress at point H (two axial stresses, one shear stress)? Draw a state of
stress diagram, indicating clearly the directions and magnitude of the three stresses.
Nothing direct in the x direction 𝜎𝑥 = 0 MPa
Bending gives direct contribution 𝜎𝑦 = 47.7 MPa in tension
74
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Torsion gives shear contribution 𝜏𝑥𝑦 = 41.8 MPa
+ simple state of stress drawing
e) Using Mohr’s circle or a direct method, calculate the principal stresses and the maximum
in-plane shear stress at H.
direct method: 𝜎𝑚𝑎𝑥,𝑚𝑖𝑛 =
𝜎𝑥 +𝜎𝑦
2
2
𝜎𝑥 −𝜎𝑦
2 = −24.2 MPa or 72.0 MPa
± √( 2 ) + 𝜏𝑥𝑦
2
𝜎𝑥 −𝜎𝑦
2 = 48.1 MPa
direct method: 𝜏𝑚𝑎𝑥 = √( 2 ) + 𝜏𝑥𝑦
f) Determine whether the part yields at point H or not, using the maximum distortion
energy (Von-Mises) criterion (Yield strength σY = 350 MPa).
Calculate Von-Mises stress 𝜎𝑉𝑀 = √𝜎12 + 𝜎22 − 𝜎1 ∙ 𝜎2 = 86.7 MPa Lower than Yield
strength, so the part does NOT yield.
Exam Question X.2 (resit 2018-2019)
The bent rod in Figure X.2 has a diameter of 15 mm and is subjected to forces of 500 N.
Point A is located on the bottom surface.
Figure X.2
a) Calculate the cross section area of the rod and the second moment of area from the rod’s
neutral axis.
𝜋
𝐴 = 𝜋 × 7.52 = 176.7 mm2 𝐼 = 4 × 7.54 = 2485 mm4
b) Calculate the axial stress at A.
𝐹
500
By definition, 𝜎 = 𝐴 = 176.7 = 2.83 MPa
c) Calculate the stress due to bending at A.
Engineering bending formula, 𝜎 =
𝑀𝑦
𝐼
=
(500∗0.11)×−7.5
2485
= −0.166 MPa in compression
d) What is the state of stress at A (two axial stresses, one shear stress)? Draw a state of stress
diagram, indicating clearly the directions and magnitude of the three stresses.
Total stress along the bar is sum as both axial and bending occur in the same
direction here, 𝜎𝑦 = 2.66 MPa in tension
e) Using Mohr’s circle or a direct method, calculate the principal stresses at A.
75
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
There is no component in the transverse direction, and no shear either. So the
principal stresses are directly 𝜎𝑥 and 𝜎𝑦 , at 0 MPa and 2.66 MPa
f) Using Mohr’s circle or a direct method, calculate the maximum in-plane shear stress that is
developed at A.
The maximum in plane shear stress is the radius of Mohr’s circle, and therefore
directly 1.33 MPa
Exam Question XI.1 (resit 2019-2020)
Using a rosette strain gauge, the following strains have been determined at point Q
on the surface of a steel machine base (Young’s modulus E = 210 GPa, Poisson’s
ratio υ = 0.31, Yield strength σY = 450 MPa), as shown in Figure XI.1.
𝜀𝐴 = 400 𝜇
𝜀𝐵 = −200 𝜇
𝜀𝐶 = 300 𝜇
Figure XI.1
a) Calculate the strain components εx, εy, and γxy.
𝜀𝑥 ′ = 𝜀𝑥 cos 2 𝜃 + 𝜀𝑦 sin2 𝜃 + 𝛾𝑥𝑦 cos𝜃sin𝜃
1
1
1
2
1
𝜀A = 𝜀𝑥 2 + 𝜀𝑦
𝜀𝐵 = 𝜀𝑥 2 + 𝜀𝑦
1
+ 𝛾𝑥𝑦 2
1
− 𝛾𝑥𝑦 2
2
𝜀𝐶 = 𝜀𝑥 = 300 𝜇
Solve the rest and
𝛾𝑥𝑦 = (𝜀𝐴 − 𝜀𝐵 ) = 600 𝜇
𝜀𝑦 = ((𝜀𝐴 + 𝜀𝐵 ) − 𝜀𝑥 ) = −100 𝜇
b) Calculate the three principal strains.
𝜀𝑥 +𝜀𝑦
2
2
2
𝜀𝑥 −𝜀𝑦
𝛾𝑥𝑦
± √( 2 ) + ( 2 ) (or Mohr’s circle if preferred)
𝜀𝑚𝑎𝑥 = 461 𝜇 𝜀𝑚𝑖𝑛 = −261 𝜇
−𝜈
𝜀𝑧 = 1−𝜈 (𝜀𝑎 + 𝜀𝑏 ) = −89.9 𝜇
c) Calculate the maximum shearing strain.
2
2
𝜀𝑥 −𝜀𝑦
𝛾𝑥𝑦
𝜏𝑚𝑎𝑥 = √( 2 ) + ( 2 ) = 721 𝜇
d) Calculate the three principal stresses.
𝜀𝑎 =
𝜎𝑎
𝐸
−
𝜈𝜎𝑏
𝐸
, 𝜀𝑏 =
𝜎𝑏
𝐸
−
𝜈𝜎𝑎
𝐸
and solve
𝜎𝑚𝑎𝑥 = 88.2 MPa, 𝜎𝑚𝑖𝑛 = −27.4 MPa and (Plane stress) 𝜎𝑧 = 0 MPa
e) Determine whether the part yields or not, using the maximum distortion energy criterion
76
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Calculate Von-Mises stress
𝜎𝑉𝑀 = √𝜎12 + 𝜎22 − 𝜎1 ∙ 𝜎2
= 104 MPa Lower than Yield strength, so the part does NOT yield.
Exam Question XI.2 (2018-2019)
Using a 60º rosette, the following strains have been determined at point Q on the surface of a
steel machine base (Young’s modulus E = 210 GPa, Poisson’s ratio υ = 0.31, Yield strength
σY = 300 MPa), as shown in Figure XI.2.
𝜀𝐴 = −500 𝜇
𝜀𝐵 = 750 𝜇
𝜀𝐶 = 250 𝜇
Figure XI.2
a) Calculate the strain components εx, εy, and γxy.
𝜀𝑥 ′ = 𝜀𝑥 cos 2 𝜃 + 𝜀𝑦 sin2 𝜃 + 𝛾𝑥𝑦 cos𝜃sin𝜃
𝜀𝐴 = 𝜀𝑥 = −500 𝜇
1
3
1
4
3
𝜀B = 𝜀𝑥 4 + 𝜀𝑦
𝜀C = 𝜀𝑥 4 + 𝜀𝑦
4
√3
4
√3
𝛾𝑥𝑦 4
+ 𝛾𝑥𝑦
−
Solve the rest and
𝛾𝑥𝑦 =
2
√3
(𝜀𝐵 − 𝜀𝐶 ) = 833 𝜇
𝜀𝑦 = (2(𝜀𝐵 + 𝜀𝐶 ) − 𝜀𝑥 )/3 = 577 𝜇
b) Calculate the three principal strains.
𝜀𝑥 +𝜀𝑦
2
2
2
𝜀𝑥 −𝜀𝑦
𝛾𝑥𝑦
± √(
) + ( ) (or Mohr’s circle if preferred)
2
2
𝜀𝑚𝑎𝑥 = 893 𝜇 𝜀𝑚𝑖𝑛 = −560 𝜇
−𝜈
𝜀𝑧 = 1−𝜈 (𝜀𝑎 + 𝜀𝑏 ) = −150 𝜇
c) Calculate the maximum shearing strain.
2
2
𝜀𝑥 −𝜀𝑦
𝛾𝑥𝑦
𝜏𝑚𝑎𝑥 = √( 2 ) + ( 2 ) = 1453 𝜇
d) Calculate the three principal stresses.
𝜀𝑎 =
𝜎𝑎
𝐸
−
𝜈𝜎𝑏
𝐸
, 𝜀𝑏 =
𝜎𝑏
𝐸
−
𝜈𝜎𝑎
𝐸
and solve
𝜎𝑚𝑎𝑥 = 167.2 MPa, 𝜎𝑚𝑖𝑛 = −65.7 MPa and (Plane stress) 𝜎𝑧 = 0 MPa
e) Determine whether the part yields or not, using the maximum distortion energy criterion
Calculate Von-Mises stress 𝜎𝑉𝑀 = √𝜎12 + 𝜎22 − 𝜎1 ∙ 𝜎2 = 160 MPa Lower than Yield
strength, so the part does NOT yield.
77
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Exam Question XI.3 (2017-2018)
The following readings have been obtained from the strain gauge rosette shown in Figure
XI.3, which is attached to the outer surface of an aluminium pressure vessel. All gauges are
in the same plane XY. Assume E=70GPa and ν = 0.3
𝜀𝐴 = 150 𝜇
𝜀𝐵 = 250 𝜇
𝜀𝐶 = 350 𝜇
Figure XI.3
Determine:
a) The values of εx, εy and γxy
𝜀𝑥 ′ = 𝜀𝑥 cos 2 𝜃 + 𝜀𝑦 sin2 𝜃 + 𝛾𝑥𝑦 cos𝜃sin𝜃 , so apply to the strain gauges and
1
3
1
4
3
4
4
𝜀A = 𝜀𝑥 4 + 𝜀𝑦
𝜀𝐵 = 𝜀𝑥 + 𝜀𝑦
√3
4
√3
𝛾𝑥𝑦
4
+ 𝛾𝑥𝑦
−
𝜀𝐶 = 𝜀𝑥 = 350 𝜇
Solve the rest and
𝛾𝑥𝑦 =
2
√3
(𝜀𝐴 − 𝜀𝐵 ) = −115.5 𝜇
𝜀𝑦 = (2(𝜀𝐴 + 𝜀𝐵 ) − 𝜀𝑥 )/3 = 150 𝜇
b) The principal strains
𝜀𝑥 +𝜀𝑦
2
2
𝜀𝑥 −𝜀𝑦
𝛾𝑥𝑦
± √( 2 ) + ( 2 )
2
𝜀𝑚𝑎𝑥 = 365 𝜇 𝜀𝑚𝑖𝑛 = 134.5 𝜇
c) The reading of gauge D
𝜀D = 𝜀𝑥 cos2 (−45°) + 𝜀𝑦 sin2 (−45°) + 𝛾𝑥𝑦 cos(−45°)sin(−45°)
𝜀𝐷 = 307 𝜇
d) The angle that the principal strain makes with the +X axis
tan2𝜃𝑝 = 𝜀
𝛾𝑥𝑦
𝑥 −𝜀𝑦
2𝜃𝑝 = −29.9° 𝜃𝑝 = −14.95° = 75.05°
Exam Question XI.4 (Resit 2017-2018)
The following readings have been obtained from the strain gauge rosette shown in Fig. XI.4,
which is attached on the free surface of a chain on a suspension bridge. Assume all gauges
are in the same plane XY. Young’s modulus for the wraught iron chain is E=193GPa and
Poisson’s ratio is ν = 0.3. The Yield stress of the wraught iron is σy=600 MPa.
78
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
𝜀𝐴 = 1000 𝜇
𝜀𝐵 = 500 𝜇
𝜀𝐶 = 300 𝜇
Figure XI.4
Determine:
a) The values of εx, εy and γxy
𝜀𝑥 ′ = 𝜀𝑥 cos 2 𝜃 + 𝜀𝑦 sin2 𝜃 + 𝛾𝑥𝑦 cos𝜃sin𝜃
1
1
1
2
1
𝜀A = 𝜀𝑥 2 + 𝜀𝑦
𝜀𝐵 = 𝜀𝑥 2 + 𝜀𝑦
1
+ 𝛾𝑥𝑦 2
1
− 𝛾𝑥𝑦 2
2
𝜀𝐶 = 𝜀𝑦 = 300 𝜇
Solve the rest and
𝛾𝑥𝑦 = (𝜀𝐴 − 𝜀𝐵 ) = 500 𝜇
𝜀𝑥 = ((𝜀𝐴 + 𝜀𝐵 ) − 𝜀𝑦 ) = 1200 𝜇
b) The stress components σx, σy and τxy
𝜀𝑎 =
𝜎𝑎
𝐸
−
𝜈𝜎𝑏
𝐸
, 𝜀𝑏 =
𝜎𝑏
𝐸
−
𝜈𝜎𝑎
𝐸
and solve
𝜎𝑎 = 273 MPa, 𝜎𝑏 = 140 MPa
𝛾𝑎𝑏 =
𝜏𝑎𝑏
𝐺
𝐸
, 𝐺 = 2(1+𝜈)
𝐺 = 74.2 GPa, 𝛾𝑎𝑏 = 37.1 MPa
c) The principal stresses and their principal planes
2𝜏𝑥𝑦
𝑡𝑎𝑛2𝜃𝑝 = 𝜎
𝑥 −𝜎𝑦
= 14.5°
2
2 = 130.3 MPa
√(𝜎𝑥 −𝜎𝑦 ) + 𝜏𝑥𝑦
±
2
2
283 MPa
Or use Mohr’s circle …
𝜀𝑚𝑎𝑥 = 365 𝜇 𝜀𝑚𝑖𝑛 = 134.5 𝜇
d) The Factor of Safety (FOS) based on Maximum-Shearing-Stress Criterion
𝜎𝑚𝑎𝑥,𝑚𝑖𝑛 =
𝜎𝑥 +𝜎𝑦
Principal stresses have same sign, so 𝑚𝑎𝑥(𝜎1 , 𝜎2 ) < 𝜎𝑌
FOS =
𝜎𝑌 /2
|𝜎1 |
=
600
283
= 2.12
79
UFMFSS-30-2 Structural Mechanics
Book I: Stress Analysis
Potentially useful Formulae
𝑦̅ =
∑ 𝐴𝑦
,𝐼
= ∑ 𝐼𝑥𝑥 + ∑ 𝐴ℎ2
∑ 𝐴 𝑁𝐴𝑥
𝑀
𝜎 𝐸
= =
𝐼𝑁𝐴 𝑦 𝑅
𝑏𝑑 3
𝑏3𝑑
, 𝐼𝑦𝑦 =
12
12
4
𝜋𝑟
𝐼𝑑𝑖𝑠𝑐 =
4
𝐼𝑥𝑥 =
1 1 1
= ∫ 𝑑𝐴
𝑅 𝐴 𝑟
𝜎=
𝑒 = 𝑟̅ − 𝑅
𝑀𝑦
𝑀(𝑟 − 𝑅)
=
𝐴𝑒(𝑅 − 𝑦)
𝐴𝑒𝑟
𝑇 𝜏 𝐺𝜃
= =
,𝑃 = 𝑇 ∙ 𝜔
𝐽 𝑅
𝐿
𝐽𝑑𝑖𝑠𝑐 =
𝜋𝑟 4
2
𝐹𝑖 𝐿𝑖
𝛿=∑
𝐸𝑖 𝐴𝑖
𝑖
𝛿=∫
𝐹𝑑𝑥
𝐸𝐴
𝐹 𝑀𝑦
−
𝐴
𝐼
𝜎𝑥 = (𝜎𝑥 )𝑐𝑒𝑛𝑡𝑟𝑖𝑐 + (𝜎𝑥 )𝑏𝑒𝑛𝑑𝑖𝑛𝑔 =
𝑑𝑦
1
= ∫ 𝑀(𝑥)𝑑𝑥 + 𝐶1
𝑑𝑥 𝐸𝐼
𝑦=
𝑣𝑚𝑎𝑥 = 𝑒 [sec (√
𝑃𝑐𝑟 =
𝜋 2 𝐸𝐼
(𝐾𝐿)2
𝜎𝑚𝑎𝑥 =
𝑄 = 𝐴𝑦̅𝑐 = ∑ 𝐴𝑦
𝐻=
𝑡𝑎𝑛2𝜃𝑝 =
𝜎𝑚𝑎𝑥,𝑚𝑖𝑛 =
𝑉𝑄∆𝑥
𝐼
2𝜏𝑥𝑦
𝜎𝑥 − 𝜎𝑦
𝜎𝑥 + 𝜎𝑦
𝜎𝑥 − 𝜎𝑦 2
2
± √(
) + 𝜏𝑥𝑦
2
2
𝜏𝑚𝑎𝑥 = √(
𝜎𝑥 − 𝜎𝑦 2
2
) + 𝜏𝑥𝑦
2
𝑃 𝐾𝐿
𝜋 𝑃
) − 1] = 𝑒 [sec ( √ ) − 1]
𝐸𝐼 2
2 𝑃𝑐𝑟
𝑃 𝑃𝑒𝑐
𝑃 𝐾𝐿
𝑃
𝑒𝑐
𝜋 𝑃
+
sec (√
) = [1 + 2 sec ( √ )]
𝐴
𝐼
𝐸𝐼 2
𝐴
𝑟
2 𝑃𝑐𝑟
𝜏=
𝑉𝑄
𝐼𝑡
𝑉𝑄
𝐼
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
+
cos2𝜃 + 𝜏𝑥𝑦 sin2𝜃
2
2
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
𝜎𝑦′ =
−
cos2𝜃 − 𝜏𝑥𝑦 sin2𝜃
2
2
𝜎𝑥 − 𝜎𝑦
𝜏𝑥′𝑦′ = −
sin2𝜃 + 𝜏𝑥𝑦 cos2𝜃
2
𝜎𝑉𝑀 = √𝜎12 + 𝜎22 − 𝜎1 ∙ 𝜎2 < 𝜎𝑌
𝜎𝑎 𝜈𝜎𝑏
𝜎𝑏 𝜈𝜎𝑎
−𝜈
−𝜈
𝜏𝑎𝑏
𝐸
(𝜎𝑎 + 𝜎𝑏 ) =
(𝜀𝑎 + 𝜀𝑏 ), 𝛾𝑎𝑏 =
−
, 𝜀𝑏 =
−
, 𝜀𝑐 =
,𝐺 =
𝐸
𝐸
𝐸
𝐸
𝐸
1−𝜈
𝐺
2(1 + 𝜈)
𝑡𝑎𝑛2𝜃𝑝 =
𝜀𝑚𝑎𝑥,𝑚𝑖𝑛 =
𝑞=
𝜎𝑥′ =
|𝜎1 | < 𝜎𝑌 , |𝜎2 | < 𝜎𝑌 or |𝜎1 − 𝜎2 | < 𝜎𝑌
𝜀𝑎 =
1
∫ 𝑑𝑥 ∫ 𝑀(𝑥)𝑑𝑥 + 𝐶1 𝑥 + 𝐶2
𝐸𝐼
𝛾𝑥𝑦
𝜀𝑥 − 𝜀𝑦
𝜀𝑥 + 𝜀𝑦
𝜀𝑥 − 𝜀𝑦 2
𝛾𝑥𝑦 2
± √(
) +( )
2
2
2
𝜀𝑥′ = 𝜀𝑥 cos 2 𝜃 + 𝜀𝑦 sin2 𝜃 + 𝛾𝑥𝑦 cos𝜃sin𝜃
𝜀𝑥 + 𝜀𝑦 𝜀𝑥 − 𝜀𝑦
𝛾𝑥𝑦
+
cos2𝜃 +
sin2𝜃
2
2
2
𝜀𝑥 + 𝜀𝑦 𝜀𝑥 − 𝜀𝑦
𝛾𝑥𝑦
𝜀𝑦′ =
−
cos2𝜃 −
sin2𝜃
2
2
2
𝛾𝑥′𝑦′
𝜀𝑥 − 𝜀𝑦
𝛾𝑥𝑦
=−
sin2𝜃 +
cos2𝜃
2
2
2
𝜀𝑥′ =
80