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Lecture 1

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Mechanics of Deformable Bodies
T-CEET222
The Lasallian Prayer
“Let us remember
that we are in the holy presence of God.”
7 March 2021
Prepared by: Engr. Jeric Sarte
2
Ang Panalanging Lasalyano
“Ating alalahanin
na ang Diyos ay ating kapiling.”
7 March 2021
Prepared by: Engr. Jeric Sarte
3
Gospel
7 March 2021
Prepared by: Engr. Jeric Sarte
4
The Lasallian Prayer
“I will continue, O my God,
to do all my actions for the love of Thee.”
Saint John Baptist de La Salle, pray for us.
Live Jesus in our hearts, forever.
7 March 2021
Prepared by: Engr. Jeric Sarte
5
Ang Panalanging Lasalyano
“Ipagpapatuloy ko, O Diyos,
ang lahat ng aking gawain
upang Ikaw ay lalong pakaibigin.”
San Juan Bautista de La Salle, ipanalangin mo kami.
Hesus, manahan ka sa aming puso, magpakailanman.
7 March 2021
Prepared by: Engr. Jeric Sarte
6
Simple Stress
7 March 2021
Prepared by: Engr. Jeric Sarte
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STRESSES IN THE MEMBERS OF A STRUCTURE
•Normal Stress
Consider the rod shown subjected to force FBC.
Whether rod will break or not under this loading
depends upon the value found for the internal
force FBC, the cross-sectional area of the rod A,
and the material of which the rod is made.
Internally, the force FBC is represented by
elemental forces distributed over the cross
sectional area, A. The magnitude of these
elemental forces is equal to the force per unit
area.
𝐹𝐡𝐢
𝜎𝐡𝐢 =
𝐴
This is called stress and is typically denoted by, s.
7 March 2021
Prepared by: Engr. Jeric Sarte
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STRESSES IN THE MEMBERS OF A STRUCTURE
•Normal Stress
In the same manner how axial forces behave,
whether tension or compression, is same with
stress. Positive sign indicates tension.
As shown in the figure, the section through the
rod to determine the internal force in the rod and
the corresponding stress is perpendicular to the
axis of the rod. The corresponding stress is
described as a normal stress.
𝜎=
7 March 2021
𝑃
π΄π‘π‘’π‘Ÿπ‘π‘’π‘›π‘‘π‘–π‘π‘’π‘™π‘Žπ‘Ÿ
Prepared by: Engr. Jeric Sarte
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STRESSES IN THE MEMBERS OF A STRUCTURE
•Normal Stress
Note that this uniform distribution of stress is
possible only if the line of action of the
concentrated loads P and P′ passes through the
centroid of the section considered. This type of
loading is called centric loading.
However, if a member is loaded axially, but
eccentrically, as shown in the figure, doing statics
suggests that a given section must be equivalent
to a force P applied at the centroid of the section
and a couple M of moment M = Pd. This
distribution of forces—the corresponding
distribution of stresses—cannot be uniform.
7 March 2021
Prepared by: Engr. Jeric Sarte
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Stress units
The SI unit for stress is pascal (Pa).
𝜎=
In engineering, we may use (metric units)
π‘˜π‘
π‘˜π‘–π‘™π‘œπ‘π‘Žπ‘ π‘π‘Žπ‘™ π‘˜π‘ƒπ‘Ž = 10 π‘ƒπ‘Ž = 2
π‘š
3
𝑃
𝑁
= 2 = π‘ƒπ‘Ž
𝐴 π‘š
In English units,
𝑁
π‘šπ‘’π‘”π‘Žπ‘π‘Žπ‘ π‘π‘Žπ‘™ π‘€π‘ƒπ‘Ž = 10 π‘ƒπ‘Ž =
π‘šπ‘š2
𝑙𝑏
1 2 π‘œπ‘Ÿ 𝑝𝑠𝑖 = 6,894.757 π‘ƒπ‘Ž
𝑖𝑛
π‘˜π‘
π‘”π‘–π‘”π‘Žπ‘π‘Žπ‘ π‘π‘Žπ‘™ πΊπ‘ƒπ‘Ž = 10 π‘ƒπ‘Ž =
π‘šπ‘š2
6
π‘˜π‘–π‘π‘ 
1 2 π‘œπ‘Ÿ π‘˜π‘ π‘– = 1000 𝑝𝑠𝑖
𝑖𝑛
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Other units may be used as well such as
1 atm = 101,325 Pa
1 torr = 133.322 Pa
1 mmHg = 133.322 Pa
1 bar = 100,000 Pa
7 March 2021
Prepared by: Engr. Jeric Sarte
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Problem 1
Assume a rod made of a steel with a maximum
allowable stress σall = 155 MPa. Can the rod safely
support the load to which it will be subjected? The
magnitude of the force FBC in the rod was 50 kN and
its diameter is 20 mm. If not, redesign the rod.
7 March 2021
Prepared by: Engr. Jeric Sarte
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Problem 1
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Prepared by: Engr. Jeric Sarte
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Problem 1
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Prepared by: Engr. Jeric Sarte
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STRESSES IN THE MEMBERS OF A STRUCTURE
•Shear Stress
A very different type of stress is
obtained when transverse forces P
and P′ are applied to a member AB.
If we pass a plane at C cutting the
member, we obtain a new plane
where internal forces are exposed
whose resultant is P. This resultant
is called shear force.
7 March 2021
Prepared by: Engr. Jeric Sarte
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STRESSES IN THE MEMBERS OF A STRUCTURE
•Shear Stress
Dividing the resultant P to the cross
sectional area of member AB gives
us the average shearing stress. Note:
this time, the area is parallel to the force.
πœπ‘Žπ‘£π‘’ =
7 March 2021
𝑃
π΄π‘π‘Žπ‘Ÿπ‘Žπ‘™π‘™π‘’π‘™
Prepared by: Engr. Jeric Sarte
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STRESSES IN THE MEMBERS OF A STRUCTURE
•Shear Stress – single shear
7 March 2021
Prepared by: Engr. Jeric Sarte
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STRESSES IN THE MEMBERS OF A STRUCTURE
•Shear Stress – double shear
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Prepared by: Engr. Jeric Sarte
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Problem 2
Two wooden planks, each 1/2 in.
thick and 9 in. wide, are joined by
the dry mortise joint shown.
Knowing that the wood used shears
off along its grain when the average
shearing stress reaches 1.20 ksi,
determine the magnitude P of the
axial load that will cause the joint
to fail.
7 March 2021
Prepared by: Engr. Jeric Sarte
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Problem 2
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Problem 2
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STRESSES IN THE MEMBERS OF A STRUCTURE
•Bearing Stress
This is the stress that is generated in
contact between two different
components/body (e.g. bolt against plate
as shown, footing against earth-ground,
etc).
Take for example the previous illustration
in the discussion of shear stress. The bolt
exerts on plate A a force P equal and
opposite to the force F exerted by the
plate on the bolt.
7 March 2021
Prepared by: Engr. Jeric Sarte
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STRESSES IN THE MEMBERS OF A STRUCTURE
•Bearing Stress
Take for example the previous illustration in the
discussion of shear stress. The bolt exerts on
plate A a force P equal and opposite to the force F
exerted by the plate on the bolt.
The force P represents the resultant of
elementary forces distributed on the inside
surface of a half-cylinder of diameter d and of
length t equal to the thickness of the plate. The
average of this equal to P divided by the
rectangular projection of the half-cylinder is called
bearing stress.
𝑃
𝑃
πœŽπ‘ = =
𝐴 𝑑𝑑
7 March 2021
Prepared by: Engr. Jeric Sarte
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Problem 3
A 40-kN axial load is applied to a
short wooden post that is
supported by a concrete footing
resting on undisturbed soil.
Determine
(a) the maximum bearing stress on
the concrete footing,
(b) (b) the size of the footing for
which the average bearing
stress in the soil is 145 kPa.
7 March 2021
Prepared by: Engr. Jeric Sarte
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Problem 3
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Problem 3
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DESIGN CONSIDERATIONS
•Ultimate Strength
of a Material
Ultimate strength is the largest stress a
material is subjected to where it breaks
or begins to carry less load. Ultimate
normal stress is determined by
π‘ƒπ‘ˆ
πœŽπ‘ˆ =
𝐴
7 March 2021
Prepared by: Engr. Jeric Sarte
Several test procedures may be done to
determine the ultimate shearing stress.
This may involved twisting of a circular
tube, etc. It is represented by the
equation
π‘ƒπ‘ˆ
πœπ‘ˆ =
𝐴
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DESIGN CONSIDERATIONS
•Allowable Load and
Allowable Stress
The maximum load that a structural member or a
machine component will be allowed to carry
under normal conditions is considerably smaller
than the ultimate load. This smaller load is the
allowable load (sometimes called the working or
design load).
The remaining portion of the load-carrying
capacity of the member is kept in reserve to
assure its safe performance.
7 March 2021
Prepared by: Engr. Jeric Sarte
The ratio of the ultimate load to the
allowable load is the factor of
safety:
πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ π‘œπ‘“ π‘†π‘Žπ‘“π‘’π‘‘π‘¦ = 𝐹. 𝑆. =
Similarly,
πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ π‘œπ‘“ π‘†π‘Žπ‘“π‘’π‘‘π‘¦ = 𝐹. 𝑆. =
π‘’π‘™π‘‘π‘–π‘šπ‘Žπ‘‘π‘’ π‘™π‘œπ‘Žπ‘‘
π‘Žπ‘™π‘™π‘œπ‘€π‘Žπ‘π‘™π‘’ π‘™π‘œπ‘Žπ‘‘
π‘’π‘™π‘‘π‘–π‘šπ‘Žπ‘‘π‘’ π‘ π‘‘π‘Ÿπ‘’π‘ π‘ 
π‘Žπ‘™π‘™π‘œπ‘€π‘Žπ‘π‘™π‘’ π‘ π‘‘π‘Ÿπ‘’π‘ π‘ 
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Problem 4
Find the stresses in members BC,
BD, and CF for the truss shown in
the figure. Indicate the tension or
compression. The cross section nal
area of each member 1600 mm2.
7 March 2021
Prepared by: Engr. Jeric Sarte
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Problem 4
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Prepared by: Engr. Jeric Sarte
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Problem 4
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Prepared by: Engr. Jeric Sarte
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Problem 5
A load P is supported as shown by
a steel pin that has been inserted
in a short wooden member hanging
from the ceiling. The ultimate
strength of the wood used is 60
MPa in tension and 7.5 MPa in
shear, while the ultimate strength
of the steel is 145 MPa in shear.
Knowing that b = 40 mm, c = 55
mm, and d = 12 mm, determine
the load P if an overall factor of
safety of 3.2 is desired.
7 March 2021
Prepared by: Engr. Jeric Sarte
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Problem 5
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Prepared by: Engr. Jeric Sarte
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Problem 5
7 March 2021
Prepared by: Engr. Jeric Sarte
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Problem 6
Two wooden members of uniform
cross section are joined by the
simple scarf splice shown. Knowing
that the maximum allowable tensile
stress in the glued splice is 75 psi,
determine
(a) the largest load P that can be
safely supported,
(b) (b) the corresponding shearing
stress in the splice.
7 March 2021
Prepared by: Engr. Jeric Sarte
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Problem 6
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Prepared by: Engr. Jeric Sarte
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Problem 6
7 March 2021
Prepared by: Engr. Jeric Sarte
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END OF PRESENTATION
ONE HEART.
ONE COMMITMENT.
ONE LIFE.
1719 – 2019
#300LaSalle
Prepared
7 Marchby:
2021
Engr. Joshua Hernandez
38
The Lasallian Prayer
“Let us remember
that we are in the holy presence of God.”
7 March 2021
Prepared by: Engr. Jeric Sarte
39
The Lasallian Prayer
“I will continue, O my God,
to do all my actions for the love of Thee.”
Saint John Baptist de La Salle, pray for us.
Live Jesus in our hearts, forever.
7 March 2021
Prepared by: Engr. Jeric Sarte
40
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