lectures on elasticity of material and stress you can also watch https://youtu.be/cQQP36f3f2g for the explanation of the lecture

1. Welcome to the physicsCentral I am sir ed and in this video I am going to talk about the elasticity of
material. If you’re intothis kindof discussion then hit the subcribe buttonand click the notification bell
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In physics, elasticityis the ability of matter to return to its original shape (regain its original dimension)
after being subjected to force. A rubber band has a certain level of elasticity, therefore we can ask
ourselves, “how much force will cause enough deformation before the rubber band break apart?”.
We can answerthisquestionusingtheHooke’s lawwhichstatesthat,the amountof force needed
to stretch/ deform an object by x units of length is proportional to the amount of deformation and a
constant of proportionality k. we can write it mathematically as,
𝐹 = −𝑘𝑥
Where k is calledthe springconstant.Thisequationwasfirstseeninthe 1660 and was firstpublishedby
the British physicist Dr. Robert Hooke in his Latin anagram entitled UT tensio, sic vis – which translated,
means "as the extension, so the force" or "the extension is proportional to the force").
Objects as subjected to different types of force causing different types of deformation.
Elongation happens when an objects length is changed due to a force that tends to stretch it.
Compression change in the dimension of the object when the type of force tends to compress it.
Shear
Torsion when the object is twisted.
We will start exploring the elongation process of materials as it is being subjected to tensile stress.
Stress is defined as the amount of force applied per unit area and it is given by the formula,
𝑠𝑡𝑟𝑒𝑠𝑠 =
𝐹𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎
=
𝐹
𝐴
Once this stress causes Elongation or have caused tensile strain there is a noticeable amount of
extension of an object under stress, and this usually expressed as a percentage of the original
length. We can rewrite the equation for strain as,
𝜀 =
∆𝐿
𝐿𝑜
=
𝐿𝑓 − 𝐿𝑜
𝐿𝑜
Where
𝜀 = 𝑠𝑡𝑟𝑎𝑖𝑛
∆𝐿 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔ℎ𝑡 = 𝐿𝑓 − 𝐿𝑜
𝐿𝑓 = 𝑓𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
𝐿𝑜 = 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔ℎ𝑡

2. Every material has its capacity to withstand stress (in our case tensile stress) before being permanently
deformed. We can describe this characteristic as modulus. For the elongation of materials, we have to
referto the youngmodulusof each material to accuratelydescribe thisphenomena.The table of values
for the young modulus of some materials are shown below.
material Tensile strength
(GPA)
Mpsi
ABS plastics 1.4 - 3.1
Acetals 2.8
Acrylic 3.2
Aluminum 69
Aluminum Alloys 70
Aluminum Bronze 120
Antimony 78
Aramid 70 - 112
Beryllium (Be) 287
Beryllium Copper 124
Bismuth 32
Bone, compact 18
Bone, spongy 76
Brass 102 - 125
Brass, Naval 100

3. Bronze 96 - 120
CAB 0.8
Cadmium 32
Carbon Fiber Reinforced Plastic 150
As we expect,differentmaterialsreactsdifferentlytotensile stress.We cannow write a general formula
for tensile stress.
𝑠𝑡𝑟𝑒𝑠𝑠 = (𝑦𝑜𝑢𝑛𝑔′𝑠𝑚𝑜𝑑𝑢𝑙𝑢𝑠)(𝑠𝑡𝑟𝑎𝑖𝑛)
𝐹⊥
𝐴
= (𝑌)(𝜀)
𝐹⊥
𝐴
= (𝑌)(
∆𝐿
𝐿𝑜
)
We can also rewrite the Hooke’s law as
𝑆𝑡𝑟𝑒𝑠𝑠
𝑆𝑡𝑟𝑎𝑖𝑛
= 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 (𝐻𝑜𝑜𝑘𝑒′𝑠 𝐿𝑎𝑤)
By following this we can write,
𝐹⊥ ∙ 𝐿𝑜
𝐴 ∙ ∆𝐿
= (𝑌) =
𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑎𝑖𝑛
This formula is used to compute the young’s Modulus this formula is just another formulation of the
Hooke’s law.

4. Problem solving
Tensile stress and tensile strain
A steel rod 2.0m long has a cross-sectional areaof 0.3𝑐𝑚2. the rod is now hung by one end to a
supportstructure anda550-kg millingmachineishungonthe otherendof the rod. Determinethe stress,
the strain and the elongation of the rod.
Set up
1. Identify the target
Stress:
We can use the formula 𝑠𝑡𝑟𝑒𝑠𝑠 =
𝐹⊥
𝐴
Strain:
We can also use, (𝑌) =
𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑎𝑖𝑛
𝑠𝑡𝑟𝑎𝑖𝑛 =
𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
(𝑌)
Elongation:
We can use the formula, 𝜀 =
∆𝐿
𝐿𝑜
2. What we know about the problem?
𝐿𝑜 = 2.0𝑚
𝐴 = 0.30𝑐𝑚2 = 3.0 × 10−5𝑚2
𝑤 = 𝑚𝑔 = 𝐹
𝑔 = 550𝑘𝑔(9.8
𝑚
𝑠2)
3. Execute
1. For stress we have,
𝑠𝑡𝑟𝑒𝑠𝑠 =
𝐹⊥
𝐴
=
5390𝑁
3.0×10−5𝑚2
= 1.8 × 108𝑃𝑎
2. For strain we have,
We use the formula,
𝑠𝑡𝑟𝑎𝑖𝑛 = 𝜀 =
𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
(𝑌)
=
1.8 × 108𝑃𝑎
20 × 1010𝑃𝑎
= 9.0 × 10−4
3. For the elongation we derive ∆𝐿 out the expression 𝜀 =
∆𝐿
𝐿𝑜
We will get ∆𝐿 = 𝜀𝐿𝑜
∆𝐿 = 𝜀𝐿𝑜
∆𝐿 = (9.0 × 10−4)(2.0𝑚) = 0.0018𝑚 = 1.8𝑚𝑚
Materialsare also subjectedtocompressive stresswhichcausesdeformationintwoorthree dimension
of the material.Tohelpyouvisualizethisconcept,imagine asponge beingsqueezedfromall
dimensions. We will seethatthere isa change involume.Hence compressive stresscreatesachange in
volume of a material. Thisdeformationcanbe expressedmathematicallyas,
𝑏𝑢𝑙𝑘 (𝑣𝑜𝑙𝑢𝑚𝑒) 𝑠𝑡𝑟𝑎𝑖𝑛 =
∆𝑉
𝑉
𝑜

5. It isalso true that differentmaterialshave diferentreactionstoforcesthattendstodeformitsvolume.
Each material have theirbulkmodulus.Bulkmodulusrelatesthe amountof stresstothe amountof
deformationof the material.We canwrite itas.
𝐵 =
𝑏𝑢𝑙𝑘 𝑠𝑡𝑟𝑒𝑠𝑠
𝑏𝑢𝑙𝑘 𝑠𝑡𝑟𝑎𝑖𝑛
(𝑏𝑢𝑙𝑘 𝑚𝑜𝑑𝑢𝑙𝑢𝑠)
A change inthe bulkstresscreatesa change involume.A change instress meansa change in pressure.
Thus we can argue that a change inpressure ∆𝑝 producesachange involume.Asistrue forgassesand
liquids. We canwrite an equationforthisfactby puttinginmindthat
𝑠𝑡𝑟𝑒𝑠𝑠 =
𝐹⊥
𝐴
= 𝑝 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
Writingthisinfact in termsof differential calculus,we have
∆𝑠𝑡𝑟𝑒𝑠𝑠 = ∆𝑝
We can nowrewrite the formulaforthe bulkmodulusas,
𝐵 =
𝑏𝑢𝑙𝑘 𝑠𝑡𝑟𝑒𝑠𝑠
𝑏𝑢𝑙𝑘 𝑠𝑡𝑟𝑎𝑖𝑛
= −
∆𝑝
∆𝑉/𝑉
𝑜
(𝑏𝑢𝑙𝑘 𝑚𝑜𝑑𝑢𝑙𝑢𝑠)
The negative signindicatesthatanincrease inthe pressure alwayscausesthe materialtoshrink.In
otherwordsif ∆𝑝 ispositive then ∆𝑉 isnegative. So𝐵 itself isapositive quantity.
The reciprocal of 𝐵 iscalledcompressibilityandisgivenbythe formula,
𝑘 =
1
𝐵
= −
∆𝑉
𝑉
𝑜
∆𝑝
= −
1
𝑉
𝑜
∙
∆𝑉
∆𝑝
(𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦)
Compressibilityisthe fractional decrease involume perfractional increase involume.
Problemsolving
A hydraulicpresscontains 0.25𝑚3 (250L) of oil.Findthe decrease involume whenitissubjectedtoan
increase inpressure ∆𝑝 = 1.6 × 102𝑃𝑎 (about160atm or 2300psi). the bulkmodulusof the oil is 𝐵 =
5.0 × 109𝑃𝑎(about5.0 × 104𝑎𝑡𝑚), and itscompressibilityis 𝑘 =
1
𝐵
= 20 × 10−6𝑎𝑡𝑚−1.
1. Identifythe target
We needtofind the decrease involume ∆𝑉 .
2. What do we have inhand?
a. ∆𝑝 = 1.6 × 102𝑃𝑎
b. 𝐵 = 5.0 × 109𝑃𝑎
c. 𝑘 =
1
𝐵
= 20 × 10−6𝑎𝑡𝑚−1.
d. 𝑉
𝑜 = 0.25𝑚3 (250L)
3. Execute
We derive aformulafor∆𝑉.
𝐵∆𝑉
𝑉
𝑜
= −
∆𝑝
1

6. ∆𝑉 = −
∆𝑝
1
𝑉
𝑜
𝐵
We have
∆𝑉 = −
(1.6 × 107𝑃𝑎)
1
(0.25𝑚3 )
(5.0 × 109𝑃𝑎)
= −8.0 × 10−4𝑚−3 𝑜𝑟 − .80𝐿
Alternativelywe canuse the conceptof compressibilitywhichwill giveus,
𝑘 =
1
𝐵
= −
∆𝑉
𝑉
𝑜
∆𝑝
= −
1
𝑉
𝑜
∙
∆𝑉
∆𝑝
So ∆𝑉 = −𝑘𝑉
𝑜∆𝑝
∆𝑉 = −𝑘𝑉
𝑜∆𝑝
∆𝑉 = −(20 × 10−6𝑎𝑡𝑚−1)(0.25𝑚3)(1.6 × 102𝑃𝑎) = −8.0 × 10−4𝑚−3 𝑜𝑟 − .80𝐿
Shearing stress
The process of parallel layers sliding past each other is known as shearing. We will notice that the
force is parallel to the area upon where it is applied.
The shear stress is expressed mathematically using the formula,
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 =
𝐹∥
𝐴
The shear strain is evident parallel to the force and perpendicular to the vertical length of the
material. Therefore we can write this using the trigonometric relationship of tangent.
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛 =
𝑥
ℎ
The shear modulus is then given by ,

7. 𝑆 =
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹∥
𝐴
𝑥
ℎ
=
𝐹∥ℎ
𝐴𝑥
(𝑠ℎ𝑒𝑎𝑟 𝑚𝑜𝑑𝑢𝑙𝑢𝑠)
Do you feel lonely or pressured?
In Some instances, you are needed to be elastic in your life just to overcome all the things you
encounter? It’s like a material that has been put in a pressurized situation to be molded into a
new one or testing your limit?