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Elasticity is a property of an object or material which will restore it to its original shape after distortion
Stress (normal stress. Axial stress) (𝛔)
Stress is defined as the force per unit area of a material.
Stress =
force
cross sectional area
Nm-2
= Pa (Pascal)
Strain (𝛆)
Strain=
𝒆𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏
𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉
Strain is defined as extension per unit length
 Axial (Longitudinal, linear, normal) Strain
𝜀𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 =
𝛿 𝑙𝑜𝑛𝑔
𝐿
 Lateral Strain:
𝜀𝑙𝑎𝑡𝑒𝑟𝑎𝑙 =
𝛿 𝑙𝑎𝑡𝑒𝑟𝑎𝑙
𝑟
 Volumetric Strain:
It is the ratio of the change in volume of a body to its original volume
Volume Strain =
Change in volume
Original volume
=
ΔV
V
Strain has no units and dimensions.
Poisson's Ratio (𝝂)
Ratio of lateral to longitudinal strain
𝜈 =
lateral strain
𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
Stress =
F
A
r
L
P
δlatera
l
L
P
δlong
Lateral and longitudinal are depend on the direction of force apply
A
F
Young’s modulus,tensile modulus, elastic modulus (E, y)
Measure of the stiffness of an elastic isotropic material
Young’s modulus=
normal stress
𝑛𝑜𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
=
(𝐹𝑜𝑟𝑐𝑒)(original 𝐿𝑒𝑛𝑔𝑡ℎ)
(𝐴𝑟𝑒𝑎)(𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛)
=
𝜎
𝜀
Units of the Young modulus E: Nm-2 or Pa
Bulk modules (B)
The bulk elastic properties of a material determine how much it will compress under a given
amount of external pressure
Units of the bulk modulus E: Nm-2 or Pa
Shear strain=
∆𝑥
𝐿
=tan (i) no dimension
Shear stress=
𝑓𝑜𝑟𝑐𝑒 𝑎𝑝𝑝𝑙𝑖𝑒𝑑
area of parallel to the force
=
𝐹
𝐴
Nm-2
= Pa
B =
𝑝𝑟𝑒𝑎𝑠𝑢𝑟𝑒(𝑠𝑡𝑟𝑒𝑠𝑠)
𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛
=
𝑝
∆𝑣/𝑣
E= 3B (1-2ν)
Bulk
Poison
Young’s
Double shear stress
=
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛
Nm-2
= Pa
Young’s
modulus(E)
Shear
modulus(G)
Bulk modulus
Under tension and
compression
Under shearing Under hydraulic
stress
Strain is=∆L/L Strain is =∆x/L Strain is=∆v/v
E=
𝑠𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
𝐿
∆𝐿
G=
𝑠𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
𝐿
∆𝑥
B=
𝑠𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=𝑝
𝑣
∆𝑣
=
𝐹
𝐴
𝑣
∆𝑣
F
F/2
F/2
F
τ=
𝑓
𝐴
τ=
𝑓
2𝐴
F
A=cross sectional area
Shear modules, modulus of rigidity
Shear Modulus of Elasticity (G)
Hook’s law
Hooke's Law is that which says that how much stress we apply on anybody that much strain will be
observed on it
Hooke's Law states that the force acting on a spring is directly proportional to its
displacement from its equilibrium position (as long as the spring does not exceed its elastic
limit)
Where F is the amount of force applied in N,
x is the displacement in the spring in m
k is the spring constant or force constant.
Figure 1: A force of 100 N is stretching a spring by 0.2 m. Calculate the force constant?
Solution:
Given: Force F = 100 N,
Extension x = 0.2 m.
the force constant is given by k = - F/X
= - 100N/0.2m = - 500 N/m.
Figure 2: A girl weighing 30 pounds stretches a spring by 60 cm. Calculate the spring constant of the
spring?
Solution:
Given: Mass m = 30 lbs. =
30
2.2
= 13.64 Kg.
Displacement x = 60 cm
the force F = ma = 13.64 × 9.8 = 133.67 N
The spring constant is given by k = -
𝐹
𝑥
= -
133.67
0.6
= - 222.786 N/m.
Spring constant, force constant (k) =- 𝑓
𝑥
Figure 1 Figure 2
elasticity
elasticity

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Grade 9 Q4-MELC1-Active and Passive Voice.pptx
Grade 9 Q4-MELC1-Active and Passive Voice.pptxGrade 9 Q4-MELC1-Active and Passive Voice.pptx
Grade 9 Q4-MELC1-Active and Passive Voice.pptx
 

elasticity

  • 1. Elasticity is a property of an object or material which will restore it to its original shape after distortion Stress (normal stress. Axial stress) (𝛔) Stress is defined as the force per unit area of a material. Stress = force cross sectional area Nm-2 = Pa (Pascal) Strain (𝛆) Strain= 𝒆𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏 𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉 Strain is defined as extension per unit length  Axial (Longitudinal, linear, normal) Strain 𝜀𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 = 𝛿 𝑙𝑜𝑛𝑔 𝐿  Lateral Strain: 𝜀𝑙𝑎𝑡𝑒𝑟𝑎𝑙 = 𝛿 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑟  Volumetric Strain: It is the ratio of the change in volume of a body to its original volume Volume Strain = Change in volume Original volume = ΔV V Strain has no units and dimensions. Poisson's Ratio (𝝂) Ratio of lateral to longitudinal strain 𝜈 = lateral strain 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 Stress = F A r L P δlatera l L P δlong Lateral and longitudinal are depend on the direction of force apply A F
  • 2. Young’s modulus,tensile modulus, elastic modulus (E, y) Measure of the stiffness of an elastic isotropic material Young’s modulus= normal stress 𝑛𝑜𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = (𝐹𝑜𝑟𝑐𝑒)(original 𝐿𝑒𝑛𝑔𝑡ℎ) (𝐴𝑟𝑒𝑎)(𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛) = 𝜎 𝜀 Units of the Young modulus E: Nm-2 or Pa Bulk modules (B) The bulk elastic properties of a material determine how much it will compress under a given amount of external pressure Units of the bulk modulus E: Nm-2 or Pa Shear strain= ∆𝑥 𝐿 =tan (i) no dimension Shear stress= 𝑓𝑜𝑟𝑐𝑒 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 area of parallel to the force = 𝐹 𝐴 Nm-2 = Pa B = 𝑝𝑟𝑒𝑎𝑠𝑢𝑟𝑒(𝑠𝑡𝑟𝑒𝑠𝑠) 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝑝 ∆𝑣/𝑣 E= 3B (1-2ν) Bulk Poison Young’s
  • 3. Double shear stress = 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛 Nm-2 = Pa Young’s modulus(E) Shear modulus(G) Bulk modulus Under tension and compression Under shearing Under hydraulic stress Strain is=∆L/L Strain is =∆x/L Strain is=∆v/v E= 𝑠𝑡𝑟𝑒𝑠𝑠 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝐹 𝐴 𝐿 ∆𝐿 G= 𝑠𝑡𝑟𝑒𝑠𝑠 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝐹 𝐴 𝐿 ∆𝑥 B= 𝑠𝑡𝑟𝑒𝑠𝑠 𝑠𝑡𝑟𝑎𝑖𝑛 =𝑝 𝑣 ∆𝑣 = 𝐹 𝐴 𝑣 ∆𝑣 F F/2 F/2 F τ= 𝑓 𝐴 τ= 𝑓 2𝐴 F A=cross sectional area Shear modules, modulus of rigidity Shear Modulus of Elasticity (G)
  • 4. Hook’s law Hooke's Law is that which says that how much stress we apply on anybody that much strain will be observed on it Hooke's Law states that the force acting on a spring is directly proportional to its displacement from its equilibrium position (as long as the spring does not exceed its elastic limit) Where F is the amount of force applied in N, x is the displacement in the spring in m k is the spring constant or force constant. Figure 1: A force of 100 N is stretching a spring by 0.2 m. Calculate the force constant? Solution: Given: Force F = 100 N, Extension x = 0.2 m. the force constant is given by k = - F/X = - 100N/0.2m = - 500 N/m. Figure 2: A girl weighing 30 pounds stretches a spring by 60 cm. Calculate the spring constant of the spring? Solution: Given: Mass m = 30 lbs. = 30 2.2 = 13.64 Kg. Displacement x = 60 cm the force F = ma = 13.64 × 9.8 = 133.67 N The spring constant is given by k = - 𝐹 𝑥 = - 133.67 0.6 = - 222.786 N/m. Spring constant, force constant (k) =- 𝑓 𝑥 Figure 1 Figure 2