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👉Stress vs strain
⚡️Part 1
🌟Stress is the ratio of applied force F to a cross section area defined as "force per unit area".
tensile stress - stress that tends to stretch or lengthen the material - acts normal to the stressed area compressive stress - stress that tends to compress or shorten the material - acts normal to the stressed area
⏺shearing stress - stress that tends to shear the material - acts in plane to the stressed area at right-angles to
compressive or tensile stress
⏺Tensile or Compressive Stress - Normal Stress
▶️Tensile or compressive stress normal to the plane is usually denoted "normal stress " or "direct stress " and can be
expressed as
σ = F n / A (1)
where
σ = normal stress (Pa (N/m 2 ), psi (lb f /in 2 ))
F n = normal force acting perpendicular to the area (N, lb f)
A = area (m 2 , in 2 )
a kip is an imperial unit of force - it equals 1000 lb f (pounds-force)
1 kip = 4448.2216 Newtons (N) = 4.4482216 kilo Newtons (kN)
▶️A normal force acts perpendicular to area and is developed whenever external loads tends to push or pull the two segments of a body.
🚧Example - Tensile Force acting on a Rod A force of 10 kN is acting on a circular rod with diameter 10 mm The stress in the rod can be calculated as
σ = (10 10 3 N) / (π ((10 10 -3 m) / 2) 2 )
= 127388535 (N/m 2 )
= 127 (MPa)
🚧Example - Force acting on a Douglas Fir Square Post A compressive load of 30000 lb is acting on short square 6 x 6 in post of Douglas fir. The dressed size of the post is 5.5 x 5.5 in and the compressive stress can be calculated as
σ = (30000 lb) / ((5.5 in) (5.5 in))
= 991 (lb/in 2 , psi)
⏺Shear Stress
▶️Stress parallel to a plane is usually denoted as "shear stress "and can be expressed as
τ = F p / A (2)
where
τ = shear stress (Pa (N/m 2 ), psi (lb f /in 2 ))
F p = shear force in the plane of the area (N, lb f)
A = area (m 2 , in 2 )
▶️A shear force lies in the plane of an area and is developed when external loads tend to cause the two segments of a body to slide over one another.
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⚡️Part 1
🌟Stress is the ratio of applied force F to a cross section area defined as "force per unit area".
tensile stress - stress that tends to stretch or lengthen the material - acts normal to the stressed area compressive stress - stress that tends to compress or shorten the material - acts normal to the stressed area
⏺shearing stress - stress that tends to shear the material - acts in plane to the stressed area at right-angles to
compressive or tensile stress
⏺Tensile or Compressive Stress - Normal Stress
▶️Tensile or compressive stress normal to the plane is usually denoted "normal stress " or "direct stress " and can be
expressed as
σ = F n / A (1)
where
σ = normal stress (Pa (N/m 2 ), psi (lb f /in 2 ))
F n = normal force acting perpendicular to the area (N, lb f)
A = area (m 2 , in 2 )
a kip is an imperial unit of force - it equals 1000 lb f (pounds-force)
1 kip = 4448.2216 Newtons (N) = 4.4482216 kilo Newtons (kN)
▶️A normal force acts perpendicular to area and is developed whenever external loads tends to push or pull the two segments of a body.
🚧Example - Tensile Force acting on a Rod A force of 10 kN is acting on a circular rod with diameter 10 mm The stress in the rod can be calculated as
σ = (10 10 3 N) / (π ((10 10 -3 m) / 2) 2 )
= 127388535 (N/m 2 )
= 127 (MPa)
🚧Example - Force acting on a Douglas Fir Square Post A compressive load of 30000 lb is acting on short square 6 x 6 in post of Douglas fir. The dressed size of the post is 5.5 x 5.5 in and the compressive stress can be calculated as
σ = (30000 lb) / ((5.5 in) (5.5 in))
= 991 (lb/in 2 , psi)
⏺Shear Stress
▶️Stress parallel to a plane is usually denoted as "shear stress "and can be expressed as
τ = F p / A (2)
where
τ = shear stress (Pa (N/m 2 ), psi (lb f /in 2 ))
F p = shear force in the plane of the area (N, lb f)
A = area (m 2 , in 2 )
▶️A shear force lies in the plane of an area and is developed when external loads tend to cause the two segments of a body to slide over one another.
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👉Strain (Deformation)
Part 2
🌟Strain is defined as "deformation of a solid due to stress".
⏺Normal strain - elongation or contraction of a line segment
Shear strain - change in angle between two line segments originally perpendicular Normal strain and can be expressed as
ε = dl / l o
= σ / E (3)
▶️where
dl = change of length (m, in)
l o = initial length (m, in)
ε = strain - unit-less
E = Young's modulus (Modulus of Elasticity) (Pa , (N/ m 2 ), psi (lb f /in 2 ))
⏺Young's modulus can be used to predict the elongation or compression of an object when exposed to a force
⚡️Note that strain is a dimensionless unit since it is the ratio of two lengths. But it also common practice to state it as the ratio of two length units - like m/m or in/in .
⏺Poisson's ratio is the ratio of relative contraction strain
🚧Example - Stress and Change of Length The rod in the example above is 2 m long and made of steel with Modulus of Elasticity 200 GPa (200 10 9 N/m 2 ) The change of length can be calculated by transforming (3) to
dl = σ lo / E
= (127 10 6 Pa) (2 m) / (200 10 9 Pa)
= 0.00127 m
= 1.27 mm
⏺Strain Energy
▶️Stressing an object stores energy in it. For an axial load the energy stored can be expressed as
U = 1/2 F n dl
where
U = deformation energy (J (N m), ft lb)
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Part 2
🌟Strain is defined as "deformation of a solid due to stress".
⏺Normal strain - elongation or contraction of a line segment
Shear strain - change in angle between two line segments originally perpendicular Normal strain and can be expressed as
ε = dl / l o
= σ / E (3)
▶️where
dl = change of length (m, in)
l o = initial length (m, in)
ε = strain - unit-less
E = Young's modulus (Modulus of Elasticity) (Pa , (N/ m 2 ), psi (lb f /in 2 ))
⏺Young's modulus can be used to predict the elongation or compression of an object when exposed to a force
⚡️Note that strain is a dimensionless unit since it is the ratio of two lengths. But it also common practice to state it as the ratio of two length units - like m/m or in/in .
⏺Poisson's ratio is the ratio of relative contraction strain
🚧Example - Stress and Change of Length The rod in the example above is 2 m long and made of steel with Modulus of Elasticity 200 GPa (200 10 9 N/m 2 ) The change of length can be calculated by transforming (3) to
dl = σ lo / E
= (127 10 6 Pa) (2 m) / (200 10 9 Pa)
= 0.00127 m
= 1.27 mm
⏺Strain Energy
▶️Stressing an object stores energy in it. For an axial load the energy stored can be expressed as
U = 1/2 F n dl
where
U = deformation energy (J (N m), ft lb)
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