Hooke's Law
Strain ε (epsilon) Calculation
In 1660 Robert Hooke (1635–1703) discovered Hooke’s law that states that the force needed to extend or compress a spring by some distance is proportional to that distance.
According to Hooke's law, elongation in material is directly proportional to the stretch tension. The above is valid when the modulus of elasticity is constant, ie when the tension does not exceed a certain limit, after which the piece begins to deform plastically. When the loaded bar tension is less the elastic limit, the rod is stretched elastically and returns to its original shape after the stress has ended. If the tension is great, this achieves the plastic zone, which means a permanent deformation to piece.
F <=== ========== ====> F
F ===> ========== <==== F
The elongation is directly proportional to the tensile force and the length of the bar and inversely proportional to the cross-sectional area and the modulus of elasticity.
In the elastic range of a material, strain is proportional to stress.
Hooke's experimental law may be given by:
Lσ = F x L0
A x E
Lσ = extension of a bar
σ = tension in general
F = force producing extension (kN)
L0 = original length of a bar
A = cross-sectional area of a bar (cm2)
E = elastic constant of the material, the Modulus of Elasticity, also named Young's Modulus (20 600 kN/cm2)
Example 1
A = 1 cm2 L0 = 100 cm F = 1 kN
Lσ = F x L0 = 0.00485 cm
A x E
Example 2
How much is the steel bar strain having a cross-sectional area of 12 cm2, when a length of 18 meters and a force of 42.5 kN.
a)
Lσ = 42.5 x 18 x 0.00485 cm
12
Lσ = 0,31 cm
b)
Lσ = 42,5 x 1800 cm
12 x 20600
Lσ = 0,31 cm
Tensile Stress
Tensile stress, or simply stress, was equated to the load per unit area or force applied per cross-sectional area perpendicular to the force measured in kN force per square cm2. Original units were different, but Hooke's idea was the same in 1687.
The load per unit area
σ = F
A
Example 1
42.5 kN normal force affects to 12 cm2 cross-sectional area. What is the tension?
σ = F
A
σ = 42,5 kN
12 cm2
σ = 3,54 kN / cm2
ε (epsilon) Elongation of the rod when squeezed or pulled
ε = Lσ / L0
ε = 0,31cm / 1800 cm
ε = 1,719 x 10-4
σ = ε x E
σ = 1,719 x 10-4 x 20600 kN/cm2
σ = 3,54 kN/cm2
Tensile strain of a bar per unit length
ε = σ / E
For the equations described above expressed by Hookes Law for elastic materials. For materials under tension, strain (ε) is proportional to applied stress σ. Removal of the stress results in a gradual return of the metal to its original shape and dimensions.
F ===> ========== <==== F
Steel bar is squezed 1:10 000 of its length. Normal tension in the bar is
σ = ε x E
σ = (1:10 000) x 20 600 kN/cm2
σ = 2.06 kN/cm2
22.5.2015*00:15 (1063 - 203)
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