1. Concepts about the science of "Strength of Materials" tasks, consistency, uniformity, priority, brief history



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Materiallar qarshiligi answers (2)

Modulus of elasticity


According to Hooke's law, the stress in a body is proportional to the corresponding strain. The constant of proportionality is called modulus of elasticity. Hooke's law is valid for small value of stress and for small deformations (strain).
The ratio of longitudinal stress to longitudinal strain, within the elastic limit to which Hooke's law is applicable, is called Young's modulus of the material i.e.,Y=longitudinal stresslongitudinal strain=(F/A)/Δl/l=Fl/AΔl.=longitudinal stresslongitudinal .The unit of Young's modulus is same as that of stress i.e., N/m2.
When uniform pressure p is applied over the whole surface of a body, it produces a uniform compression. If the pressure is small, the compression is proportional to the pressure. The ratio of the pressure to the volume strain is called Bulk modulus of the material,B=−p/(ΔV/V)
The ratio of shearing (or tangential) stress to sharing strain is called shear modulus or modulus of rigidity i.e.,η=Fs/Ax/l=FslxA=FsAθ
Poisson's ratio
Consider pulling of a wire by a force F The longitudinal dimension (length) of the wire is increased from l to l+Δl+Δ. Its lateral dimension (diameter) is decreased from D to D−ΔD−Δ. Within elastic limits there is a complete proportionality between the lateral strain and the longitudinal strain and the ratio between the lateral strain and longitudinal strain is called Poisson's ratio i.e.,σ=ΔD/DΔl/l=lDΔDΔlPoisson's ratio is a dimensional quentity.

Relationship among Modulii of elasticity


Young's modulus, Bulk modulus and shear modulus are related byY=9ηB/3B+η
σ=3B−2η/6B+2η.

Stress strain curve


Consider pulling of a wire. When the strain is small (say less than 1%), the stress is proportional to the strain. This is the region where Hooke's law is valid. The point on the stress-strain curve upto which stress is proportional to strain is called proportional limit. The body return to its original state after removal of external force.
If the strain is increased a little bit, the stress is not proportional to the strain. This region is called plastic region. The body does not return to its original state in the plastic region.

The fracture point is the point at which the wire break. The stress at fracture point is called breaking stress.
20.Analysis of potential energy in torsion, stress state and shaft decay (potential energy, types of stress, distribution, impact surfaces, resistance, plastic, brittle, shear, break).
The potential energy accumulated in the twisting bar is determined similarly to the tension-compression deformation.
Potential energy stored in an element
.
Assuming the connection (7.3).
.
Integrating this expression along the length, the full potential energy of the bar is found:
(7.14)
If the torsion and shaft length are constant
(7.15)
In the last connection, the following expression obtained on the basis of (7.9) was used

In the transverse sections of the twisting shaft, tensile stresses are formed, and they change from zero in the center of the section to their maximum values at the edges of the section (5.4) based on the linear law of connection According to the law of the pair of tensile stresses, the same tensile stresses are applied in the longitudinal sections of the beam.
On the surface of the rotating shaft, the cross-sections consisting of longitudinal and transverse components are affected only by tensile stresses Hence, an element separated by such generators is in a state of pure displacement.
Both normal and tensile stresses are exerted on sites inclined to these structures. This is exactly what was seen in 4.2 above. It is known that in this case, the greatest tensile and compressive normal stresses affect the surfaces of pure sliding (relative to the axis of the shaft in torsion) at 45 and 135 degrees
Thus, in the twisting of shafts with a circular cross-section, both the experimental stresses acting on the transverse and longitudinal sections, and the normal stresses on the oblique sections located at an angle of 45 degrees to them can be dangerous. Therefore, the way the shaft wears out depends on the ability of its material to resist shearing and breaking.
For example, steel shafts, which belong to the category of plastic materials, are eroded by cross-sectional shearing under the influence of tensile stresses. Wood shafts are cracked and eroded under the influence of tensile stresses in longitudinal sections along the fibers. If the shaft material shows poor resistance to tensile stresses, such as cast iron, the fracture path is perpendicular to the direction of the principal tensile stresses, i.e., on a helical surface inclined at 45 degrees to the shaft.

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