NOTEBOOK OF MECHANIC OF MATERIALS
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NOTEBOOK OF MECHANIC OF MATERIALS Class: Mechanic of Materials Name: Héctor Abraham Torres Chávez ID: 1569872 Hour: M5 Teacher: Ing. Pablo Ernesto Tapia Date: 5 of Jun of 2012 Introduction The mechanic of materials enhances the study of the forces was initiated in statics. The field of statics covers, fundamentally, the relations among the forces that act over a solid, not deformable. In constant with statics, the mechanic of materials studies and establishes the relation among the external beds and their effects in the interior of the solids, besides does not suppose that the solids are not deformable and any minimum deformation have great . Axial Tension UNITS Distance Mass Time Force Gravity Stress SI M Kg S N (Kgm/π 2 ) 9.8m/π 2 Pa (N/π2 Normal (σ) Axial Bearing Flexion Shear (Ζ¬) Torsion Cross English Ft Slugs S Lbm 32.2 (ft/π 2 ) Psi (lb/ππ2 ) TENSION COMPRESSION π =π΄ Normal Stress Note to obtain a normal stress the axial load and the area P= Axial Load must be perpendicular A =Cross Sectional Area Types of Vanes ο· Bearing Stress or Crushing The stress as opposed to the stress of compression that exists in the interior of the bodies under external loads is the one that is produce don the surface of contact of the bodies. Is expressed as π = π΄π Bearing Stress P= Force acting over the normal stress Ap= Normal Projected Area Shearing Stress A shearing stress is defined as a stress which is applied parallel to the cross section of a material, as opposed to a normal stress which is applied longitudinally. π΄= ππ4 4 V= Shearing Force A= Cross Area, parallel to the V π π=π΄ π π΄ =π∗π€ π=π΄ π = π/2π΄ π΄ = ππ ∗ β Strain When a solid body is exposed to external loads it deforms. That means that changes in size and form are presented The term deformation is general and includes either, changes of length as of angles To define the strain of extension the case of axial deformation is examinated. The original length L of the bar and its final length L are used to define the strain E as the ratio of the total deformation and the original length. V= Poisson’s Ratio π = π/πΏ S= Deformation GRAPH OF UNIT DEFORMATION ο· Simple Deformation Is the deformation due to axial loads, It can be shortening or lengthening of an element Torsion Torsion is due to a pair of forces acting on the end of an element, generally with a circular cross section, equilibrated a point of its length. When an element is supporting a torsion load, a shear stress and angular deformation appears. ∅ = Angular Deformation π = Moment of Torsion πΏ= Length of the Element π = Shear Stress π· = Outer diameter π = inner diameter πΊ = Shear Modulus of Elasticity πΌπ = Polar Moment of Inertia π =πΉ∗π πΌπ = ππ 4 32 πΌπ = π(π· 4 − π4 ) 32 16π π = ππ3 16ππ· π = π(π·4 −π4 ) ππΏ ∅ = πΊπΌπ Second Part TOPICS ο· Stress due the flexion ο· Centroids ο· Complex Form ο· Moment of Inertia ο· Steiner’s theorem (theorem of the parallel axes) Stress due the flexion I= Moment of Inertia of Area Centroids Is a geometric center of the object’s shape, it’s a center of a mass or the center of gravity. In formally it’s the average of all points weighted by a local density, it’s the point where the area will be able to balance, being supported only on that point. Simple Forms: π π=2 π= π= β π= 2 π 2 π 2 π π=2 β π=3 Complex Form The product of the total area by the distance to the center of the total area is equal to the sum of the product of the area per each component by the distance of its centroid to the same axes of reference. This principal is use when the complex form is composing of several simple forms. π΄π₯ = ∑ π΄ππ₯π π₯= ∑ π΄ππ₯π π΄π π΄π¦ = ∑ π΄ππ¦π π¦= ∑ π΄ππ¦π π΄π Moment of Inertia Properties that indicates the stiffness of a beam. Is the opposition of a body to be bend due to its cross area the deflection of a beam is inversely proportional to the moment of inertia of body. πΌπ₯ = πΌπ¦ = πΌπ₯ = πΌπ¦ = πΌπ₯ = πΌπ¦ = πΌπ₯ = πβ3 12 βπ3 12 ππ 4 64 ππ 4 64 ππ·4 64 ππ·4 πβ3 36 64 − − ππ 4 64 ππ 4 64 Steiner’s theorem (theorem of the parallel axes) Can be used to determine the moment of inertia of area about any axis, given the moment of inertia of area in the object about the parallel axis trough the objet centroid and the perpendicular distance between the axes. πΌπ₯ = ∑ πΌπ₯π + ∑ π΄π ππ₯π 2 πΌπ¦ = ∑ πΌπ¦π + ∑ π΄π ππ¦π 2
