Stress & Strain

Elements

  • Mechanics treats a wide range of situations by viewing bodies as collections of small elements.
  • We view a body as composed of many small (theoretically infinitesimal) cubes of material.  An element is defined as one such cube of material in the body. 
              Generally, failure initiates at some local region in a body because conditions in those elements become critical. By contrast, overall deformation of a body is always due to the cumulative effect of deformation that occurs in elements throughout a body. So the response to forces is related to what happens in the small elements that form the body.  

Internal Force

  1. We apply Mechanics of Materials to structures, machines, or mechanisms in which forces at many levels are relevant.
  2. If we view a body as composed of two adjacent parts that meet at an imaginary plane or line passing through the body, then the internal force (moment) is defined as the force (moment) exerted by the two adjacent parts on each other.
  3. The individual elements exert forces on each other. We add up the forces across all the elements at the interface. We define the internal force Pas the net or resultant force transmitted between two adjacent portions of a body over all elements at the separating cross-section.
  4. The forces acting between the elements at an internal surface are statically equivalent to a single force in some direction, plus a couple or moment. These internal loads(force and/or couple) can be found from applying equilibrium to a portion of the body bounded by the internal surface.


Normal Stress σ

We define stress, σ, as the force per unit area on an element.
 Normal stress: the force acts perpendicularly (normally) to the element face.
  1. Force per area or stress is an effective way of describing forces on elements, since the force is essentially uniform on neighboring, identical, infinitesimal elements. 

                     The forces acting on small neighboring elements are equal. Say the force on an element of area ΔA is Δ P. Then, if four of those elements are combined into one with area 4Δ A, its force must be 4Δ P. Note that the force per area 4ΔP/4ΔA=ΔP/ΔA is the same. We can summarize all such combinations of element areas and forces by saying that the force per area ΔP/ΔA is locally uniform.
    We define stress, , as the force per area on an element.                      σ = ΔP/ΔA
    Stress alone, without reference to force or area individually,
    describes forces acting on elements.
             Stress has units of Force/Area : N/m2 =Pa 
  2. Draw stresses on an exposed internal surface of a body or just acting on a single element.
              Sometimes we depict stress by showing it acting on the surface of an element still embedded in the rest of the body (as above).Like forces, which obey Newton’s 3rd Law, an equal and opposite stress acts on an element in the adjoining body across their common surface. 

              Sometimes we depict stress by drawing only an element, in 2-D or 3-D, and the stresses on its faces. Usually, the element is aligned with x-y-zaxes. Notice that the element is in euilibrium, because the two forces on it, each equal to the stress times the element area, are equal and opposite. 
  3. Stress on an element can be very complex.          Stress, even on a single element, can become complicated. The stress can be normal stress perpendicularly to the element face or shear stress that act parallel to element faces . Additionally normal stresses act on faces of different orientations .Stresses can act on elements with faces not aligned with x-y-zplanes.
  4. Stresses will vary from element to element gradually across a surface. But, the element forces,
    when combined, are statically equivalent to the internal force and moment. That is why internal loads are sometimes referred to as stress resultants. 

           Mechanics of Materials addresses commonly occurring situations in which the distribution of stress across an internal surface can be found from the internal load and the shape of the surface.
  5. In special circumstances,the normal stress may be uniform across an entire surface.
         a common loading: two opposite axial forces. Imagine forces of magnitude P were applied to a deformable rectangular sheet of rubber.
    Lines have been drawn on the sheet to show the deformation of individual elements when equal and opposite forces are applied at the two dots.Here the sheet is deformed while forces act. Focus on the elements at two cross-sections: S1 and S2 .

    In a region such as S2, the deformation is nonuniform, so the stress will be nonuniform.Because the internal force on the sheet is the sum of the forces across all the elements,the average normal stress,σavg , can be found even if the stress is nonuniform:          σavg =P/A  

    We observe that the elements deform identically in the central portion of the sheet (S1), away from the applied forces. Since the material is uniform, the forces on all elements must be equal.
    So, in the central portion of the bar where the element forces are uniform and the stress at each point is also equal to the average stress:         σ=P/A 
  6. The local normal stress is uniform and equal to P/A under certain conditions.
               The normal stress is uniform (so σ=P/A ), under these conditions:
     
    St. Venant’s principle refers to the observation that the stress distribution becomes uniform as one moves away from the concentrated applied forces, subject to the above conditions. 
  7. Stresses that act to shorten an element are designated as compressive.    

    Normal stresses shown so far, which elongate a body, are tensile. Normal stresses may act instead to shorten a body (compressive). Neighboring elements then press on each other, as in this structural post. Besides giving the magnitude of the normal stress, we either specify that the stress is tensile or compressive, or we use a positive sign to denote tension and negative to denote compression.


Normal Strain ε

  • Normal strain, ε, is defined as the elongation per initial length, and can refer to an infinitesimal element or to a finite segment of a body.
  • Normal strain, , describes the element level intensity of deformation due to elongation, and it is defined as the increase in length of an element, due to deformation, divided by the element’s original length.
  • For an infinitesimal element of length Δ L that elongates by Δδ , we define normal strain, ε , by:
    ε =Δδ/Δ L
    Because the elements in this sheet can be seen to elongate by different amounts, the deformation
    (strain) in this sheet varies from element to element.
     
    The deformation is uniform over the segment of length L. If that segment elongates by δ, then the strain,ε , at each element in that segment is equal to the segment’s average normal strain:    ε =δ/L . We can still define an average normal strain ε avg   for the loaded part of the body. If the length Lb elongates by δb , then ε avg=δb/Lb
                   Strain is a pure number with no units, since the units of δ and L are both, say, meters. For machines and structures,strains are usually small (10-5 to 10-3 ). To avoid exponents with small numbers, we sometimes use percent strain or micro-strain: 1% strain 0.01 and 1 micro-strain 10-6 . In some situations, e.g., rubber and some biological tissues, strains can be large ( 1 or more). 
  • Like with stress, the definition of strain is particularly helpful for two reasons:
    • Strain describes the deformation of an element, without the need to specify the size of the element.
    • Provided the deformation in some region is uniform, one can compute the strain (equal in all elements in the region) by ε =δ/L , where L is the length of the elongating region and δ is its elongation.
  • Strains corresponding to shortening of a body are designated as compressive.Compressive strain is still change in length (shortening) per initial length. 

            We can specify the strain with a positive number, and state that it is compressive, or we give the strain a negative sign, provided we agree that negative strain means compression.
  • Don’t confuse strain with elongation and don’t confuse stress with strain.
             Elongate a 100 mm long strip (A) of rubber by 10 mm. Elongate a 20 mm long strip (B) of the same rubber by 5 mm. Compare their strains:
    ε A=δA/LA =10mm/100mm = 0.1               ε B=δB / LB = 5mm/20mm =0.25      
               Strip A has a lower strain even though its elongation is greater. 

    In engineering, stress quantifies the intensity of force per area that seeks to elongate or shorten an element. Strain quantifies the actual elongation or shortening, normalized by length. As we see next, the same stress can produce very different strains in two bodies of identical dimensions,if they are composed of different materials.



Shear Stress & Shear Strain

We saw at the start of the chapter that a small square element deforms in general into a parallelogram. So far, we have only considered elements with normal strain, those that elongate or shorten, but remain rectangular. Now, we consider shear strain,which quantifies deformation in which element edges no longer form right angles. 
  • Shearing corresponds to internal surfaces that slide parallel to themselves and elements that distort.
           Take a rectangular block of rubber, attach metal plates to its top and bottom,and etch a grid of square elements on its side face.Move the upper plate relativen to the lower plate in the direction parallel to the dimension a,maintaining the plates parallel. The etched grid distorts as shown here.
         
           Not all elements deform the same, but elements located away from the left and right edges distort identically. Their lower and upper faces do not lengthen. The side faces are no taller, but they have tilted.
  • Shear strain is quantified by the angle change at an element corner when it is deformed.
    Shear strain,γ , describes the element level intensity of deformation due to shape change, and it is defined as the tangent of the angle change, due to deformation, between two lines that are originally perpendicular.
    • element edges, which were originally perpendicular, make angles that are greater than and less than 90° or π/2 radians when deformed. Let β be the change from the initial right angle.
      The greater is β, the greater is the distortion or shear of the element. The shear strain, γ, is defined in terms of the angular distortion β .
      γ = tanβ
      If β in radians is small (β<< 1 ), then β≈ β,so γ≈ β. Also, when β is small, the change in length of the vertical elements is negligible.
  • Shear strain can be related to the shear displacement divided by the thickness of the sheared layer.
  • Using trigonometry, we can find the angle β in terms of the relative shear displacement. Let u be the displacement (motion) of a plate in the shear direction.β is related to the relative shear displacement,Δushear (i.e.,utop-ubottom ) and the height, h, perpendicular to the shearing.
  • Besides two oppositely acting shear forces, there must be additional loads to balance moments.
                 While one can readily picture the shear strain, it is more difficult to picture the forces on the plates that shear the block. Here are several possibilities (only the second and third are in equilibrium).

    Assume the forces to be as in the third case above. Consider the internal forces at a surface parallel to F0.
           There must be a horizontal shear force V equal to F0 . We call this a shear force because it acts parallel to, rather than normal to, the face on which it acts.
                  Now cut across the block vertically. There must be a vertical shear force, although its magnitude is unclear because it depends on how the moment is balanced.
    So elements that shear have forces on both horizontal and vertical faces.
  • Shear stress,τ -It is defined as the shear force per unit area.
                  It describes the elementlevel intensity of internal shear force, and
    it is defined as the shear force exerted by two adjacent elements on each other,
    divided by the area on which the force acts.
    The shear stress is rarely uniform, but the average value can be found in terms of the shear force, V, and the area, A, on which it acts.
  • Shear stress, or shear force per area on an element, must be equal on horizontal and vertical planes.
              The shear stress just found is shown as τ1 acting on the top and bottom faces of a small element. There are likewise equal and opposite shear stresses,τ2 , on the vertical faces.
            For this element to be in equilibrium, the moment about the center must be zero. If the element is dx by dy by 1 thick, then ΣM=-τ1(dx)(1)(dy) + τ2(dx)(1)(dy)=0 .So, the shear stresses must be equal:τ1=τ2 . A single shear stress,τ , describes the shear force per area on horizontal and vertical faces.
  • For a linear elastic material, the shear strain is recovered upon unloading and it is proportional to the shear stress.
          Shear modulus, G, is defined as the proportionality between shear stress and shear strain for a linear elastic material.
    G = τ / γ


Shear and Bearing Stress in Pin Joints
Pins commonly fail in shear, so it is important to be able to calculate the shear stress in pins.
  • Pins often connect two or more members which can exert transverse forces on the pin.
           A pin often joins two or more members and allows them to pivot with respect to each other. The forces between members and the pin typically act in the direction perpendicular to the axis of the pin. The pin joint in this elliptical exercise machine features a typical symmetricarrangement: the body that is forked or clevised (ski) straddles the second body (arm) and both bodies contact the pin.
  • Calculate the forces exerted by members on the pin using methods from Statics.

                  From equilibrium of the system, we would find the magnitude of the force, F0, which the pin and center body (arm, here) exert on one another. (We would combine x and y components if the statics analysis was done with components.) The pin is in equilibrium, so each half of the clevis applies equal forces of magnitude F0/2 to the pin. We can draw the joint in the plane of the forces. We are interested in forces between the pin and the connected bodies and within the pin. With this arrangement of transverse forces, the pin is said to be in “double shear.”
  • Bearing stresses are the forces per area at the surfaces where the members and the pin press on each other.
            The members and the pin contact each other on cylindrical surfaces. We say the bodies bear (press) on each other. The forces per area, the contact or bearing stresses, are sometimes computed. Here we show the contact force, F0/2, between the pin and one side of the clevis as a distributed force. (Analogous contact forces act between the center member and the pin.)
  • Calculation of bearing stress is highly approximate because the area of contact is estimated very roughly.
           In general, the bearing stress σbearing , bearing force P bearing and bearing area A bearing are related by
    σbearing = P bearing / A bearing
           The bodies actually contact over less than the half-cylindrical surface wd2. The area of contact is usually estimated as the projected area wd. So the bearing stress between the clevis and the pin is approximately
    σbearing = P bearing / A bearing = Fo/2wd
    The stresses and strains due to such contact are highly nonuniform, and this estimate is very approximate. If such stresses were considered at all critical, they should be calculated with more accurate methods
  • Shear stresses act between two portions of the pin across an internal plane perpendicular to the
    pin axis.
              In general, we can look at any cross-section through the pin and determine internal forces and moments.The exaggerated deformation of the pin makes it clear that there is bending. Here we focus on one cross-section and in particular on the shear force at that cross-section. Later we will see why the shear force is maximum at this cross-section.
  • Draw a free body diagram of a portion of  the pin on which external forces and the shear force
    act and calculate the shear force and then shear stress.
              Consider the cross-section of the pin between the left half of the clevis and the arm. We find the internal loads in the pin at this cross-section by isolating a portion of the pin from one end to the cross- section.The internal force V is a shear force since it acts parallel to the internal cross-section.M is a  bending moment, which we do not study further here.
    To balance transverse forces, the shear force V at the chosen cross-section must equal F0/2. Again, this situation of a pin symmetrically loaded is referred to as “double shear.” V acts across the circular area πd2/4. So the average shear stress is
                                   τ = V/A = V/πd2/4 = 4Fo/2πd2
  • A pin on which only two opposing forces acts undergoes “single shear.”
            Another common configuration of a pin has only two members contacting the pin. The two forces exerted on the pin must be equal and opposite. For equilibrium, one or both members also exert a moment on the pin. To minimize the bending, the members are usually near each other and the pin is short. By equilibrium, the shear force V=Fo . This loading of the pin is referred to as “single shear.” V again acts across the circular area πd2/4. So the average shear stress is

    τ= V/A = 4V/πd2 = 4Fo/πd2


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