Stresses and strains

As a follow-on to the discussion of statics we need to consider whether a material subject to a given set of forces will break and if not, how much will it bend?  To determine this, the first step is to compute the stress in the material. There are two flavors of stress, the normal stress which is the stress in the direction perpendicular to an imaginary plane in the material, and the shear stress which is the stress in the direction parallel to that same imaginary plane. Of course, this imaginary plane could be in any direction, so the magnitude of the normal and shear stresses depends strongly on the choice of said imaginary plane. How should one choose said plane and what is the relationship between normal and shear stresses? We’ll discuss that shortly, but first we’ll individually define normal then shear stress.

 

          

where  F  is  the  force  (either  tension  or  compression)  acting  perpendicular  to  an  imaginary  plane surface passing through a piece of material and A is the cross section area. It is called “normal” not in the sense of being “typical” or “standard” but in the sense of being perpendicular or orthogonal to the cross-section of the material. Stress is defined as positive if the material is in tension (i.e. the material  is being  pulled  apart)  and  negative  if  the  material  is  in  compression (i.e. being  squeezed together).  From the definition it is clear that stress has units of force/area, i.e. the same as pressure. The units are typically N/m2 or lbf/in2. Often the unit of “kips” (kilopounds per square inch = 1000 lbf/in2) is used to report stress.

In  order  to  characterize  the  deformation  of  a  material  in  response  to  stress  we  define  another property  called  strain  (ε) which is  the  fractional  amount  of  elongation  (increase  in  length)  or contraction (decrease  in  length) in  a  material  caused  by  a  stress.   For  example,  if  under  a  given amount of tensile stress, a steel bar stretches from a length (L) of 1.00 inch to 1.01 inch (a change in length, ΔL, of 0.01 inch) the strain = (1.0 – 1.00)/1.00 = 0.01. In other words, 

       

For most materials (other than gooey ones, i.e. Silly Putty™, Play-Doh™, …) the amount of strain before failure is relatively small (i.e. less that 0.1, meaning that the material deforms less than 10% before failing.)

An elastic material has a linear relationship between stress and strain, i.e. 

        

where E is called the elastic modulus, i.e. the slope of the plot of σ vs. ε in the elastic region shown in Figure 10. Note that since ε is dimensionless, E also has units of force per unit area.

The strength of a material is generally reported in terms of the maximum stress it can withstand. For  a  sufficiently  small  stress,  materials  return  to  their  original  length  or  shape  after  the  stress  is removed. The smallest stress for which the material does not return to its original length or shape after the stress is removed is called the yield stress (σ yield). Beyond this stress, generally the slope of the σ vs. ε plot generally (but not always) becomes smaller. There is often an increase in slope as ε is increased  still  further,  up  to  a  maximum σ called  the  “ultimate  stress”,  beyond  which σ actually decreases as ε increases, leading finally to fracture (breakage) of the material at which point it can no longer  hold  any  stress.  A  material’s  yield  stress  may  be  (and  usually  is)  different  in  tension, compression and shear.

Note that we can write Equation 20 in the form F/A = (ΔL/L)E or F = (EA/L)(ΔL), which looks just like the force on a linear spring, F = kx, with k = EA/L.  (One might wonder what happened to the – sign, that is, isn’t F = -kx? Stress is usually defined as positive in tension where as for the spring  F  is  defined  as  positive  in  compression.)   So  E  and  the  material  dimensions  A  and  L determine its “spring constant.”  Figure 10 shows examples of stress vs. strain (σ vs. ε) relationships for different materials.  A ductile material such as steel will yield significantly before fracture whereas a brittle material such as a ceramic or concrete will fail without significant yielding, that is, the σ vs. ε curve is nearly linear up to the failure point. This doesn’t mean that ceramics are necessary weak, in fact they may have higher E than ductile materials, but they are unforgiving to over-stressing (really, over-straining.)

E  and  yield  or  ultimate  stress  have  the  same  units  (Pa  or  lbf/in2)  but  there  is  no  particular relationship between E and yield or ultimate stress. Materials can be hard (high E) but break easily (low yield  or  ultimate  stress)  or  vice  versa. 

      

    

direction stress is applied relative to the material. Many engineering materials are anisotropic, i.e. they are not isotropic. A typical example of such materials is graphite-epoxy composites composed of fibers of graphitic carbon (which have very high yield stress in the plane of the graphite sheets, and low yield stress in the direction perpendicular to this plane) that are bonded to an epoxy polymer, which has relatively low yield stress in tension but high yield stress in compression and shear. The result is a material that has very high stress for its weight. The Boeing 787 uses composites for most of  the  structure;  this  has  the  advantage  of  high ratio  of  yield  stress to  weight,  no  possibility  of corrosion, and relative ease of forming into any desired shape.  (One could also say that a much older material, i.e. wood, is also an anisotropic composite material.)

Of course, in any design one must employ a material with a yield stress greater than the actual stress in  the  material;  the  ratio  of  the yield  stress  of  the  material  to  the  actual  predicted  stress  in  the material is call the factor of  safety.  For example, if a material has a yield stress in tension of 10,000 lbf/in2 and  the  system  is  designed  such  that  the  maximum  tension  is  2,500  lbf/in2 ,  the  factor  of safety for this particular design is (10,000 lbf/in2/2,500 lbf/in2) = 4, at least in tension.