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311ch9

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  • 1. <ul><li>I) Stresses and Strains </li></ul><ul><li>In statics we assumed rigid bodies </li></ul><ul><li>In strength of materials, we acknowledge that bodies are deformable , not rigid. </li></ul><ul><li>We will study the stresses applied forces produce in a body and the accompanying strains . </li></ul>
  • 2. <ul><li>A.) Axial Tensile and Compressive </li></ul><ul><li>Stresses </li></ul><ul><li>Consider a 2” x 4” piece of wood with </li></ul><ul><li>force P applied at each end. </li></ul>800 lb 800 lb 2” 4” A B
  • 3. <ul><li>Anywhere you cut this bar across its section, in order to satisfy equilibrium , the 800 lb force must act on that section. </li></ul><ul><li> F x = 0 = - 800 lb + P A = 0 </li></ul><ul><li>P A = 800 lb </li></ul>P A 800 lb A B
  • 4. <ul><li>We assume that the force is distributed evenly throughout the section so that an equal portion of the 800 lb force acts on each square inch of the cross-section </li></ul>800 lb 2” 4” 1” 1”
  • 5. <ul><li>Since we have 8 square makes, the amount of force on each square inch is: </li></ul><ul><li>800 lb = 100 lb = 100 psi </li></ul><ul><li>8in 2 in 2 </li></ul>
  • 6. <ul><li>Which is the definition of stress : </li></ul><ul><li> = P </li></ul><ul><li>A </li></ul><ul><li>  = stress = unit stress </li></ul><ul><li>= average stress </li></ul><ul><li>= engineering stress </li></ul><ul><li> P = applied force </li></ul><ul><li> A= constant cross-sectional area over which the stress develops </li></ul>
  • 7. <ul><li>  t = Tensile Stress (produced by </li></ul><ul><li>Tensile Forces) </li></ul><ul><li>  c = Compressive Stress (produced by </li></ul><ul><li>Compressive Forces) </li></ul>
  • 8. <ul><li>Normal Stress = stress acting on a plane perpendicular to (or normal to) the line of action of the applied force (as in our example). </li></ul><ul><li>we will discuss stresses on inclined planes in week 4. </li></ul>
  • 9. <ul><li>B. TENSILE AND COMPRESSIVE </li></ul><ul><li>STRESSES - Application </li></ul><ul><li>1.) Analyze the capacity of existing </li></ul><ul><li>member </li></ul><ul><li>P (all) =   (all) A </li></ul><ul><li>A= cross-sectional area, perpendicular to the direction of the force </li></ul>
  • 10. <ul><li>P (all) = axial load capacity (max allowable axial load) </li></ul><ul><li>= amount of load the member can </li></ul><ul><li> safely carry </li></ul><ul><li> (all) = allowable axial stress=amount of stress which is judged acceptable for the given material </li></ul>
  • 11. <ul><li>Example: 2 x 4 wood with tensile force applied. Find the axial load capacity of the 2 x 4. </li></ul><ul><li>A = 3.5” x 1.5” = 5.25in 2 ,  (all) = 400 psi </li></ul><ul><li> (all) = 400 psi </li></ul><ul><li>P t(all) =  (all) A=(400 psi)(5.25in 2 ) = 2,100 lb </li></ul>
  • 12. <ul><li>2. Design a member to support a known </li></ul><ul><li>load. </li></ul><ul><li>A (req) = P_ </li></ul><ul><li>  (all) </li></ul><ul><li>A (req) = required cross-sectional area of the axially loaded member being designed. </li></ul><ul><li>P = applied axial load </li></ul><ul><li> (all) = allowable axial stress </li></ul>
  • 13. <ul><li>Example: Find what diameter steel rod is required to support a 2100# tension force. </li></ul><ul><li>P = 2100 lb </li></ul><ul><li> (all) = 24,000 psi (= lb/in 2 ) </li></ul><ul><li>A (req) = P = 2,100 lb = .0875 in 2 </li></ul><ul><li> (all) 24,000 lb/in 2 </li></ul>diameter = d 2100 lb 2100 lb
  • 14. <ul><li>A (req) = .0875 in 2 =  (r) 2 =  (d req ) 2 </li></ul><ul><li>4 </li></ul><ul><li>(d req ) 2 = .0875 in 2 (4) = 0.1114 in 2 </li></ul><ul><li>  </li></ul><ul><li> d req =(0.1114 in 2 ) 1/2 = .333” = 3/8” </li></ul><ul><li> </li></ul><ul><li>Therefore a 3/8” diameter steel rod supports as much tensile force as 2” x 4”. </li></ul>
  • 15. <ul><li>Note : App. A,B,C*,D* give cross-sectional areas for different steel shapes. E gives cross-sectional areas for lumber. </li></ul>
  • 16. <ul><li>3.) Cautions concerning design and </li></ul><ul><li>analysis: </li></ul><ul><li>a.) Some materials have a higher allowable compressive stress than tensile stress (  t(all) <  c(all) ) </li></ul><ul><li>Examples: Wood, concrete, cast iron </li></ul><ul><li>(refer to App. F, G, 722, 723) </li></ul>
  • 17. <ul><li>b.) The length of a member effects how much compressive force it can sustain. Will discuss in depth in Week 14 (columns). </li></ul><ul><li>C all < T all </li></ul><ul><li>For now, we will assume compressive members are short enough to ignore effect of length. </li></ul>
  • 18. <ul><li>c). Holes for bolts etc. must be subtracted from the gross cross-sectional area to obtain the net cross-sectional area (A net ). </li></ul><ul><li>d.) Design Manuals have been published by Steel (AISC), Wood (AITC, NDS), Concrete (ACI), and Aluminum associations. </li></ul>
  • 19. <ul><li>e.) Assuming even force distribution over the cross-sectional area is usually valid, but not always . </li></ul>
  • 20. <ul><li>C. STRESS ON NET AREA </li></ul><ul><li>Ex. 2 x 4 with bolt holes for 2-1/2” bolts </li></ul><ul><li>Determine allowable tensile load </li></ul>9/16” dia. holes 3 1/2” 1 1/2” T T
  • 21. <ul><li>T all =  t,all A net </li></ul><ul><li> t(all) = 400 psi </li></ul><ul><li>A net = A - A holes </li></ul><ul><li>= (3.5” x 1.5”) - (2)(9/16”)(1.5”) </li></ul><ul><li>= 3.56 in 2 </li></ul><ul><li>T all = (400 psi)(3.56 in 2 ) = 1425 lb </li></ul>
  • 22. <ul><li>Note : 3.5” and 1.5” are “dressed dimensions” (see APP.E) </li></ul><ul><li>2 x 4 is actually 1.5” x 3.5” </li></ul><ul><li>2 x 8 is actually 1.5” x 7.25” </li></ul>
  • 23. <ul><li>D.) Bearing Stress (  p ) </li></ul><ul><li>-A contact pressure between separate bodies. </li></ul><ul><li>-A type of compressive stress </li></ul>
  • 24. <ul><li>Example: Wood rafter in a building </li></ul>2 X 4 2 X 8 2,000 lb
  • 25.  p = P A bear P = 2000 lb A bear = 1 1/2” x 71/4” = 10.88 in 2  p = 2,000 lb = 184 psi 10.88 in 2
  • 26. <ul><li>E.) Shear Stress </li></ul><ul><li> Normal stress (tensile and compressive) acts in a direction perpendicular to the surface on which it acts. It is produced by a force whose line of action is perpendicular to the surface on which the stress is produced. </li></ul>
  • 27. <ul><li>Shear stress acts in a direction parallel to the surface on which it acts. It is produced by a force whose line of action is parallel to the surface on which the stress is produced. </li></ul>
  • 28. T T Tension -pulling apart V V Shear -sliding past
  • 29. <ul><li>Example: Nail through boards </li></ul>Shear Plane V/5 V/5 V V Shear Plane
  • 30. <ul><li> v = V </li></ul>A  v = shear stress (psi) V = shear force (lb) A= cross-sectional area parallel to the force.
  • 31. <ul><li>Example: Find the shear stress in the 1” diameter bolt shown. </li></ul>1” diameter bolt 4,000 lb Steel Plates 4,000 lb Shear Plane
  • 32. <ul><li> v = V </li></ul><ul><li>A </li></ul><ul><li>V = 4,000 lb </li></ul><ul><li>A =  d 2 =  (1”) 2 = .785 in 2 </li></ul><ul><li>4 4 </li></ul><ul><li>  v = 4,000 lb = 5,096 psi </li></ul><ul><li>.785 in 2 </li></ul>
  • 33. <ul><li>F.) TENSILE AND COMPRESSIVE </li></ul><ul><li> STRAINS AND DEFORMATIONS </li></ul><ul><li>In picking a member size, we not only </li></ul><ul><li>need to know it can handle the load </li></ul><ul><li>without failing due to too much stress </li></ul><ul><li>(i.e. fracture or “break”), we must also </li></ul><ul><li>be sure it will not deflect (or deform) </li></ul><ul><li>excessively. </li></ul>
  • 34. <ul><li>Example: Dock with wooden ladder for </li></ul><ul><li> a footbridge. </li></ul><ul><li>This is an example of deformation or </li></ul><ul><li>deflection due to bending stress which </li></ul><ul><li>we will cover later. </li></ul>
  • 35. <ul><li>Similarly, when a steel rod is in Tension, </li></ul><ul><li>it will deform, but it is not as noticeable. </li></ul><ul><li> = deformation = the amount a body is </li></ul><ul><li>lengthened by a Tensile force and </li></ul><ul><li>shortened by a compressive force. </li></ul>L  T T
  • 36. <ul><li>To permit comparison with acceptable </li></ul><ul><li>values, the deformation is usually </li></ul><ul><li>converted to a unit basis, which is the </li></ul><ul><li>strain . </li></ul><ul><li> =   </li></ul><ul><li> L </li></ul><ul><li> = strain (= unit strain) </li></ul><ul><li> = deformation that occurs over length L </li></ul><ul><li>L = original length of member </li></ul>
  • 37. <ul><li>Example: </li></ul>
  • 38. <ul><li>Example: 3/8” cable, 100’ long stretches </li></ul><ul><li> 1” before freeing a Jeep which </li></ul><ul><li> is stuck in the mud. </li></ul><ul><li> Find - the strain in the cable. </li></ul>100’ Jeep Bronco
  • 39. <ul><li> =   </li></ul><ul><li> L </li></ul><ul><li> = 1” </li></ul><ul><li>L = 100’ (12”/1) = 1200” </li></ul><ul><li> = 1” = 0.0008333 in/in </li></ul><ul><li>1200” </li></ul><ul><li>We’ll come back to see if this is will break the cable. </li></ul>
  • 40. <ul><li>G.) Shear Strain – SKIP ! </li></ul><ul><li>The effect of shear on a body is to cause the body to rotate thru an angle  . </li></ul>P P L V V  v 
  • 41. <ul><li>SKIP ! </li></ul><ul><li>tan  =  v _ _ </li></ul><ul><li>L </li></ul><ul><li> = shear strain (radians) </li></ul><ul><li> v = shear deformation (in) </li></ul><ul><li>L = Distance over which shear deformation </li></ul><ul><li>occurs (in) </li></ul><ul><li> </li></ul><ul><li> </li></ul>
  • 42. <ul><li>SKIP ! </li></ul><ul><li>Since  is always a very small angle, and </li></ul><ul><li>for small angles: </li></ul><ul><li>tan  =  </li></ul><ul><li>Then:  =  v /L </li></ul>
  • 43. <ul><li>Review of Stress and Strain </li></ul><ul><li>Axial Stress and Strain </li></ul><ul><li> = P </li></ul><ul><li> A </li></ul><ul><li> =  L  </li></ul><ul><li>Shear Stress and Strain </li></ul><ul><li> = V_  </li></ul><ul><li> A </li></ul><ul><li> =  v /L - SKIP ! </li></ul>
  • 44. <ul><li>H.) The Relationship Between Stress </li></ul><ul><li> and Strain </li></ul><ul><li>As you apply load to a material, the strain increases constantly (or proportionately) with stress. </li></ul>
  • 45. <ul><li>Example: In a tension test you apply a gradually increasing load to a sample. You can determine the amount of strain (  that occurs in a sample at any given stress level (  . </li></ul><ul><li> (ksi)  (in/in) </li></ul><ul><li> 0 0.000 </li></ul><ul><li> 3 0.001 </li></ul><ul><li> 6 0.002 </li></ul><ul><li> 9 0.003 </li></ul><ul><li> 12 0.004 </li></ul>
  • 46. Stress ,  (ksi) Strain ,  in/in x 0.001)           
  • 47. <ul><li>Since the stress is proportional to the strain, ratio of stress to strain is constant . </li></ul><ul><li> /  </li></ul><ul><li> (ksi)  (in/in)  (ksi) </li></ul><ul><li> 0 0 0 </li></ul><ul><li> 3 0.001 3000 </li></ul><ul><li> 6 0.002 3000 </li></ul><ul><li> 9 0.003 3000 </li></ul><ul><li> 12 0.004 3000 </li></ul>
  • 48. <ul><li>This constant ratio of stress to strain is called the Modulus of Elasticity (E). </li></ul><ul><li>E =  /  </li></ul><ul><li>The Modulus of Elasticity is always the same for a given material (see p. 617). We call it a material constant . </li></ul>
  • 49. <ul><li>Knowing E for a given material and : </li></ul><ul><li>E =  /  </li></ul><ul><li>1.) We can find how much stress is in the material if we know the strain: </li></ul><ul><li>  = E  </li></ul><ul><li>2.) We can find how much strain is in the material if we know the stress: </li></ul><ul><li>  =  E </li></ul>
  • 50. <ul><li>CAUTION ! </li></ul><ul><li>If the tension test continues, the stress will reach a level called the Proportional Limit (  PL ). If the stress is increased above  PL , the strain will increase at a higher rate. </li></ul>
  • 51.  Stress  ), ksi Strain (  ), in/in  PL
  • 52. <ul><li>Strain is directly proportional to stress as long as the stress does not exceed the proportional limit (  PL ) of the material. That is: </li></ul><ul><li>  = E = constant, ONLY if   PL </li></ul><ul><li> </li></ul><ul><li>Anytime you use E to find stress or strain, you must check to make sure   PL . </li></ul>
  • 53. <ul><li>Ex. Given: Previous Truck cable strain </li></ul><ul><li> Find: Stress in the steel cable </li></ul><ul><li> = 1” </li></ul><ul><li>L = 1200” </li></ul><ul><li> = 1” = 0.0008333 in/in </li></ul><ul><li>1200” </li></ul><ul><li>  E ( as long as  PL ) </li></ul><ul><li> </li></ul><ul><li>E= 30,000,000 psi (for steel) </li></ul>
  • 54. <ul><li> E  = 30,000,000 psi (.0008333 in/in) </li></ul><ul><li>= 24,990 psi (pretty high) </li></ul><ul><li>CHECK: is  <  PL ? </li></ul><ul><li> = 24,990 psi <  PL = 34,000 psi (OK) </li></ul>
  • 55. <ul><li>For Shear Stress and Strains: </li></ul><ul><li>G =  v = Modulus of Rigidity </li></ul><ul><li>  </li></ul>
  • 56. <ul><li>Substituting  = P and  =  </li></ul><ul><li> A L </li></ul><ul><li>into E =   Assuming  <  pL ) </li></ul><ul><ul><li> </li></ul></ul><ul><ul><li>E = P/A </li></ul></ul><ul><ul><li>  /L </li></ul></ul><ul><ul><li>Note: E, G are constant </li></ul></ul>
  • 57. <ul><li>Solving for  </li></ul><ul><li>  = P </li></ul><ul><li>L AE </li></ul><ul><li>  = PL IF ,  <  PL </li></ul><ul><li> AE </li></ul>
  • 58. <ul><li>Example 2: </li></ul><ul><li>How much will a 1” diameter by 150” long steel bar stretch if a 1000 lb axial tension load is applied to it ? </li></ul><ul><li> = P = 1000 lb = 1,273 psi <  Y  36,000 psi </li></ul><ul><li>A  (0.5”) 2 </li></ul><ul><li> =  = __1,273 psi___ = 0.00004244 in/in </li></ul><ul><li> E 30,000,000 psi </li></ul><ul><li> L  0.00004244 in/in)(150”) = 0.006366 ” </li></ul>

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