311 C H12

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311 C H12

  1. 1. <ul><li>IV.) Torsion on Circular Shafts </li></ul><ul><li>A.) Torque (T) </li></ul>
  2. 3. <ul><li>B.) Torsional Shearing Stress (  v ) </li></ul><ul><li>-Under Pure Torsional Shear, experiments have shown that for circular shafts of constant cross-section: </li></ul><ul><li>1.) Plane section remains plane. </li></ul><ul><li>2.) A straight line radius remains a straight line. </li></ul>
  3. 4. T A B B’ O FIXED END
  4. 5. T Plane Sections Straight Line Radius  L T
  5. 8. <ul><li>From Hooke’s Law, Stress is directly proportional to strain. </li></ul>T  v
  6. 9. <ul><li>Consider a very small square area of a circular cross-section. </li></ul>T  v r c Area=a
  7. 10. <ul><li>The Stress on area “a” is: </li></ul><ul><li> =   V (r) </li></ul><ul><li> c </li></ul><ul><li>The Force on area “a” is: </li></ul><ul><li>F =  a  =   V r(a) </li></ul><ul><li> c </li></ul><ul><li>The moment produced by this force on area “a” is: </li></ul><ul><li>M = F(r)  =   V ra(r) =  V ar 2 </li></ul><ul><li> c c </li></ul>
  8. 11. <ul><ul><ul><li>The total moment produced by the forces on all the small areas gives you the torqe resistance of the cross-section: </li></ul></ul></ul><ul><li>T =  M =   V ar 2 =  V (  ar 2 ) c c </li></ul><ul><li>From Statics, recall that the Polar Moment of Inertia (J) is defined as: </li></ul><ul><li> J =  ar 2 </li></ul><ul><li>Therefore: T =  V J </li></ul><ul><li> c </li></ul>
  9. 12. <ul><li>Therefore the torque resistance of a section of a given diameter at a maximum stress level  V is: </li></ul><ul><li> T =  V J </li></ul><ul><li> c </li></ul><ul><li>Or, the stress produced by a torque T, on a section of a given diameter is: </li></ul><ul><li>  V = Tc </li></ul><ul><li> J </li></ul>
  10. 13. <ul><li>B.) Angle of twist,  </li></ul><ul><li>Shearing strain is defined in Ch.9 </li></ul><ul><li>in terms of the shear deformation,  v as: </li></ul><ul><li> v =  v /L </li></ul> L T
  11. 14. <ul><li>In the sketch shown, the shear deformation is the distance BB’, therefore: </li></ul><ul><li> v =BB’/L </li></ul>T O B B’  c
  12. 15. <ul><li>From geometry, we know that the arc length BB’ = c  therefore: </li></ul><ul><li> </li></ul><ul><li> v = c  </li></ul><ul><li>L </li></ul> c T O B B’
  13. 16. <ul><li>Since Hooke’s Law states: </li></ul><ul><li>G=  v /  v </li></ul><ul><li>we can substitute in the </li></ul><ul><li>above expression for  v : </li></ul><ul><li>G=  v __ =  v L </li></ul><ul><li> c  /L c  </li></ul> c T O B B’
  14. 17. <ul><li>Solving for  : </li></ul><ul><li> =  v L (One formula for  </li></ul><ul><li> G c </li></ul><ul><li> V = Tc </li></ul><ul><li> J </li></ul><ul><li> =  v L = ( T c ) L = TL ( 2 nd formula for  </li></ul><ul><li> G c (J)G c JG </li></ul>
  15. 18. Summary of Torsion Formulas <ul><li> = Angle of twist (radians) </li></ul><ul><li> V = Shear Stress (psi) </li></ul><ul><li>L = Length (inches) </li></ul><ul><li>T = Torque (in-lb) </li></ul><ul><li>G = Modulus of Rigidity (psi) </li></ul><ul><li>c = Outside radius of shaft (inches) </li></ul><ul><li>J = Polar Moment of Inertia (in 4 ) </li></ul><ul><li> </li></ul> =  v L = TL G c JG  V = Tc J
  16. 19. <ul><li>A solid steel circular shaft is 4.5 feet long and has a diameter of 5 inches. If the shaft is fixed at one end a torque of 15,000 ft-lb is applied at the free end, find: </li></ul><ul><li> 1.) the maximum shear stress in the shaft and show where it occurs (on the sketch provided). </li></ul><ul><li>2.) the shear stress at the center of the cross-section (point O). </li></ul><ul><li>3.) the angle of twist at the free end. </li></ul>

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