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Algebraic Substitution

Example 1




Let




                         answer




Example 2




Let
                                  Example 3
Let




                                                                                  x=asin                        x=atan            x=asec

                                                                                  dx=acos                       dx=asec2          dx=asec    tan

                                                                                        a2−x2=acos                  x2+a2=asec         x2−a2=atan

                                                                                Example 1




                                                                                First see whether the quadratic fits the pattern by completing the square.

                                                                                x2 + 12x + 45 = (x + 6)2 + 9.




                                                                                It does. Set x = 3tan w − 6. then dx = 3sec2 w and w = arctan(          ).




                                                                                Example 2




                                                                                First see whether the quadratic fits the pattern by completing the square.

                                                                                9 + 8x − x2 = 25 − (x − 4)2.



                                                                                It does. Substitute x = 5sin w + 4 , then dx = 5cos w and w = arcsin(        ).




                                                            answer


                                                                                Example 1 Evaluate the following integral.

Trigonometric substitutions are often useful for integrals containing factors
of the form (a2−x2)n (x2+a2)n or(x2−a2)n The exact substitution used
depends on the form of the integral:


  (a2−x2)n                 (x2+a2)n              (x2−a2)n
Problem 310


Show that the hollow circular shaft whose inner diameter is half the outer
diameter has a torsional strength equal to 15/16 of that of a solid shaft of the
same outside diameter.


Solution 310


Hollow circular shaft:




Solid circular shaft:




                                                  ok!


Problem 311


An aluminum shaft with a constant diameter of 50 mm is loaded by torques
applied to gears attached to it as shown in Fig. P-311. Using G = 28 GPa,
determine the relative angle of twist of gear D relative to gear A.
Problem 311




                                                                                                                  answer


                                                                                 Problem 314


                                                                                 The steel shaft shown in Fig. P-314 rotates at 4 Hz with 35 kW taken off at A,
Rotation of D relative to A:
                                                                                 20 kW removed at B, and 55 kW applied at C. Using G = 83 GPa, find the
                                                                                 maximum shearing stress and the angle of rotation of gear A relative to gear
                                                                                 C.




                       answer


Problem 312
                                                                                 Solution 314
A flexible shaft consists of a 0.20-in-diameter steel wire encased in a
stationary tube that fits closely enough to impose a frictional torque of 0.50
lb·in/in. Determine the maximum length of the shaft if the shearing stress is
not to exceed 20 ksi. What will be the angular deformation of one end
relative to the other end? G = 12 × 106 psi.


Solution 312




                                                                                 Relative to C:




If θ = dθ, T = 0.5L and L = dL
For AB


∴                                          answer



                                                                                  For BC




                                                                                  For CD




                       answer                                                     Use d = 69.6 mm answer


Problem 315


A 5-m steel shaft rotating at 2 Hz has 70 kW applied at a gear that is 2 m from   Part (b)
the left end where 20 kW are removed. At the right end, 30 kW are removed
and another 20 kW leaves the shaft at 1.5 m from the right end. (a) Find the
uniform shaft diameter so that the shearing stress will not exceed 60 MPa.
(b) If a uniform shaft diameter of 100 mm is specified, determine the angle
by which one end of the shaft lags behind the other end. Use G = 83 GPa.


Solution 315




                                                                                                      answer




Part (a)

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Torsubsion

  • 1. Algebraic Substitution Example 1 Let answer Example 2 Let Example 3
  • 2. Let x=asin x=atan x=asec dx=acos dx=asec2 dx=asec tan a2−x2=acos x2+a2=asec x2−a2=atan Example 1 First see whether the quadratic fits the pattern by completing the square. x2 + 12x + 45 = (x + 6)2 + 9. It does. Set x = 3tan w − 6. then dx = 3sec2 w and w = arctan( ). Example 2 First see whether the quadratic fits the pattern by completing the square. 9 + 8x − x2 = 25 − (x − 4)2. It does. Substitute x = 5sin w + 4 , then dx = 5cos w and w = arcsin( ). answer Example 1 Evaluate the following integral. Trigonometric substitutions are often useful for integrals containing factors of the form (a2−x2)n (x2+a2)n or(x2−a2)n The exact substitution used depends on the form of the integral: (a2−x2)n (x2+a2)n (x2−a2)n
  • 3. Problem 310 Show that the hollow circular shaft whose inner diameter is half the outer diameter has a torsional strength equal to 15/16 of that of a solid shaft of the same outside diameter. Solution 310 Hollow circular shaft: Solid circular shaft: ok! Problem 311 An aluminum shaft with a constant diameter of 50 mm is loaded by torques applied to gears attached to it as shown in Fig. P-311. Using G = 28 GPa, determine the relative angle of twist of gear D relative to gear A.
  • 4. Problem 311 answer Problem 314 The steel shaft shown in Fig. P-314 rotates at 4 Hz with 35 kW taken off at A, Rotation of D relative to A: 20 kW removed at B, and 55 kW applied at C. Using G = 83 GPa, find the maximum shearing stress and the angle of rotation of gear A relative to gear C. answer Problem 312 Solution 314 A flexible shaft consists of a 0.20-in-diameter steel wire encased in a stationary tube that fits closely enough to impose a frictional torque of 0.50 lb·in/in. Determine the maximum length of the shaft if the shearing stress is not to exceed 20 ksi. What will be the angular deformation of one end relative to the other end? G = 12 × 106 psi. Solution 312 Relative to C: If θ = dθ, T = 0.5L and L = dL
  • 5. For AB ∴ answer For BC For CD answer Use d = 69.6 mm answer Problem 315 A 5-m steel shaft rotating at 2 Hz has 70 kW applied at a gear that is 2 m from Part (b) the left end where 20 kW are removed. At the right end, 30 kW are removed and another 20 kW leaves the shaft at 1.5 m from the right end. (a) Find the uniform shaft diameter so that the shearing stress will not exceed 60 MPa. (b) If a uniform shaft diameter of 100 mm is specified, determine the angle by which one end of the shaft lags behind the other end. Use G = 83 GPa. Solution 315 answer Part (a)