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03-02ChapGere.0009

# 03-02ChapGere.0009 - 212 A-PDF Torsion CHAPTER 3 Split DEMO...

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Problem 3.5-5 A steel tube ( G 11.5 10 6 psi) has an outer diameter d 2 2.0 in. and an inner diameter d 1 1.5 in. When twisted by a torque T , the tube develops a maximum normal strain of 170 10 6 . What is the magnitude of the applied torque T ? Solution 3.5-5 Steel tube 212 CHAPTER 3 Torsion G 11.5 10 6 psi d 2 2.0 in. d 1 1.5 in. e max 170 10 6 S HEAR STRAIN ( FROM E Q . 3-33) max 2 e max 340 10 6 S HEAR S TRESS ( FROM TORSION FORMULA ) Also, max G max t max Tr I P Td 2 2 I P 1.07379 in. 4 I P 32 ( d 2 2 d 1 4 ) 32 [ (2.0 in.) 4 (1.5 in.) 4 ] Equate expressions: S OLVE FOR TORQUE 4200 lb-in. 2(11.5 10 6 psi)(1.07379 in. 4 )(340

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03-02ChapGere.0009 - 212 A-PDF Torsion CHAPTER 3 Split DEMO...

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