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  03-02ChapGere.0003
  
  # 03-02ChapGere.0003 - f between the ends of the bar Solution...
  
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  Solution 3.4-11 Tapered tube 206 CHAPTER 3 Torsion t thickness (constant) d A , d B average diameters at the ends d B 2 d A (approximate formula) I P d 3 t 4 T T A B L A NGLE OF TWIST Take the origin of coordinates at point O . I P ( x ) [ d ( x ) ] 3 t 4 td A 3 4 L 3 x 3 d ( x ) x 2 L ( d B ) x L d A B L d B = 2 d A d A O x d (x) d x L For element of length dx : For entire bar: f 2 L L d f 4 TL 3 Gtd A 3 2 L L dx x 3 3 TL 2 Gtd A 3 d f Tdx GI P ( x ) Tdx G ¢ td 3 A 4 L 3 x 3 4 TL 3 Gtd A 3 # dx x 3 Problem 3.4-12 A prismatic bar AB of length L and solid circular cross section (diameter d ) is loaded by a distributed torque of constant intensity t per unit distance (see figure). (a) Determine the maximum shear stress max in the bar. (b) Determine the angle of twist
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  Unformatted text preview: f between the ends of the bar. Solution 3.4-12 Bar with distributed torque A B t L t 5 intensity of distributed torque d 5 diameter G 5 shear modulus of elasticity (a) M AXIMUM SHEAR STRESS (b) A NGLE OF TWIST f 5 # L d f 5 32 t p Gd 4 # L x dx 5 16 tL 2 p Gd 4 d f 5 T ( x ) dx GI p 5 32 tx dx p Gd 4 T ( x ) 5 tx I P 5 p d 4 32 T max 5 tL t max 5 16 T max p d 3 5 16 tL p d 3 A B t L dx x A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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