03-02ChapGere.0003 - f between the ends of the bar....

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Solution 3.4-11 Tapered tube 206 CHAPTER 3 Torsion t 5 thickness (constant) d A , d B 5 average diameters at the ends d B 5 2 d A (approximate formula) I P 5 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 ) 5 p [ d ( x )] 3 t 4 5 p td A 3 4 L 3 x 3 d ( x ) 5 x 2 L ( d B ) 5 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 5 # 2 L L d f 5 4 TL 3 p Gtd A 3 # 2 L L dx x 3 5 3 TL 2 p Gtd A 3 d f 5 Tdx GI P ( x ) 5 Tdx G ¢ p td 3 A 4 L 3 x 3 5 4 TL 3 p 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 t 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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This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Institute of Technology.

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