03-02ChapGere.0010

# 03-02ChapGere.0010 - 5 2 e max t max 5 G g max 5 2 G e max...

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Problem 3.5-7 The normal strain in the 45° direction on the surface of a circular tube (see figure) is 880 3 10 2 6 when the torque T 5 750 lb-in. The tube is made of copper alloy with G 5 6.2 3 10 6 psi. If the outside diameter d 2 of the tube is 0.8 in., what is the inside diameter d 1 ? Solution 3.5-7 Circular tube with strain gage SECTION 3.5 Pure Shear 213 T 5 360 N . m G 5 78 GPa A LLOWABLE STRESSES Tension: 90 MPa Compression: 70 MPa Shear: 40 MPa Allowable tensile strain: e max 5 220 3 10 2 6 D IAMETER BASED UPON ALLOWABLE STRESS The maximum tensile, compressive, and shear stresses in a bar in pure torsion are numerically equal. Therefore, the lowest allowable stress (shear stress) governs. d 5 0.0358 m 5 35.8 mm d 3 5 45.837 3 10 2 6 m 3 t max 5 16 T p d 3 d 3 5 16 T p t allow 5 16(360 N . m) p (40 MPa) t allow 5 40 MPa D IAMETER BASED UPON ALLOWABLE TENSILE STRAIN T ENSILE STRAIN GOVERNS d min 5 37.7 mm d 5 0.0377 m 5 37.7 mm 5 53.423 3 10 2 6 m 3 d 3 5 16(360 N . m) 2 p (78 GPa)(220 3 10 2 6 ) t max 5 16 T p d 3 Ê d 3 5 16 T p t max 5 16 T 2 p G e max g max
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Unformatted text preview: 5 2 e max ; t max 5 G g max 5 2 G e max T T = 750 lb-in. d 2 = 0.8 in. Strain gage 45 ° d 2 5 0.80 in. T 5 750 lb-in. G 5 6.2 3 10 6 psi Strain gage at 45 8 : e max 5 880 3 10 2 6 M AXIMUM SHEAR STRAIN g max 5 2 e max M AXIMUM SHEAR STRESS t max 5 T ( d 2 / 2) I P Ê I P 5 Td 2 2 t max 5 Td 2 4 G e max t max 5 G g max 5 2 G e max I NSIDE DIAMETER Substitute numerical values: d 1 5 0.60 in. 5 0.4096 in. 4 2 0.2800 in. 4 5 0.12956 in. 4 d 1 4 5 (0.8 in.) 4 2 8(750 lb-in.)(0.80 in.) p (6.2 3 10 6 psi)(880 3 10 2 6 ) Ê d 2 4 2 d 1 4 5 8 Td 2 p G e max Ê d 1 4 5 d 2 4 2 8 Td 2 p G e max I P 5 p 32 ( d 2 4 2 d 1 4 ) 5 Td 2 4 G e max T T 45 ° d 2 d 1 Solution 3.5-6 Solid circular bar of steel A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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