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02-04ChapGere.0002

# 02-04ChapGere.0002 - the wire is 20 3 10 2 6/°C and the...

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Problem 2.6-3 A standard brick (dimensions 8 in. 4 in. 2.5 in.) is compressed lengthwise by a force P , as shown in the figure. If the ultimate shear stress for brick is 1200 psi and the ultimate compressive stress is 3600 psi, what force P max is required to break the brick? Solution 2.6-3 Standard brick in compression SECTION 2.6 Stresses on Inclined Sections 123 A 2.5 in. 4.0 in. 10.0 in. 2 Maximum normal stress: s x P A Maximum shear stress: ult 3600 psi ult 1200 psi Because ult is less than one-half of ult , the shear stress governs. 24,000 lb P max 2(10.0 in. 2 )(1200 psi) t max P 2 A or P max 2 A t ult t max s x 2 P 2 A P 2.5 in. 8 in. 4 in. P 2.5 in. 8 in. 4 in. Problem 2.6-4 A brass wire of diameter d 2.42 mm is stretched tightly between rigid supports so that the tensile force is T 92 N (see figure). What is the maximum permissible temperature drop T if the allowable shear stress in the wire is 60 MPa? (The coefficient of thermal expansion for
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Unformatted text preview: the wire is 20 3 10 2 6 /°C and the modulus of elasticity is 100 GPa.) Solution 2.6-4 Brass wire in tension T d T T d T d 5 2.42 mm a 5 20 3 10 2 6 / 8 C E 5 100 GPa t allow 5 60 MPa Initial tensile force: T 5 92 N Stress due to initial tension: Stress due to temperature drop: s x 5 E a ( D T ) (see Eq. 2-18 of Section 2.5) Total stress: s x 5 T A 1 E a ( ¢ T ) s x 5 T A A 5 p d 2 4 5 4.60 mm 2 M AXIMUM SHEAR STRESS Solve for temperature drop D T : S UBSTITUTE NUMERICAL VALUES : 5 120 MPa 2 20 MPa 2 MPa / 8 C 5 50 8 C ¢ T 5 2(60 MPa) 2 (92 N) / (4.60 mm 2 ) (100 GPa)(20 3 10 2 6 / 8 C) ¢ T 5 2 t max 2 T / A E a Ê t max 5 t allow t max 5 s x 2 5 1 2 B T A 1 E a ( ¢ T ) R Probs. 2.6-4 and 2.6-5 A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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