02-04ChapGere - 122 CHAPTER 2 Axially Loaded Members...

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122 CHAPTER 2 Axially Loaded Members Stresses on Inclined Sections Problem 2.6-1 A steel bar of rectangular cross section (1.5 in. 3 2.0 in.) carries a tensile load P (see figure). The allowable stresses in tension and shear are 15,000 psi and 7,000 psi, respectively. Determine the maximum permissible load P max . Solution 2.6-1 Rectangular bar in tension PP 1.5 in. 2.0 in. Problem 2.6-2 A circular steel rod of diameter d is subjected to a tensile force P 5 3.0 kN (see figure). The allowable stresses in tension and shear are 120 MPa and 50 MPa, respectively. What is the minimum permissible diameter d min of the rod? Solution 2.6-2 Steel rod in tension s allow 5 120 MPa t allow 5 50 MPa Maximum shear stress: t max 5 s x 2 5 P 2 A Maximum normal stress: s x 5 P A P 5 3.0 kN Ê A 5 p d 2 4 Because t allow is less than one-half of s allow , the shear stress governs. Solve for d : d min 5 6.18 mm t max 5 P 2 A Ê or Ê 50 MPa 5 3.0 kN (2) ¢ p d 2 4 P P 1.5 in. 2.0 in. A 5 1.5 in. 3 2.0 in. 5 3.0 in. 2 Maximum Normal Stress: s x 5 P A s allow 5 15,000 psi t allow 5 7,000 psi Because t allow is less than one-half of s allow , the shear stress governs. P max 5 2 t allow A 5 2(7,000 psi) (3.0 in. 2 ) 5 42,000 lb Maximum shear stress: t max 5 s x 2 5 P 2 A P = 3.0 kN P d P P d
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Problem 2.6-3 A standard brick (dimensions 8 in. 3 4 in. 3 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 5 2.5 in. 3 4.0 in. 5 10.0 in. 2 Maximum normal stress: s x 5 P A Maximum shear stress: s ult 5 3600 psi t ult 5 1200 psi Because t ult is less than one-half of s ult , the shear stress governs. 5 24,000 lb P max 5 2(10.0 in. 2 )(1200 psi) t max 5 P 2 A Ê or Ê P max 5 2 A t ult t max 5 s x 2 5 P 2 A P 2.5 in. 8 in. 4 in. P 2.5 in. 8 in. Problem 2.6-4 A brass wire of diameter d 5 2.42 mm is stretched tightly between rigid supports so that the tensile force is T 5 92 N (see figure). What is the maximum permissible temperature drop D T if the allowable shear stress in the wire is 60 MPa? (The coefficient of thermal expansion for 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
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124 CHAPTER 2 Axially Loaded Members Problem 2.6-5 A brass wire of diameter d 5 1/16 in. is stretched between rigid supports with an initial tension T of 32 lb (see figure).
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This homework help was uploaded on 03/19/2008 for the course EM 319 taught by Professor Kennethm.liechti during the Spring '08 term at University of Texas.

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02-04ChapGere - 122 CHAPTER 2 Axially Loaded Members...

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