Mechanics_of_Materials_Chap_02-03 - 106 CHAPTER 2 Axially...

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106 CHAPTER 2 Axially Loaded Numbers Problem 2.5-3 A rigid bar of weight W 5 750 lb hangs from three equally spaced wires, two of steel and one of aluminum (see figure). The diameter of the wires is 1 / 8 in. Before they were loaded, all three wires had the same length. What temperature increase D T in all three wires will result in the entire load being carried by the steel wires? (Assume E s 5 30 3 10 6 psi, a s 5 6.5 3 10 2 6 ±°F, and a a 5 12 3 10 2 6 ±°F.) Solution 2.5-3 Bar supported by three wires W = 750 lb SAS S 5 steel A 5 aluminum W 5 750 lb E s 5 30 3 10 6 psi E s A s 5 368,155 lb a s 5 6.5 3 10 2 6 ± 8 F a a 5 12 3 10 2 6 ± 8 F L 5 Initial length of wires d 1 5 increase in length of a steel wire due to temperature increase D T 5 a s ( D T ) L A s 5 p d 2 4 5 0.012272 in. 2 d 5 1 8 in. d 2 5 increase in length of a steel wire due to load W ±2 d 3 5 increase in length of aluminum wire due to temperature increase D T 5 a a ( D T ) L For no load in the aluminum wire: d 1 1 d 2 5 d 3 or Substitute numerical values: N OTE : If the temperature increase is larger than D T , the aluminum wire would be in compression, which is not possible. Therefore, the steel wires continue to carry all of the load. If the temperature increase is less than D T , the aluminum wire will be in tension and carry part of the load. 5 185 8 F Ê ¢ T 5 750 lb (2)(368,155 lb)(5.5 3 10 2 6 / 8 F) ¢ T 5 W 2 E s A s ( a a 2 a s ) Ê a s ( ¢ T ) L 1 WL 2 E s A s 5 a a ( ¢ T ) L 5 WL 2 E s A s W Rigid Bar W 2 W 2 d 3 d 1 d 2
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SECTION 2.5 Thermal Effects 107 Problem 2.5-4 A steel rod of diameter 15 mm is held snugly (but without any initial stresses) between rigid walls by the arrangement shown in the figure. Calculate the temperature drop D T (degrees Celsius) at which the average shear stress in the 12-mm diameter bolt becomes 45 MPa. (For the steel rod, use a 5 12 3 10 2 6 /°C and E 5 200 GPa.) Solution 2.5-4 Steel rod with bolted connection 15 mm 12 mm diameter bolt R 5 rod B 5 bolt P 5 tensile force in steel rod due to temperature drop D T A R 5 cross-sectional area of steel rod From Eq. (2-17) of Example 2-7: P 5 EA R a ( D T ) Bolt is in double shear. V 5 shear force acting over one cross section of the bolt t 5 average shear stress on cross section of the bolt A B 5 cross-sectional area of bolt t 5 V A B 5 EA R a ( ¢ T ) 2 A B V 5 P / 2 5 1 2 EA R a ( ¢ T ) S UBSTITUTE NUMERICAL VALUES : t 5 45 MPa d B 5 12 mm d R 5 15 mm a 5 12 3 10 2 6 / 8 C E 5 200 GPa ¢ T 5 24 8 C Ê ¢ T 5 2(45 MPa)(12 mm) 2 (200 GPa)(12 3 10 2 6 / 8 C)(15 mm) 2 ¢ T 5 2 t d B 2 E a d R 2 A R 5 p d R 2 4 Ê where d R 5 diameter of steel rod A B 5 p d B 2 4 Ê where d B 5 diameter of bolt Solve for ¢ T : ¢ T 5 2 t A B EA R a 15 mm 12 mm diameter bolt B R Problem 2.5-5 A bar AB of length L is held between rigid supports and heated nonuniformly in such a manner that the temperature increase D T at distance x from end A is given by the expression D T 5 D T B x 3 / L 3 , where D T B is the increase in temperature at end B of the bar (see figure).
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Mechanics_of_Materials_Chap_02-03 - 106 CHAPTER 2 Axially...

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