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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 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 T in all three wires will result in the entire load being carried by the steel wires? (Assume E s 30 10 6 psi, s 6.5 10 6 /°F, and a 12 10 6 /°F.) Solution 2.5-3 Bar supported by three wires W = 750 lb S A S S steel A aluminum W 750 lb E s 30 10 6 psi E s A s 368,155 lb s 6.5 10 6 / F a 12 10 6 / F L Initial length of wires 1 increase in length of a steel wire due to temperature increase T s ( T ) L A s d 2 4 0.012272 in. 2 d 1 8 in. 2 increase in length of a steel wire due to load W /2 3 increase in length of aluminum wire due to temperature increase T a ( T ) L For no load in the aluminum wire: 1 2 3 or Substitute numerical values: N OTE : If the temperature increase is larger than 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 T , the aluminum wire will be in tension and carry part of the load. 185 F ¢ T 750 lb (2)(368,155 lb)(5.5 10 6 F) ¢ T W 2 E s A s ( a s ) s ( ¢ T ) L WL 2 E s A s a ( ¢ T ) L WL 2 E s A s W S A S Rigid Bar S A S W 2 W 2 3 1 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 T (degrees Celsius) at which the average shear stress in the 12-mm diameter bolt becomes 45 MPa. (For the steel rod, use 12 10 6 /°C and E 200 GPa.) Solution 2.5-4 Steel rod with bolted connection 15 mm 12 mm diameter bolt R rod B bolt P tensile force in steel rod due to temperature drop T A R cross-sectional area of steel rod From Eq. (2-17) of Example 2-7: P EA R ( T ) Bolt is in double shear. V shear force acting over one cross section of the bolt average shear stress on cross section of the bolt A B cross-sectional area of bolt t V A B EA R ( ¢ T ) 2 A B V P 2 1 2 EA R ( ¢ T ) S UBSTITUTE NUMERICAL VALUES : 45 MPa d B 12 mm d R 15 mm 12 10 6 / C E 200 GPa ¢ T 24 C ¢ T 2(45 MPa)(12 mm) 2 (200 GPa)(12 10 6 C)(15 mm) 2 ¢ T 2 t d B 2 E d R 2 A R d R 2 4 where d R diameter of steel rod A B d B 2 4 where d B diameter of bolt Solve for ¢ T : ¢ T 2 t A B EA R 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 T at distance x from end A is given by the expression T T B x 3 / L 3 , where T B is the increase in temperature at end B of the bar (see figure). Derive a formula for the compressive stress c in the bar. (Assume that the material has modulus of elasticity E and coefficient of thermal expansion .) L A T T B B x 0
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108 CHAPTER 2 Axially Loaded Numbers Solution 2.5-5 Bar with nonuniform temperature change At distance x : R EMOVE THE SUPPORT AT END B OF THE BAR : Consider an element dx at a distance x from end A .
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