# T change in temperature of the member l the original

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T: Change in temperature of the member. L: The original length of the member. T : The change in length of the member. Example 15: The A-36 steel bar shown is constrained to just fit between two fixed supports when T 1 =60º F . If the temperature is raised to T 2 =120º F determine the average normal thermal stress developed in the bar. For steel =6.6×10 -6 1/ºF , E=29×10 3 Ksi . 0 y F F A -F B =F T - F =0 T = × T×L T =E× × T =29×10 3 ×6.6×10 -6 ×(120-60) =11.484 Ksi in 5 . 0 in 5 . 0 in 20 A F B F A B T F

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1 Example 16: A 2014-T6 aluminum tube having a cross sectional area of 600 mm 2 is used as a sleeve for an A-36 steel bolt having a cross sectional area of 400 mm 2 . When the temperature is T 1 =15º C , the nut hold the assembly in a snug position such that the axial force in the bolt is negligible. If the temperature increases T 2 =80º C , determine the average normal stress in the bolt and sleeve. For aluminum =23×10 -6 1/ºC , E=73.1 GPa, for steel =12×10 -6 1/ºC , E=200 GPa. 0 y F F sl -F b =0 F sl =F b =F F sl T sl F b T b ) ( ) ( ) ( ) ( [ × T×L+ AE FL ] b =[ × T×L- AE FL ] sl 12×10 -6 ×0.15×(80-15)+ 9 6 10 200 10 400 15 . 0 F = 23×10 -6 ×0.15×(80-15)- 9 6 10 1 . 73 10 600 15 . 0 F 0.0052949×10 -6 F=0.00010725 F=20255 N b = b A F = 6 10 400 20255 b =50.637655 MPa sl = sl A F = 6 10 600 20255 sl =33.758436 MPa mm 150 T sl ) ( F sl ) ( T b ) ( F b ) ( Position Initial Position Final b F sl F
2 Example 17: The rigid bar shown is fixed to the top of the three posts made of A-36 steel and 2014-T6 aluminum. The posts each have a length of 250 mm when no load is applied to the bar, and the temperature is T 1 =20ºC . Determine the force supported by each posts if the bar is subjected to a uniform distributed load of 150 KN/m and the temperature is raised to T 2 =80ºC . For steel =12×10 -6 1/ºC , E=200 GPa , for aluminum =23×10 -6 1/ºC , E=73.1 GPa . Steel Aluminum Steel 0 y F 2F st +F al =90000 …………….(1) =( st ) T -( st ) F =( al ) T -( al ) F [ × T×L- AE L F st ] st =[ × T×L- AE L F al ] al 12×10 -6 ×0.25×(80-20)- =23×10 -6 ×0.25×(80-20)- 1.20956×10 -9 F al -0.994718×10 -9 F st =0.000165 ………………..(2) From equations (1) and (2) F st =-16444.7 N F al =122888.8 N mm 300 mm 300 m KN / 150 mm 250 mm 40 mm 60 mm 40 st F st F al F KN 90 6 . 0 150 Position Initial Position Final T st ) ( F st ) ( T al ) ( F al ) ( 9 2 3 10 200 ) 10 40 ( 4 25 . 0 st F 9 2 3 10 1 . 73 ) 10 60 ( 4 25 . 0 al F

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3 Example 18: The rigid bar AD is pinned at A and attached to the bars BC and ED as shown. The entire system is initially stress-free and the weights of all bars are negligible. The temperature of bar BC is lowered 25ºK and that of the bar ED is raised 25ºK . Neglecting any possibility of lateral buckling, find the normal stresses in bars BC and ED . For BC , which is brass, assume E=90 GPa , =20×10 -6 1/ºK and for ED , which is steel, take =12×10 -6 1/ºK , E=200 GPa . The cross-sectional area of BC is 500 mm 2 , of ED is 250 mm 2 .
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• Winter '15
• MAhmoudali

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