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Expansion pressure from disimilar metals | |||
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Posted by: dansensel ® 05/26/2005, 10:00:34 Author Profile Mail author Edit |
I have a small problem for all of you materials people: If I wrap a steel bandclamp around an aluminum rod (1" diameter) and then heat this assembly to 1100°F, what type of tension can be expected as a result of the thermal stresses. Only the thermal stresses are needed. Thanks. |
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Re: Expansion pressure from disimilar metals | |||
Re: Expansion pressure from disimilar metals -- dansensel | Post Reply | Top of thread | Forum |
Posted by: zekeman ® 05/27/2005, 15:02:06 Author Profile Mail author Edit |
Coefficient of thermal expansion of aluminum is about 13 microinches/inch per deg F For steel depending on its composition has a range of : 6 to 10 microinches /inch per deg F Therefore, taking the worst case of 6 for steel, over 1130 degree temp rise assuming start temp at 70 deg F, the difference in strain between the steel and Al is (13-6)*1130*10^-6=7.91*10^-3 In order to accommodate this expansion the band is stretched under tension and the aluminum rod is compressed by the same total force on the band. The usual and conservative assumption (a good one) is that the differntial strain is manifested in the band only. Since the modulus of elasticity of steel is 30,000,000, the resultant stress in the steel band is 30*10^6*7.91*10^-3= 237,300 psi which is a problem. At the other end of the range for steel the tensile stress is 30*10^6*(13-10)*1130*10^-3= 101,700 psi which is still a problem at elevated temps. |
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Re: Expansion pressure from disimilar metals | |||
Re: Re: Expansion pressure from disimilar metals -- zekeman | Post Reply | Top of thread | Forum |
Posted by: dansensel ® 05/27/2005, 15:36:22 Author Profile Mail author Edit |
Thank you very much kind sir. |
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