A 250mm long brass bar (CTE = 20.5X10^-6/ 0C) is setup as shown below. Initially

Question

A 250mm long brass bar (CTE = 20.5X10^-6/ 0C) is setup as shown below. Initially, when the temperature is 15C, there is a total clerance of 0.25mm between the right end (left end is secured to the left side support) of the bar and the right side support. At what temperature, the end of the bar will be in contact with the frame? When temperature of the bar rises to 90C, is there any stress developed in the bar? If yes, what is its magnitude? These are my answers, down below. My teacher said the first part was right but the second part was wrong. Can someone please explain the correct answers to me and what i am doing wrong. Thanks 250x20.5x10^-6=5125 250,000/5125= 48.78(Temperature change) 48.78+15=63.78 (Temperature at which the bar will be contact with the frame) 90-15=75 250x20.5x10^-6x75= 3.84mm (stress developed after temperature has risen to 90 degrees) 3.84-0.25= 3.59mm (Magnitude) 100x10^9x3.59/250= 1436000000 pa or 1.436 GPa

Explanation / Answer

extension in bar = l??T to be in contact, extension = 25 mm 25=250 x 20.5X10^-6 x ?T ?T=48.78 ? temperature = 150+48.78=198.78 at 900 degrees, ?T=900-150=750 extension= 250 x 20.5X10^-6 x 750 =3.84mm but only 0.25 mm extension is permissibleso the rod is in stress equal in magnitude that causes 3.84-0.25=3.59mm extension stress=younng's modulus X strain=100X10^9 X 3.59/250 =1436000000 pa =1.436 Giga Pascal... The temperatures would be 150 and 900, instead of 15 and 90.

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