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  • Hi all,

    I found following two questions in a book; the solutions has confused me. Pleas help with the correct approach:

    1.a steel rod 20 mm dia is passed through a brass tube of 25mm internal dia and 30 mm external dia. the tube is 80 cm long and is closed by thin rigid washers and fastened by nuts, screwed to the rod. the nuts are tightened until the compressive force in the tube is 5kN. cal the stresses in the rod and tube.

    Es = 200 GPa; Eb= 80 GPa

    2. a steel bolt of 16 mm dia passes centrally through a copper tube of internal dia 20 mm and external dia 30 mm. the length of the whole assembly is 500 mm. after tight fitting of the assembly, the nut is overtightened by quarter of a turn. what are the stresses induced in the nut and the bolt if pitch of nut is 2mm.

    Es = 200 GPa ; Ec = 120 GPa


    the key difference in the approach according to the book is that, for the first problem the elongation in rod and compression in tube is taken to be same.however for the second problem,the degree of elongation and compression is not taken equal. is approach correct for the second qtn ?

    Requesting an urgent response!


    Regards!!
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  • Tapan_mech

    Member • Nov 15, 2011

    I'd like to request please help in clearing the doubt!
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  • Tapan_mech

    Member • Nov 15, 2011

    I'd like to request please help in clearing the doubt!
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  • Ramani Aswath

    Member • Nov 15, 2011

    In the first case the cross sectional areas of the two materials are not too different. The total length is also more. It is likely that both materials are in the elastic deformation zone where Hook's law applies. In the second case the two areas are quite different and the length is also less. Probably one of the materials is into the plastic deformation region.
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  • Tapan_mech

    Member • Nov 15, 2011

    Firstly, Thanks a lot for the reply!


    Coming to the question, assuming the fact that in both the problems the deformation is in the elastic zone( else we can't apply Hook's law in second case and solution can't be found), how else could we come to the conclusion that elongation and compression in second case will not be the same ?
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  • Tapan_mech

    Member • Nov 15, 2011

    According to me. the elongation of bolt and compression of tube be equal (being rigid composite assembly)and their magnitude should be equal to the over tightened length traversed by the nut ie quarter of a turn.
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  • Tapan_mech

    Member • Nov 16, 2011

    Request you to please correct me if am wrong..
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  • Ramani Aswath

    Member • Nov 16, 2011

    Tapan_mech
    Request you to please correct me if am wrong..
    No, you are not wrong. While what I wrote may happen, it does not address the basic question of how the elongation and compression can be diferent in the second case.
    Let us look at it again.
    In the second case it is said 'the nut is overtightened by quarter of a turn'. With a 2mm pitch this is 0.5mm. In efeect the length of the Iron rod has been forcefully reduced to 499.5mm. The Copper outer tube will reisist this by pushing back. However the length of the copper tube will get reduced by an amount x mm defined by the compressive stress making its length (500 - x) mm
    The compressive load on the copper tube acts as a tensile load on the 499.5 mm steel rod expanding it by mm, making the total length 499.5 + (0.5 - x) = (500 - x)mm, the same as that of the copper tube. However, x need not be equal to (0.5 - x). This will be decided by the properties of the two materials and the bearing areas.

    In the first case what is specified is the load andnot a fixed strain of the central rod.
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  • Ramani Aswath

    Member • Nov 16, 2011

    Correction. Edit is deactivated. In the above post the line: 'The compressive load on the copper tube acts as a tensile load on the 499.5 mm steel rod expanding it by mm', should read as
    The compressive load on the copper tube acts as a tensile load on the 499.5 mm steel rod expanding it by (0.5 - x )mm
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  • Nav IN

    Member • Jul 30, 2015

    how to solve the first sum
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