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Railways NTPC (Technical Ability) Engineering Mechanics and Strength of Materials Question Bank Engineering Mechanics

  • question_answer
    A solid shaft of diameter 100 mm, length 1000 mm is subjected to a twisting moment ?T? The maximum shear stress developed in the shaft is 60 N/mm. A hold of 50 mm diameter is now drilled throughout the length of shaft. To develop a maximum shear stress of 60 N/ mm in the hollow shaft, the torque ?T? must he reduced by:                             

    A) T/4                              

    B) T/8            

    C) T/12                             

    D) T/16  

    Correct Answer: D

    Solution :

    \[d\text{ }=\text{ }100\text{ }mm,\text{ }l\text{ }=\text{ }1000\text{ }mm\] \[{{\tau }_{\max }}=\frac{16T}{\pi {{d}^{3}}}=60\,\,\text{N/m}{{\text{m}}^{\text{2}}}\] \[T={{T}_{s}}=\frac{\pi {{d}^{3}}}{16}\times 60\] For the hollow shaft, \[{{\tau }_{\max }}=\frac{16T}{\pi {{d}^{3}}(1-{{k}^{4}})}=\frac{16T}{\pi {{d}^{3}}\left[ 1-{{\left( \frac{50}{100} \right)}^{4}} \right]}\] \[=\frac{16}{15}\times \frac{16T}{\pi {{d}^{3}}}\] \[T=\frac{16}{15}\times \frac{\pi {{d}^{3}}}{16\times 60}=\frac{15}{16}{{T}_{s}}=\frac{15}{16}T\] Reduction in torque \[=T\left( 1-\frac{15}{16} \right)=\frac{T}{16}\]


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