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kondaur [170]
3 years ago
8

A rectangular aluminum bar (~1.52.1 cm in cross section) and a circular steel rod (~1 cm in radius) are each subjected to an axi

al force of 20 kN. Assuming that both are 30 cm long in their unloaded configuration, find (a) the stress in each, (b) the extensional strain in each, and (c) the amount of lengthening in each. Let E¼70 GPa for aluminum and 200 GPa for steel.
Physics
1 answer:
garri49 [273]3 years ago
4 0

Answer:

(a) Aluminium= 6.3492*10^{7} N/m^{2}  , steel=6.366*10^{7} N/m^{2}

(b) Aluminium=0.0907*10^{-2}  , steel=0.0318*10^{-2}

(c) Aluminium= 0.2721 cm, Steel= 0.009549cm

Explanation:

(a)

For aluminium bar, stress, \sigma is given by  

\sigma= \frac {F}{A} where A is cross-sectional area of bar and F is force applied

Therefore, \sigma= \frac {20*10^{3}}{1.5*10^{-2}*2.1*10^{-2}}=6.3492*10^{7} N/m^{2}

For steel rod,  

\sigma= \frac {F}{A}= \frac {20*10^{3}}{ \pi R^{2}}=\frac {20*10^{3}}{ \pi 0.01^{2}}=6.366*10^{7} N/m^{2}

(b)

Strain, \epsilon in aluminium is given by

\epsilon = \frac { \sigma}{E}where E is Young’s Modulus

\epsilon= \frac {6.39*10^{7}}{70*10^{9}}=0.0907*10^{-2}

For steel rod

\epsilon = \frac { \sigma}{E}

\epsilon= \frac {6.37*10^{7}}{200*10^{9}}=0.0318*10^{-2}

(c)

Strain, \epsilon is given by  

\epsilon = \frac {Elongation}{original length} hence the change in length is product of original length and \epsilon

For aluminium rod,

Elongation= 30*0.0907*10^{-2}=0.2721 cm

For steel rod

Elongation= 30*0.0318*10^{-2}= 0.009549 cm

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So  

   t  = t_c  + t_s

    t =  0.121 + 0.235

    t =  0.356 \ s

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