A tie bar $(20 m m \times 10 m m)$ carries a tensile load of $1 k N$ as shown in the figure below. Under this load, the maximum intensity of stress over the mean value will increase by

20 %

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40 %

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60 %

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80 %

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Solution

The correct option is C $60 %$
Mean value of intensity of stress
$= \frac{1000}{20 \times 10} = 5 N / m m^{2}$
Maximum value of intensity of stress
$= 5 [1 + \frac{6 e}{d}] = 5 [1 + \frac{6 \times 1}{10}]$
$= 8 N / m m^{2}$
% increase in stress
$= (\frac{8 - 5}{5}) \times 100 = 60 %$

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A tie bar (20mm×10mm) carries a tensile load of 1 kN as shown in the figure below. Under this load, the maximum intensity of stress over the mean value will increase by

The correct option is C 60%Mean value of intensity of stress =100020×10=5 N/mm2 Maximum value of intensity of stress =5[1+6ed]=5[1+6×110] =8 N/mm2 % increase in stress =(8−55)×100=60%

A tie bar $(20 m m \times 10 m m)$ carries a tensile load of $1 k N$ as shown in the figure below. Under this load, the maximum intensity of stress over the mean value will increase by

The correct option is C $60 %$
Mean value of intensity of stress
$= \frac{1000}{20 \times 10} = 5 N / m m^{2}$
Maximum value of intensity of stress
$= 5 [1 + \frac{6 e}{d}] = 5 [1 + \frac{6 \times 1}{10}]$
$= 8 N / m m^{2}$
% increase in stress
$= (\frac{8 - 5}{5}) \times 100 = 60 %$