Question

Consider a steel (youngs modulus e = 200 gpa) column hinged on both sides. Its height is 1.0 m and cross-section is 10 mm 20 mm. The lowest euler critical buckling load (in n) is

Answer 1

Hey mate ^_^

=======
Answer:
=======

$Subject$: Mechanics of Materials      

$Topic$: Euler’s Theory of Columns

Lowest Euler critical buckling load

$= \frac{\pi^{2}EI }{L^{2} }$

$= \frac{(3.14)^{2} \times 200 \times 10^{9} \times 0.2 \times (0.01)^{3} }{12 \times 1}$

$= \frac{0.98596 \times 40 \times 0.001 \times 0.01 \times 10 ^{9} }{12}$

$= \frac{0.98596 \times 0.04 \times 0.01 \times 10^{9} }{12}$

$= \frac{0.0394384 \times 0.01 \times 10^{9} }{12}$

$= \frac{0.000394384 \times 10^{9} }{12}$

$= 3286.53$

Therefore,

The lowest euler critical buckling load (in n) is $3286.53$

=====
$Note$:
=====

The final answer must be,

$3285\: to \:3295$

#Be Brainly❤️

Answer 2

$\huge \bf{ \red{hey}}$

$\huge{ \mathfrak { \pink{here \: is \: answer}}}$

$\bf{= \frac{\pi^{2}ei }{l^{2} }}$

$\bf{= \frac{(3.14)^{2} \times 200 \times 10^{9} \times 0.2 \times (0.01)^{3} }{12 \times 1}}$

$\bf{= \frac{0.98596 \times 40 \times 0.001 \times 0.01 \times 10 ^{9} }{12}}$

$\bf{= \frac{0.98596 \times 0.04 \times 0.01 \times 10^{9} }{12}}$

$\bf{= \frac{0.0394384 \times 0.01 \times 10^{9} }{12}}$

$\bf{= \frac{0.000394384 \times 10^{9} }{12}}$

$\huge{\red{= 3286.53=3286.53 }}$

$\huge \boxed{ \boxed{ \purple{hope \: it \: helps}}}$

$\huge{ \green{thanks}}$