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A rectangular steel bar 37.5 mm wide and 50 mm thick is pinned at each end and subjected to axial compression. The bar has a length of 1.75 m. The modulus of elasticity is 200 Gpa. What is the critical buckling load

User Wood
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1 Answer

3 votes

Answer:

The critical buckling load is
\mathbf{P_o = 141.61 \ kN}

Step-by-step explanation:

Given that:

the width of the rectangular steel = 37.5 mm = 0.0375 m

the thickness = 50 mm = 0.05 m

the length = 1.75 m

modulus of elasticity = 200 Gpa = 200 10⁹ × Mpa

We are to calculate the critical buckling load
P_o

Using the formula:


P_o = (\pi ^2 E I)/(L^2)

where;


I = (0.0375^3*0.05)/(12)


I = 2.197 * 10^(-7)


P_o = (\pi ^2 *200*10^9 * 2.197*10^(-7))/(1.75^2)


P_o = 141606.66 \ N


\mathbf{P_o = 141.61 \ kN}

The critical buckling load is
\mathbf{P_o = 141.61 \ kN}

User MHG
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