The steel bar A B of the frame is assumed to be pin...

The steel bar $A B$ of the frame is assumed to be pin connected at its ends for $y-y$ axis buckling. If $P=18 \mathrm{kN}$ determine the factor of safety with respect to buckling about the $y-y$ axis due to the applied loading. $E_{\mathrm{st}}=200 \mathrm{GPa}$ $\sigma_{Y}=360 \mathrm{MPa}$

Key Concepts

Material Properties

Material properties, particularly the modulus of elasticity, are essential in buckling analysis because they determine the stiffness of the material. A higher modulus of elasticity indicates a stiffer material that can sustain higher loads before buckling, while the yield stress provides insight into the material's plastic behavior, although in pure buckling analysis, the elastic modulus is primarily used.

Boundary Conditions and Effective Length

The boundary conditions of a column, such as being pin-connected at its ends, play a critical role in determining its buckling behavior. These conditions influence the effective length of the column used in Euler's formula. For example, pin-connected ends typically yield an effective length equal to the actual length, which simplifies the computation of the critical buckling load.

Euler Buckling Theory

Euler Buckling Theory provides a method to calculate the critical load at which a slender column becomes unstable and buckles laterally. It relates the material’s modulus of elasticity, the moment of inertia of the cross-section, and the effective length of the column to determine the load at which buckling occurs. This theory is fundamental in structural stability analysis for columns and slender bars subjected to axial compression.

Factor of Safety

The factor of safety in the context of buckling is defined as the ratio of the critical load (the load at which buckling is predicted to occur) to the applied load. This concept is used to determine the margin of safety in a design, ensuring that the structure will not buckle under expected loads. It is a crucial aspect of structural design to prevent unexpected failure.

Transcript

00:01 Here the load p applied on the frame is given as p is equal to 18 kilo newton.

00:09 The yung's modulus of steel bar is given as 200 giga pascal and the yield stress is given as sigma y is equals to 360 megapascal.

00:24 Now to determine the factor of safety with respect to buckling about y y axis.

00:28 So let us consider forces in member ac and ab.

00:32 Forces in member ac and ab.

00:34 This is so this is force p is equal to 18 kilo newton and this is y direction and this force is f c f ac this angle is theta and this force is f ab ab so and theta is equal to 10 inverse 2 divided by 4 so let us first calculate f ac c so net force in x direction which is be equal to 0 and we can write f ac c sine theta is equal to force p and from here we get f ac c is equal to f ac is equal to p divided by sine theta let us put the value of p p is given as 18 divided by sine theta is given as 10 inverse 10 inverse 2 divided by 10 x divided by 4 and from this we get f ac is equal to 30 kilo newton this is 3 now net force in y direction will be equals to 0 so we will write f ac coss theta is equals to f a b and from this we will get f a b f a b is equal to f ac which is 30 multiplied by cost theta which is equal to 4 by 5 4 divided by 5 and this will equal to 24 kilo newton.

02:36 So the value of force in ab member is 24 kilo newton.

02:41 Local carried by member.

02:43 Now to determine the critical buckling load, we need to find the critical buckling load.

02:51 So pcr will be equals to pi square, ei divided by effective length square.

03:00 So let us calculate it pi square, the value of e...

Please give Ace some feedback