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Chapter 12

Q. 12.1

A column carrying compressive load has circular cross-section of diameter = d (Figure 12.7). Determine its core.

12.7

Step-by-Step

Verified Solution

Let the load P act at a point [(d cosθ )/2, (d sinθ )/2]on the circumference of the circle. Therefore, [(d sin θ)/2]

∈_x=\frac{d}{2} \cos \theta \text { and } ∈_y=\frac{d}{2} \sin \theta

From Eq. (12.3), the expression for NA is given by

\sigma_{x x}=\frac{P}{A}+\frac{M_x y}{\bar{I}_{x x}}+\frac{M_y x}{\bar{I}_{y y}}            (12.3)

\frac{P}{A}+\frac{M_x y}{\bar{I}_{x x}}+\frac{M_y x}{\bar{I}_{y y}}= 0           (1)

where

M_x=P  \in_y=\frac{P d}{2} \sin \theta, \quad M_y=P \in_x=\frac{P d}{2} \cos \theta

For the circular cross-section of diameter d, we get

A=\frac{\pi d^2}{4}, \quad \bar{I}_{x x}=\bar{I}_{y y}=\frac{\pi d^4}{64}=\left\lgroup \frac{d^2}{16} \right\rgroup A

Therefore,

K_x^2=K_y^2=\frac{d^2}{16}

where K_x \text { and } K_y are the radii of gyration about x and y axes, respectively.
Thus, Eq. (1) becomes

x\left\lgroup \frac{M_y}{K_y^2} \right\rgroup+y\left\lgroup \frac{M_x}{K_x^2} \right\rgroup =-P

Putting values of M_x \text { and } M_y , we get

x\left[\frac{(P d / 2) \cos \theta}{d^2 / 16}\right]+y\left[\frac{(P d / 2) \sin \theta}{d^2 / 16}\right]=-P

or          x \cos \theta+y \sin \theta=-\frac{d}{8}

so          (-\cos \theta) x+(-\sin \theta) y=\frac{d}{8}                (2)

Comparing the above Eq. (2) with the canonical form of a straight line, we get to know that this line is tangential to a circle with centre at O and radius = d/8 = r/4. This is shown in Figure 12.8.
Evidently as point of application varies with θ along the circumference of the circle, the NA will move around tangentially along the circle of radius d/8. Hence, the required core is a concentric circle of radius d/8 which will always have tensile stress if P is tensile and will always have comprehensive stress if P is compressive.

12.8