We show the coordinate system for the triangle in Figure 12.10.

The equation to the neutral axis is

\frac{P}{A}+\frac{M_x y}{\bar{I}_{x x}}+\frac{M_y x}{\bar{I}_{y y}}=0     [refer to Eq. (12.3)]          (1)

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

where

M_x=P \in_y \text { and } M_y=P \in_x ; \quad \bar{I}_{x x}=\bar{I}_{y y}=\frac{\sqrt{3}}{96} b^4

Therefore,

1. When load is at A , \in_x=0, \in_y=b / \sqrt{3} then Eq. (1) becomes

\frac{x \in_x}{\bar{K}_y^2}+\frac{y \in_y}{\bar{K}_x^2}=-1

with \bar{K}_x^2=\bar{K}_y^2=b^2 / 24 . Substituting, we get

\frac{b}{\sqrt{3}} \frac{24}{b^2} y=-1

or          y=-\frac{b}{8 \sqrt{3}}              (2)

2. When load is at B, \in_x=-b / 2, \in_y=-b / 2 \sqrt{3} , then Eq. (1) becomes

\left\lgroup -\frac{b}{2} \cdot \frac{24}{b^2} \right\rgroup x+\left\lgroup -\frac{b}{2 \sqrt{3}} \cdot \frac{24}{b^2} \right\rgroup y=-1

or            \frac{x}{(b / 12)}+\frac{y}{(b / 2 \sqrt{3})}=1            (3)

3. When load is acting at C, \in_x=b / 2, \in_y=-b / 2 \sqrt{3} then Eq. (1) becomes

\frac{x}{(b / 12)}+\frac{y}{(-b / 4 \sqrt{3})}=1                 (4)

Considering Eqs. (2)–(4) and also from the geometry of the main equilateral triangle, we can represent the core as shown in Figure 12.11.

Here the core is marked as another equilateral triangle with O (i.e., the centroid of the main equilateral triangle) as its centroid and length of each side as a/4.