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## Q. 12.4

A square pillar is subjected to a compressive force Q = 3500 kN and a bending moment M = 85 kN m. Calculate the dimension of the square cross-section, if the allowable stresses in tension and compression are 18 MPa and 6 MPa, respectively. Neglect the weight of the pillar.

## Verified Solution

Let each side of the square cross-section be b mm. Then
Compressive stress due to Q is found as

$\sigma_1=\frac{3500\left(10^3\right)}{b^2}=\frac{3 \cdot 5\left(10^6\right)}{b^2} MPa$

Normal stress due to M (maximum) is

$\sigma_2=\pm \frac{M}{z}=\pm \frac{6 M}{b^3}=\pm \frac{(6)(85)\left(10^6\right)}{b^3}=\pm \frac{510\left(10^6\right)}{b^3} MPa$

Maximum compressive stress is

$\left(\sigma_{ C }\right)_{\max }=\frac{3.5 \times 10^6}{b^2}+\frac{510 \times 10^6}{b^3} MPa$               (1)

and maximum tensile stress is

$\left(\sigma_{ t }\right)_{\max }=\frac{510\left(10^6\right)}{b^3}-\frac{(3.5)\left(10^6\right)}{b^2} MPa$           (2)

Therefore, from Eq. (1)

$\frac{3.5 \times 10^6}{b^2}+\frac{510 \times 10^6}{b^3}=18$

$\Rightarrow 18 b^3-3.5 \times 10^6 b-510 \times 10^6=0$

$\Rightarrow b^3-194.44\left(10^3\right) b-28.33\left(10^6\right)=0$

Solving, we get b = 500.9874 mm ⇒ b ≥ 500.9874 mm. Similarly, from Eq. (2), we get

$\frac{510\left(10^6\right)}{b^3}-\frac{(3.5)\left(10^6\right)}{b^2}=6$

$\Rightarrow \quad 510\left(10^6\right)-3.5\left(10^6\right) b=6 b^3$

$\Rightarrow \quad 6 b^3+3.5\left(10^6\right) b-510\left(10^6\right)=0$

$\Rightarrow b^3+583.33\left(10^3\right) b-85\left(10^6\right)=0$

Solving, we get b = 140.918 mm ⇒ b ≥ 140.918 mm.
Thus, the required dimension is maximum of the two values, that is, b = 500.99 mm ≈ 501 mm.