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Chapter 12

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.

Step-by-Step

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.