A masonry pier has a cross-section 3 m by 2 m, and is subjected to a load of 1000 kN, the line of the resultant being 1.80 m from one of the shorter sides, and 0.85 m from one of the longer sides. Find the maximum tensile and compressive stresses produced. (Cambridge)

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Accepted Answer

P represents the line of action of the thrust. The bending moments are

(0.15)(1000 × 10³) = 150 kNm about OX

(0.30)(1000 × 10³) = 300 kNm about OY

Now,

[latex]I_x = \frac{1}{12}(3)(2)^3 = 2 m^4 \\ I_y = \frac{1}{12}(2)(3) = 4.5 m^4[/latex]

The cross-sectional area is

A = (3)(2) = 6 m²

For a point whose coordinates are (x, y) the compressive stress is

[latex]\sigma = -\frac{P}{A}\left(1 + \frac{Ae_x x}{I_y} + \frac{Ae_y y}{I_x} ight)[/latex]

which gives

[latex]\sigma = - \frac{1000 imes 10^3}{6}\left(1 + \frac{x}{2.5} + \frac{9y}{20} ight)[/latex]

The compressive stress is a maximum at B, where x = 1.5 m and y = 1 m. Then

[latex]\sigma_B = \frac{10^6}{6}\left(1 + \frac{3}{5} + \frac{9}{20} ight) = -0.342 MN/m^2[/latex]

The stress at D, where x = - 1.5 m and y = - 1 m, is

[latex]\sigma_D = \frac{10^6}{6}\left(1 - \frac{3}{5} - \frac{9}{20} ight) = +0.008 MN/m^2[/latex]

which is the maximum tensile stress.

Question 12.P.2: A masonry pier has a cross-section 3 m by 2 m, and is subjec......

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