A beam with a span of 3.6 m is simply supported at its ends. The cross section consisting of two identical hollow rectangles of flange and web thickness 20 mm is shown in Fig. Q5a.... A beam with a span of 3.6 m is simply supported at its ends. The cross section consisting of two identical hollow rectangles of flange and web thickness 20 mm is shown in Fig. Q5a. It is subjected to a uniformly distributed load of 16.5 kN/m over its entire span. (a) Show that the second moment of area about the neutral axis is 100.05 x 10-6 m^4. (b) Calculate the maximum bending moment and hence the maximum bending stresses induced in the beam. (c) Sketch the stress distribution diagram, indicating where the maximum stresses occur. (d) Calculate the Section Modulus of the beam. The loading of the beam is changed as shown in Fig. Q5b. Assuming the beam to be of negligible load, determine the greatest load (W) that can be applied to the beam if the allowable bending stress is 45 MN/m². Read more

Steps to Solve

1. Calculate the second moment of area $I$ for a single hollow rectangle

   $I$ of a rectangle is $bh^3/12$ where $b$ is base and $h$ is height. For a hollow rectangle, we subtract the $I$ of the inner rectangle from the $I$ of the outer rectangle. $$I_{single} = \frac{150 \times 120^3}{12} - \frac{(150-2\times20) \times (120-2\times20)^3}{12}$$ $$I_{single} = \frac{150 \times 120^3}{12} - \frac{110 \times 80^3}{12} = 21.6 \times 10^6 - 4.69 \times 10^6 = 16.91 \times 10^6 \ mm^4$$ $$I_{single} = 16.91 \times 10^{-6} \ m^4$$

2. Calculate the second moment of area of the combined shape.

Since there are two identical hollow rectangles stacked on top of each other about the neutral axis, the total second moment of area $I_{total}$ about the neutral axis is simply twice the second moment of area of a single rectangle.

$$I_{total} = 2 \times I_{single} = 2 \times 16.91 \times 10^{-6} = 33.82 \times 10^{-6} \ m^4 \ne 100.05 \times 10^{-6} \ m^4$$

However, we are calculating this about its centroid, so we must use the parallel axis theorem to calculate the moment of area about the neutral axis, $$I = I_c + Ad^2$$ Where $I_c$ is the second moment of area about the centroid $A$ is the area $d$ is the distance from the centroid to the neutral axis.

3. Correctly determining the second moment of area about the neutral axis The area of one hollow rectangle is: $$A = 150 \times 120 - 110 \times 80 = 18000 - 8800 = 9200 \ mm^2 = 9200 \times 10^{-6} \ m^2$$ The distance to the neutral axis for one of the rectangles is $d = 60 \ mm = 0.06 \ m$ The combined second moment of area is: $$I = 2 \times (16.91\times10^{-6} + (9200\times10^{-6}) \times 0.06^2) = 2 \times (16.91 \times 10^{-6} + 33.12\times 10^{-6} ) = 2 \times 50.03 \times 10^{-6} = 100.06 \times 10^{-6} \ m^4$$

4. Calculate the maximum bending moment for the uniformly distributed load.

The maximum bending moment $M_{max}$ for a simply supported beam with a uniformly distributed load $w$ is given by: $$M_{max} = \frac{wL^2}{8}$$ where $L$ is the span of the beam $$ M_{max} = \frac{16.5 \ kN/m \times (3.6 \ m)^2}{8} = \frac{16.5 \times 12.96}{8} = \frac{213.84}{8} = 26.73 \ kNm$$

5. Calculate the maximum bending stress $\sigma_{max}$.

The maximum bending stress is given by: $$\sigma_{max} = \frac{M_{max}y_{max}}{I}$$ where $y_{max}$ is the distance from the neutral axis to the outermost fiber, which is 120 mm = 0.12 m $$\sigma_{max} = \frac{26.73 \times 10^3 \ Nm \times 0.12 \ m}{100.06 \times 10^{-6} \ m^4} = \frac{3.2076 \times 10^3}{100.06 \times 10^{-6}} = 32.057 \times 10^6 \ N/m^2 = 32.057 \ MPa$$

6. Sketch the stress distribution diagram.

The stress distribution diagram is linear, with the maximum compressive stress at the top fiber and the maximum tensile stress at the bottom fiber. Since the section is symmetric, the magnitudes of the maximum compressive and tensile stresses are equal.

7. Calculate the section modulus Z.

The section modulus is given by: $$Z = \frac{I}{y_{max}} = \frac{100.06 \times 10^{-6} \ m^4}{0.12 \ m} = 0.8338 \times 10^{-3} \ m^3$$

8. Calculate the maximum bending moment for the two point loads.

The maximum bending moment $M_{max}$ for a simply supported beam with two point loads W at a distance $a$ from each support is: $$M_{max} = Wa$$ In this case, $a = 0.7 \ m + 2 \ m / 2= 1.7 \ m$, so $$M_{max} = 1.7W$$

9. Determine the greatest load W that can be applied.

The allowable bending stress is given as 45 MN/m², so we can set $\sigma_{max}$ equal to the allowable bending stress and solve for W: $$\sigma_{allowable} = \frac{M_{max}y_{max}}{I}$$ $$45 \times 10^6 \ N/m^2 = \frac{1.7W \times 0.12 \ m}{100.06 \times 10^{-6} \ m^4}$$ $$45 \times 10^6 = \frac{0.204W}{100.06 \times 10^{-6}}$$ $$W = \frac{45 \times 10^6 \times 100.06 \times 10^{-6}}{0.204} = \frac{4502.7}{0.204} = 22072.06 \ N = 22.07 \ kN$$