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**Given: **

A simply supported beam of length L=5m is given in Figure 1a. The beam is subjected to uniform distributed load equal to 3.2kN/m. If the beam has a rectangular cross-section, as shown in Figure 1b, calculate the absolute maximum bending stress in the beam and draw the bending stress diagram.

**Solution:**

The bending stress of a beam element is calculated using the bending equation, i.e. *M / I = σ / y*, where *M *is the bending moment,* I* is the moment of inertia about the neutral axis (N.A.) of the section, *σ* is the bending stress and *y* is the distance from the neutral axis to the point of bending stress. In other words, we can write that:

**σ = M y / I (1)**

The above relation shows that the bending stress will be maximized if the distance *y* becomes maximum, i.e. at the top or bottom of the section. The latter is translated to a distance of 5cm from the neutral axis of the section. It is pointed out that the absolute maximum bending stress concerns the section of the maximum bending moment.

Moreover, the given beam is simply supported with uniform loading on the entire span. Thus, the maximum bending moment will occur at the mid-span of the beam and is calculated following the formula M_{max} = q L^{2}/8, where *q* is the uniform load on the beam and *L* is the span of the beam. The maximum bending moment is:

**M _{max} = 3.2 x 5^{2} / 8 = 10kNm (2)**

For a rectangular section, the moment of inertia I_{xx}, about the neutral axis, is calculated as:

**I _{xx} = bd^{3}/12 = 6 x 12^{3} / 12 = 0.864 x 10^{3}cm^{4} = 0.864 x 10^{-5} m^{4} (3)**

Using Eq. 1, the absolute maximum bending stress is calculated as:

**σ _{max} = (M_{max}) (y_{max}) / (I_{xx}) = 10kNm x 0.06m / (0.864 x 10^{-5}m^{4}) = 0.694 x 10^{5} kN/m^{2} = 69.4 MPa**

The bending stress diagram is shown in Figure 2. The bending stress above the neutral axis (N.A.) is compressive (negative) while the bending stress below the neutral axis is tensile (positive).

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