**Problem 2-1**

A simply supported beam is subjected to the loading as shown in figure 2-1(a). Determine the absolute maximum bending stress in the beam if the beam has a rectangular cross-section as shown in figure 2-1(b). Also draw bending stress diagram.

**Figure 2-1(a)**

**Figure 2-1(b)**

The bending stress is calculated by using the bending equation given below

M / I = σ / y

Where;

M = bending moment

I = moment of inertia about neutral axis of the section

σ = bending stress

y = the distance from neutral axis to the point of bending stress

we can write;

σ = M y / I

The above relation shows that bending stress will be maximum when the distance y is maximum i.e. at the top or bottom of the section means at a distance of 5 cm from the neutral axis of this section..

Absolute maximum bending stress will occur at the section where the bending moment is maximum.

The given beam is simply supported with uniform load on the entire span. In this case the maximum bending moment will occur at the mid-span of the beam and can be calculated by the following formula

M_{max} = w L^{2}/8

Where; w = uniform load on the beam (in this problem it is 2 kN/m)

L = beam span (equal to 4 m)

Therefore M_{max} = 2 x 4^{2 }/ 8 = 4 kNm

For a rectangular section, Moment of inertia about the neutral axis, I_{xx} = bd^{3}/12

In this section, b = 5 cm, d = 10 cm. The neutral axis is passing through the centroid of the section.

Therefore I_{xx} = 5 x 10^{3} / 12 = 0.416 x 10^{3} cm^{4} = 0.416 x 10^{-5} m^{4}

You can also use our moment of inertia calculator.

Absolute maximum bending stress = (M_{max}) (y_{max}) / (I_{xx});

substituting the values we get;

Absolute maximum bending stress
= 4 kNm x 0.05 m / (0.416 x 10^{-5}m^{4}) = 0.48 x 10^{5 }kN/m^{2} = **48 MPa **

(1 MPa = 1 Mega Pascal = 10^{6} N/m^{2})

The bending stress diagram will be as given below in Fig 2-1(c)

**Figure 2-1(c)**

The bending stress above the neutral axis (N.A.) will be compressive which is shown as negative whereas the bending stress below the neutral axis will be tensile and shown as positive.

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