Question 6.1: A beam of rectangular cross-section carries a total load of ...

From the overall free-body diagram of the beam shown in Figure 6.10(b), let us take moment about point B. It gives R_{ A }=10  kN . Now if we take a section at a distance x from the left end and consider the equilibrium of its left side as shown in Figure 6.10(c), we can write

M_x=R_{ A } x-w \frac{x^2}{2}            (1)

Putting R_{ A }=10  kN and w = 20/3.6 kN/m, we get

M_x=10 x-\frac{20}{3.6} \frac{x^2}{2}

It is evident from the load-distribution pattern that shear force will vary along the length and will be zero at midpoint (refer to Example 5.1 in the previous chapter). So, maximum bending moment will occur there. Putting x = 3.6/2 = 1.8 m in Eq. (1), we obtain

M_x=M_{\max }=10 \times 1.8-\frac{20}{3.6} \cdot \frac{(1.8)^2}{2}=9  kN m

Now bending stress from Eq. (6.5) is

\sigma_x=\frac{M y}{I}           (6.5)

\sigma=\frac{M y}{I}

Hence, the maximum bending stress at the bottom-most fibre will be tensile and maximum bending stress at topmost fibre will be compressive. The magnitude for both is given by M_{\max } / Z where Z is the section modulus. Equating this with allowable stress 7 MPa and remembering Z = bh³/6 for a rectangular cross-section having width b and depth h, we get

\frac{M_{\max }}{b h^2 / 6}=7 \times 10^6

or      \frac{9 \times 10^6 \times 6}{b h^2}=7 \times 10^6

or      \frac{9 \times 10^6 \times 6}{b \times(2 b)^2}=7 \times 10^6 \quad(\text { as } h=2 b)

or b = 124.5 mm and h = 249 mm. So, depth is 249 mm and width is 124.5 mm.