Question 20.5: A simply supported beam AB has a span of 5 m and carries a u...

The shear force at a section of the beam will be a maximum with the head or tail of the load at that section. Initially, before writing down an expression for shear force, we require the support reaction at $\mathrm{A}, R_{\mathrm{A}}$. Thus, with the head of the load at a section a distance $x$ from $\mathrm{A}$, the reaction, $R_{\mathrm{A}}$, is found by taking moments about $\mathrm{B}$.

Thus

$R_{\mathrm{A}} \times 5-0.6 \times 5 \times 2.5-1.5 x\left(5-\frac{x}{2}\right)=0$

whence

$R_{\mathrm{A}}=1.5+1.5 x-0.15 x^{2}$           (i)

The maximum shear force at the section is then

$S(\max )=-R_{\mathrm{A}}+0.6 x+1.5 x$     (ii)

or, substituting in Eq. (ii) for $R_{\mathrm{A}}$ from Eq. (i)

$S(\max )=-1.5+0.6 x+0.15 x^{2}$     (iii)

Equation (iii) gives the maximum shear force at any section of the beam with the load moving from left to right. Then, when $x=0, S(\max )=-1.5  \mathrm{kN}$ and when $x=5 \mathrm{~m}$, $S(\max )=+5.25  \mathrm{kN}$. Furthermore, from Eq. (iii) $S(\max )=0$ when $x=1.74 \mathrm{~m}$.

The maximum shear force for the load travelling from right to left is found in a similar manner. The final diagram of maximum shear force is shown in Fig. 20.12(b) where we see that reversal of shear force may take place within the length cd of the beam; cd is sometimes called the focal length.