Question 20.P.16: Determine the equation of the influence line for the reactio...

Release the beam at A and apply a load W as shown in Fig. S.20.16.

Then, R_{B} = 2W.

Using Macauley’s method

\begin{aligned}E I\left(\frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}\right) &=W x-2 W[x-2] \\E I\left(\frac{\mathrm{d} v}{\mathrm{~d} x}\right) &=\frac{W x^{2}}{2}-W[x-2]^{2}+\mathrm{C}_{1} \\E I v &=\frac{W x^{3}}{6}-\frac{W[x-2]^{3}}{3}+\mathrm{C}_{1} x+\mathrm{C}_{2}\end{aligned}

The boundary conditions are: υ = 0 when x = 2 m and 4 m. These give

\mathrm{C}_{1}=\frac{-10 W}{3}, \quad \mathrm{C}_{2}=\frac{16 W}{3}

so that

v=\left(\frac{W}{E I}\right)\left\{\left(\frac{x^{3}}{6}\right)-\frac{[x-2]^{3}}{3}-\left(\frac{10 x}{3}\right)+\frac{16}{3}\right\}

For the R_{A} IL, v = 1 when x = 0

so that

1=\left(\frac{W}{E I}\right)\left(\frac{16}{3}\right)

and

\frac{W}{E I}=\frac{3}{16}

Then

R_{\mathrm{A}}=\left(\frac{3}{16}\right)\left\{\left(\frac{x^{3}}{6}\right)-\frac{[x-2]^{3}}{3}-\left(\frac{10 x}{3}\right)+\frac{16}{3}\right\}

When a udl of 30 kN/m covers the span AB

R_{\mathrm{A}}=30\left(\frac{3}{16}\right) \int_{0}^{2}\left[\left(\frac{x^{3}}{6}\right)-\left(\frac{10 x}{3}\right)+\frac{16}{3}\right] \mathrm{d} x

i.e.

R_{A} = 26.25 kN.