Question 20.P.18: Derive the equation for the unit influence line for the vert...

Remove the support at B and apply a vertical load P at B. Then, using Macauley’s method

E I \frac{\mathrm{d}^{2} υ}{\mathrm{~d} x^{2}}=\frac{2}{3} P x-P[x-6]   (i)

(Note: the vertical reaction at A is 2P/3)

E I \frac{\mathrm{d} υ}{\mathrm{~d} x}=\frac{P}{3} x^{2}-\frac{P}{2}[x-6]^{2}+C_{1}    (ii)

E I υ=\frac{P}{9} x^{3}-\frac{P}{6}[x-6]^{3}+C_{1} x+C_{2}    (iii)

When x = 0, υ = 0 so that, from Eq. (iii), C_{2} = 0. Also, when x = 18 m, υ = 0 so that, from Eq. (iii)

0=\frac{P \times 18^{3}}{9}-\frac{P}{6}[18-6]^{3}+18 C_{1}

which gives

C_{1} = – 20P

Eq. (iii) then becomes

E Iυ=P\left(\frac{x^{3}}{9}-\frac{1}{6}[x-6]^{3}-20 x\right)  (iv)

For the influence line υ = -1 at x = 6 m. Then, from Eq. (iv)

\frac{P}{E I}=-\frac{1}{96}

The equation for the influence line is then

υ=-\frac{1}{96}\left(\frac{x^{3}}{9}-\frac{1}{6}[x-6]^{3}-20 x\right)  (v)

The area under the curve between A and B is given by

\text { Area }=\frac{1}{96} \int_{0}^{6}\left(\frac{x^{3}}{9}-20 x\right) \mathrm{d} x=3.375

The reaction at B is then = 24 × 3.375 = 81 kN.