Question 20.P.17: Derive the equation for the unit influence line for the reac...
Derive the equation for the unit influence line for the reaction at the support B in the continuous beam shown in Fig. P.20.17 and calculate its value if a uniformly distributed load of intensity 20 kN/m covers the span AB; the flexural rigidity of the beam is EI. Also sketch the unit influence lines for the shear force and bending moment at the point D.
Remove the support at B and apply a vertical load P at B. Then, measuring x from the left hand support and using Macauley’s method (Section 13.2)
E I \frac{\mathrm{d}^{2} υ}{\mathrm{~d} x^{2}}=\frac{P}{2} x-P[x-4] (i)
E I \frac{\mathrm{d}υ}{\mathrm{~d} x}=\frac{P}{4} x^{2}-\frac{P}{2}[x-4]^{2}+C_{1} (ii)
E I υ=\frac{P}{12} x^{3}-\frac{P}{6}[x-4]^{3}+C_{1} x+C_{2} (iii)
When x = 0, υ = 0 so that, from Eq. (iii), C_{2} = 0. Also, when x = 8 m, υ = 0 and again from Eq. (iii)
0=\frac{P \times 8^{3}}{12}-\frac{P}{6}[8-4]^{3}+8 C_{1}
which gives
C_{1} = -4P
Eq. (iii) then becomes
E I υ=P\left(\frac{x^{3}}{12}-\frac{1}{6}[x-4]^{3}-4 x\right) (iv)
For the unit influence line υ = – 1 at x = 4 m. Note that R_{B} will be in the opposite direction to P. Therefore, from Eq. (iv)
-E I=P\left(\frac{4^{3}}{12}-4 \times 4\right)
which gives
\frac{P}{E I}=-\frac{3}{32}
The equation for the influence line is then
υ=\frac{-3}{32}\left(\frac{x^{3}}{12}-\frac{1}{6}[x-4]^{3}-4 x\right) (v)
Note that Eq. (v) gives the equation for the complete unit influence line as opposed to Eq. (vi) of Ex. 20.9.
The area under the curve between A and B is given by
\text { Area }_{\mathrm{AB}}=\frac{3}{32} \int_{0}^{4}\left(4 x-\frac{x^{3}}{12}\right) \mathrm{d} x=2.5
The reaction at B is therefore = 20 × 2.55 = 0 kN
The influence lines for the shear force and bending moment at D are shown in Fig. S.20.17.
