Question 17.24: Two beams are hinged together at C as shown in Fig. 17.27. U...
Two beams are hinged together at C as shown in Fig. 17.27. Using the principle of virtual work, determine reaction at D.
Let reaction R_{D} is given virtual displacement (\delta y)_{D} as shown in Fig. 17.27(a). Let the beams turned for virtual angles \delta \theta_{1} and \delta \theta_{2} at F and A, respectively.
Using principles of virtual work,
\left[R_{D} \cdot(+\delta y)_{D}\right]+\left[20 \cdot(-\delta y)_{E}\right]+\left[10 \cdot(-\delta y)_{B}\right]=0 ….. (1)
From right–angled triangles of beam CDF,
\delta \theta_{1}=\frac{(\delta y)_{C}}{4}=\frac{(\delta y)_{D}}{2}=\frac{(\delta y)_{E}}{1}
i.e. (\delta y)_{C}=4 . \delta \theta_{1},(\delta y)_{D}=2 . \delta \theta_{1},(\delta y)_{E}=\delta \theta_{1}
From right–angled triangles of beam AC,
\delta \theta_{2}=\frac{(\delta y)_{C}}{3}=\frac{(\delta y)_{B}}{2}
(\delta y)_{C}=3 . \delta \theta_{2},(\delta y)_{B}=2 . \delta \theta_{2}
since (\delta y)_{C} is common virtual distance for both beams,
thus, (\delta y)_{C} = 3.\delta \theta_{2} = 4.\delta \theta_{1}
i.e. \delta \theta_{1}=\frac{3}{4} \cdot \delta \theta_{2}
(\delta y)_{B}=2 . \delta \theta_{2},(\delta y)_{D}=2 . \delta \theta_{1}=2\left(\frac{3}{4} \cdot \delta \theta_{2}\right)
=\frac{3}{2} \cdot \delta \theta_{2}
and (\delta y)_{E}=\delta \theta_{1}=\frac{3}{4} \cdot \delta \theta_{2}
Substituting the values of virtual distances in equation (1),
R_{D}\left(\frac{3}{2} \cdot \delta \theta_{2}\right)-20\left(\frac{3}{4} \cdot \delta \theta_{2}\right)-10\left(2 \delta \theta_{2}\right)=0
R_{D}=23.33 k N
