Question 17.21: Using virtual work, determine support reactions for simple s...

Let the beam is turned by virtual angle, $\delta \theta$ about A so that the virtual distance at B, C, D and E are $(\delta y)_{B}$, $(\delta y)_{C}$, $(\delta y)_{D}$ and $(\delta y)_{E}$, respectively, as shown in Fig. 17.24 (a).
Using the principle of virtual work,

$R _{ E \cdot}(+\delta y)_{ E }+10 \cdot(-\delta y)_{ D }+20 \cdot(-\delta y)_{ C }+10(-\delta y)_{ B }=0$

$R _{ E } \cdot(\delta y)_{ E }-10 \cdot(\delta y)_{ D }-20 \cdot(\delta y)_{ C }-10(\delta y)_{ B }=0$                                                                      ….. (1)

From right angle triangles,

$\tan \delta \theta=\delta \theta=\frac{(\delta y)_{E}}{4}=\frac{(\delta y)_{D}}{3}=\frac{(\delta y)_{C}}{2}=\frac{(\delta y)_{B}}{1}$

$(\delta y)_{E}=4 . \delta \theta,(\delta y)_{D}=3 . \delta \theta,(\delta y)_{C}=2 . \delta \theta,(\delta y)_{B}=\delta \theta$

Substituting virtual distances in equation (1),

$R _{ E }(4 . \delta \theta)-10(3 . \delta \theta)-20(2 . \delta \theta)-10(\delta \theta)=0$

$R _{ E }=20  kN$

Since, $R _{ A }+ R _{ E }=40  kN$

thus $R _{ A }=40-20=20  kN$