Question 17.27: Using virtual work, determine reaction of roller bearing for...

Using the principle of virtual work,

\left[R_{D}(+\delta y)_{D}\right]+\left[30(-\delta y)_{M}\right]+\left[30(-\delta y)_{L}\right]+\left[(8 \times 2)(+\delta y)_{N}\right]=0

R_{D} \cdot(\delta y)_{D}-30(\delta y)_{M}-30(\delta y)_{L}+16(\delta y)_{N}=0                                                                             ….. (1)

From the right–angled triangles,

\delta \theta=\frac{(\delta y)_{N}}{1}=\frac{(\delta y)_{L}}{2}=\frac{(\delta y)_{M}}{10}=\frac{(\delta y)_{D}}{12}

from equation (1),

R _{ D } \cdot(12 . \delta \theta)-30(10 . \delta \theta)-30(2 . \delta \theta)+16 .(\delta \theta)=0

R _{ D }=28.67  kN