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Principle of virtual work

A virtual displacement $ \delta$$ \bf s$ of a point is any arbitrary infinitesimal change in the position of the point consistent with the constraints imposed on the motion of the point. This displacement can be just imagined.

Virtual work $ \delta$U done by a force is defined as $ \bold$FT$ \delta$$ \bf s$.

Virtual work $ \delta$U done by a couple is defined as $ \bold$MT$ \delta$$ \varphi$.

The principle of virtual work (pvw) can be used in statics for solution of equilibrium problem. The following is valid:

The necessary and sufficient condition for the equilibrium of a particle is zero virtual work done by all working forces acting on the body during any virtual displacement $ \delta$$ \bf s$ consistent with the constraints imposed on the particle.

The necessary and sufficient condition for the equilibrium of a rigid body is zero virtual work done by all external forces acting on the particle during any virtual displacement $ \delta$$ \bf s$ consistent with the constraints imposed on the body.

When using the principle of virtual work for a system of connected rigid bodies (mechanism) we must keep in mind that no virtual work is done by internal forces, by reactions in smooth constraints, or by forces normal to the direction of motion. The virtual work is done by reactions when friction is present.

Exercise 3.12.1   Sample problem Using pvw determine the magnitude of a force Z for equilibrium of a crank-slider mechanism in the position given by the angle $ \varphi$ = 30o. Given is M = 50 Nm, Q = 35 N, r = 0.1 m.

Figure 3.81: Exercise 3.12.1. Equilibrium of the crank-slider mechanism
\includegraphics[width=90mm]{SVW91}

Solution
First we denote the position of points of action of applied forces Q, Z and the position of the crank by coordinates $ \varphi$, z, y. According to pvw we can write

- M $\displaystyle \delta$$\displaystyle \varphi$ - Q $\displaystyle \delta$y - Z $\displaystyle \delta$z = 0 

where

y = $\displaystyle {\frac{r}{2}}$ sin$\displaystyle \varphi$ ,        $\displaystyle \delta$y = $\displaystyle {\frac{r}{2}}$ cos$\displaystyle \varphi$ $\displaystyle \delta$$\displaystyle \varphi$ 

z = r cos$\displaystyle \varphi$ + r $\displaystyle \sqrt{4-\sin^2\varphi}$ + b ,        $\displaystyle \delta$z = - r sin$\displaystyle \varphi$ $\displaystyle \left(\vphantom{1+\frac{\cos\varphi}{\sqrt{4-\sin^2\varphi}}}\right.$1 + $\displaystyle {\frac{\cos\varphi}{\sqrt{4-\sin^2\varphi}}}$$\displaystyle \left.\vphantom{1+\frac{\cos\varphi}{\sqrt{4-\sin^2\varphi}}}\right)$ $\displaystyle \delta$$\displaystyle \varphi$  .

For $ \delta$$ \varphi$ $ \neq$ 0 we have

Z = $\displaystyle {\frac{M+Q \frac r2 \cos\varphi}{r \sin\varphi \left(1+\frac{\cos\varphi}{\sqrt{4-\sin^2\varphi}}\right)}}$ = $\displaystyle {\frac{50+35 \cdot 0,05 \cdot \cos{30^\circ}}{0,1 \cdot 0,5\left(1+
\frac{\cos{30^\circ}}{\sqrt{4-\frac 14}}\right)}}$ 

The result is

Z = 711.92 N 

Exercise 3.12.2   Equilibrium of a mechanism Using pvw determine the magnitude of a couple M acting on the crank when the position of a mechanism (see Fig. 3.82) is given by $ \varphi$ = 30o. We know that F = 300 N, $ \alpha$ = 45o, Z = 900 N, r = 0.04 m.

Figure 3.82: Exercise 3.12.2. Equilibrium of a mechanism
\includegraphics[width=60mm]{SVW93}

Solution
M = 42.8 Nm

Exercise 3.12.3   Equilibrium of a car hood A car hood (see Fig. 3.83) is in equlibrium position given by $ \varphi$ = 30o. Determine the stiffness k of a spring the free length of which is l0 = 0.07 m. It is known that Z = 50 N, r = 0.1 m. Use pvw.

Figure 3.83: Exercise 3.12.3. Equilibrium of a mechnism
\includegraphics[width=60mm]{SVW95}

Solution
k = 8333 Nm-1

Exercise 3.12.4   Equilibrium of a mechanism of a front wheel suspension A car wheel suspension (see Fig. 3.84) is loaded by a force Z = 2500 N. The spring has a free length l0 = 0.1 m. Using pvw determine the stiffness k of the spring. The length r = 0.28 m and the angles are $ \varphi$ = 60o, $ \alpha$ = 55o. Determine the stiffness of a spring the free length of which is l0 = 0.07 m. Use pvw.

Figure 3.84: Exercise 3.12.4. Equilibrium of a mechanism of a front wheel suspension
\includegraphics[width=60mm]{SVW96}

Solution
k = 34509 Nm-1

Exercise 3.12.5   Equilibrium of a bridge The equlibrium position of a draw bridge (see Fig. 3.85) is given by $ \varphi$ = 30o. Using pvw determine the value of a couple M acting on drum. It is known that r = 0.1 m, l = 4.5 m, Q = 5000 N.

Figure 3.85: Exercise 3.12.5. Equilibrium of a bridge
\includegraphics[width=90mm]{SVW97}

Solution
M = 250 Nm


next up previous contents
Next: Solutions of exercises Up: Statics Previous: Work and potential energy   Contents
marcel 2001-08-14