How to Analyze Beams Using Virtual Work?

Create the Virtual Deflected Shape

This section will discuss how to get the reaction at $A$ using the virtual work method.

The first step is to allow $R_A$ to perform virtual work on the beam and allow it to create an imaginary deflected shape. To develop such deflection, think of the structure as a lever with support $B$ as the pivot point. As $R_A$ acts upwards, the beam rotates at hinge support $B$, producing the deflected shape below.

Get Total Work

Next, we solve the "net" work $W_{tot}$ due to external $W_e$ and internal forces $W_i$. In our example:

• The external work $W_e$ is from the actual loads and their imaginary displacement.
• The internal work $W_i$ is from the $R_A$ and its imaginary displacement.

From these, the sum of external work is equal to

$W_e=-40\left({\delta }_C\right)-80\left({\delta }_D\right)+16\left({\delta }_E\right)+4[\frac{1}{2}(3)\left({\delta }_E\right)]$

Recall that work is positive when the force (moment) and displacement (rotation) are in the same direction and negative if otherwise.

For the uniform load in segment $BE$, work is force times the covered area on the deflected shape.

On the other hand, the internal work is equal to

$W_i=R_A\left({\delta }_A\right)$

Now that we have $W_e$ and $W_i$, we can solve for the $W_{tot}$:

$W_{tot}=W_e+W_i$

$W_{tot}=-40\left({\delta }_C\right)-80\left({\delta }_D\right)+16\left({\delta }_E\right)+4[\frac{1}{2}(3)\left({\delta }_E\right)]+R_A\left({\delta }_A\right)$

Apply Work-Energy Theorem and Equilibrium

With the total work obtained, we can apply the static equilibrium principle and work-energy theorem to get the following:

$0=-40\left({\delta }_C\right)-80\left({\delta }_D\right)+16\left({\delta }_E\right)+4[\frac{1}{2}(3)\left({\delta }_E\right)]+R_A\left({\delta }_A\right)$

Notice that this expression has many unknowns; however, we can express the displacements in terms of another by doing ratio and proportion on the deflected shape. Our goal here is now to express all imaginary translations in one variable (say ${\delta }_A$):

$\frac{{\delta }_A}{8}=\frac{{\delta }_C}{5}=\frac{{\delta }_D}{2}=\frac{{\delta }_E}{3}$

${\delta }_C=\frac{5}{8}{\delta }_A$

${\delta }_D=\frac{2}{8}{\delta }_A$

${\delta }_E=\frac{3}{8}{\delta }_A$

Substituting these values to the original equation, we get the following:

$R_A\left({\delta }_A\right)+[-40\left(\frac{5}{8}{\delta }_A\right)-80\left(\frac{2}{8}{\delta }_A\right)+16\left(\frac{3}{8}{\delta }_A\right)+4[\frac{1}{2}(3)\left(\frac{3}{8}{\delta }_A\right)]]=0$

We can cancel out all ${\delta }_A$ which will make $R_A$ the only remaining unknown quantity; hence:

$R_A=36.75kN$

We can double-check your answer by taking a moment summation at $B$ to solve for the reaction at $A$ and compare it with our answer.

Create the Virtual Deflected Shape

To start, cut the beam at $F$, so we can "expose" the internal forces that act on the beam: $V_F$ and $M_F$.

Just like we did to reaction $R_A$, allow $V_F$ to do "work" to the structure. At the left section, the shear rotates the beam downward about the roller, while at the right, the shear displaces it upward about the hinge.

Get Total Work

Next, we solve the "net" work $W_{tot}$ due to external $W_e$ and internal forces $W_i$. In our example:

• The external work $W_e$ is from the actual loads and their imaginary displacement.
• The internal work $W_i$ comes from two sources: the $V_F$ and its imaginary displacement and $M_F$ and its imaginary rotation.

From these, the sum of external work is equal to

$W_e=40\left({\delta }_C\right)-80\left({\delta }_D\right)+16\left({\delta }_E\right)+4[\frac{1}{2}(3)\left({\delta }_E\right)]$

Recall that work is positive when the force (moment) and displacement (rotation) are in the same direction and negative if otherwise.

On the other hand, the internal work is equal to

$W_i=V_F\left({\delta }_{F_1}\right)+V_F\left({\delta }_{F_2}\right)-M_F\left(d{\theta }_A\right)+M_F\left(d{\theta }_B\right)$

Since we are concerned with the shear, we must cancel $M_F$. That only happens when $d{\theta }_A$ and $d{\theta }_B$ are equal. In other words, the slope of the deflection at $A$ must be equal to that of $B$.

$d{\theta }_A=d{\theta }_B=d\theta$

$W_i=V_F\left({\delta }_{F_1}\right)+V_F\left({\delta }_{F_2}\right)-M_F(d\theta )+M_F(d\theta )$

$W_i=V_F\left({\delta }_{F_1}\right)+V_F\left({\delta }_{F_2}\right)$

Now that we have $W_e$ and $W_i$, we can solve for the $W_{tot}$:

$W_{tot}=W_e+W_i$

$W_{tot}=40\left({\delta }_C\right)-80\left({\delta }_D\right)+16\left({\delta }_E\right)+4[\frac{1}{2}(3)\left({\delta }_E\right)]+V_F\left({\delta }_{F_1}\right)+V_F\left({\delta }_{F_2}\right)$

Apply Work-Energy Theorem and Equilibrium

With the total work obtained, we can apply the static equilibrium principle and work-energy theorem to get the following:

$0=40\left({\delta }_C\right)-80\left({\delta }_D\right)+16\left({\delta }_E\right)+4[\frac{1}{2}(3)\left({\delta }_E\right)]+V_F\left({\delta }_{F_1}\right)+V_F\left({\delta }_{F_2}\right)$

Notice that this expression has many unknowns; however, we can express the displacements in terms of another by doing ratio and proportion on the deflected shape. Our goal here is now to express all imaginary translations in one variable (say $d\theta$):

$\tan (d\theta )\approx d\theta =\frac{{\delta }_C}{3}=\frac{{\delta }_{F_1}}{5}$

$\tan (d\theta )\approx d\theta =\frac{{\delta }_{F_2}}{3}=\frac{{\delta }_D}{2}=\frac{{\delta }_E}{3}$

Remember that for relatively small angles, the tangent of such is almost the same as the angle.

${\delta }_C=3d\theta$

${\delta }_{F_1}=5d\theta$

${\delta }_{F_2}=3d\theta$

${\delta }_D=2d\theta$

${\delta }_E=3d\theta$

Substituting these values to the original equation, we get the following:

$[V_F(5d\theta )+V_F(3d\theta )]+[40(3d\theta )-80(2d\theta )+16(3d\theta )+4[\frac{1}{2}(3)(3d\theta )]]=0$

We can cancel out all $d\theta$ which will make $V_F$ the only remaining unknown quantity; hence:

$V_F=-3.25kN$

We can double-check your answer using shear equations or diagrams.

Create the Virtual Deflected Shape

To start, cut the beam at $F$, so we can "expose" the internal forces that act on the beam: $V_F$ and $M_F$ just like we did above.

Just like we did to reaction $R_A$ and $V_F$, allow $M_F$ to do "work" to the structure. At the left section, the moment rotates the beam counterclockwise about the roller, while at the right, the shear displaces it clockwise about the hinge.

Get Total Work

Next, we solve the "net" work $W_{tot}$ due to external $W_e$ and internal forces $W_i$. In our example:

• The external work $W_e$ is from the actual loads and their imaginary displacement.
• The internal work $W_i$ comes from two sources: the $V_F$ and its imaginary displacement and $M_F$ and its imaginary rotation.

From these, the sum of external work is equal to

$W_e=-40\left({\delta }_C\right)-80\left({\delta }_D\right)+16\left({\delta }_E\right)+4[\frac{1}{2}(3)\left({\delta }_E\right)]$

Recall that work is positive when the force (moment) and displacement (rotation) are in the same direction and negative if otherwise.

On the other hand, the internal work is equal to

$W_i=-V_F\left({\delta }_{F_1}\right)+V_F\left({\delta }_{F_2}\right)+M_F\left(d{\theta }_A\right)+M_F\left(d{\theta }_B\right)$

Since we are concerned with the shear, we must cancel $M_F$. That only happens when ${\delta }_{F_1}$ and $delt{a}_{F_2}$ are equal.

${\delta }_{F_1}={\delta }_{F_2}={\delta }_F$

$W_i=-V_F\left({\delta }_F\right)+V_F\left({\delta }_F\right)+M_F\left(d{\theta }_A\right)+M_F\left(d{\theta }_B\right)$

$W_i=M_F\left(d{\theta }_A\right)+M_F\left(d{\theta }_B\right)$

Now that we have $W_e$ and $W_i$, we can solve for the $W_{tot}$:

$W_{tot}=W_e+W_i$

$W_{tot}=[M_F\left(d{\theta }_A\right)+M_F\left(d{\theta }_B\right)]+[-40\left({\delta }_C\right)-80\left({\delta }_D\right)+16\left({\delta }_E\right)+4[\frac{1}{2}(3)\left({\delta }_E\right)]]$

Apply Work-Energy Theorem and Equilibrium

With the total work obtained, we can apply the static equilibrium principle and work-energy theorem to get the following:

$0=[M_F\left(d{\theta }_A\right)+M_F\left(d{\theta }_B\right)]+[-40\left({\delta }_C\right)-80\left({\delta }_D\right)+16\left({\delta }_E\right)+4[\frac{1}{2}(3)\left({\delta }_E\right)]]$

Notice that this expression has many unknowns; however, we can express the displacements in terms of another by doing ratio and proportion on the deflected shape. Our goal here is now to express all imaginary translations in one variable (\delta_F):

$\tan \left(d{\theta }_A\right)\approx d{\theta }_A=\frac{{\delta }_C}{3}=\frac{{\delta }_F}{5}$

$\tan \left(d{\theta }_B\right)\approx d{\theta }_B=\frac{{\delta }_F}{3}=\frac{{\delta }_D}{2}=\frac{{\delta }_E}{3}$

Remember that for relatively small angles, the tangent of such is almost the same as the angle.

${\delta }_C=\frac{3}{5}{\delta }_F$

${\delta }_D=\frac{2}{3}{\delta }_F$

${\delta }_E={\delta }_F$

Substituting these values to the original equation, we get the following:

$[M_F\left(\frac{1}{5}{\delta }_F\right)+M_F\left(\frac{1}{3}{\delta }_F\right)]+[-40\left(\frac{3}{5}{\delta }_F\right)-80\left(\frac{2}{3}{\delta }_F\right)+16\left({\delta }_F\right)+4[\frac{1}{2}(3)\left({\delta }_F\right)]]=0$

We can cancel out all $d\theta$ which will make $M_F$ the only remaining unknown quantity; hence:

$M_F=103.75kN•m$

We can double-check your answer using moment equations or diagrams.