In this post, let's explore how to analyze a beam example using work and energy principles. Using this method, we'll illustrate how to solve for reactions, shear, and moment.
Type of Analysis: Classical Approach, Linear Analysis, Static Loads, Plane Beam, Determinate Beam
Preparation
Before analyzing a structure, we'll need to make some preparations first. That includes setting up our references and finding their determinacy.
Set-Up References
An excellent structural analysis must have a uniform mathematical understanding of the structure. It ensures that other people can easily understand your results.
We first place a Cartesian grid with its origin defined by our preference. In this case, let's assign the origin \(A(0, 0)\) at the leftmost point of the beam. Consequently, the x-axis will run along the length of the beam with the y-axis perpendicular to it at the origin.
We also need to identify the location of all points of interest: the location of supports, change in loads, and differences in the beam's cross-sectional properties. Starting from \(A\) and going toward the other end, we identify the following points of interest:
\(A(0.0m, 0.0m)\). The roller support.
\(C(3.0m, 0.0m)\). The \(40kN\) downwardconcentrated load.
\(D(6.0m, 0.0m)\). The \(80kN\) downwardconcentrated load.
\(B(8.0m, 0.0m)\). The hinge support and the start of the \(4kN/m\) uniform distributed load ↓.
\(E(11.0m, 0.0m)\). The \(16kN\) downwardconcentrated load and the end of the \(4kN/m\) uniform distributed load ↓.
You can label each joint according to your preference. The most important thing is that its coordinates must be defined appropriately.
Determinacy
We need to find the structure's determinacy \(D\) to know our approach.
For a 2D beam, it is:
\(D=r-(3+e_c)\)
This beam example have three reaction components: \(R_A\), \(B_v\), and \(B_h\), and no internal connections \(e_c = 0\); hence \(D=0\)
\(D=3-(3+0)=0\)
A determinacy of zero indicates that the structure can be analyzed using only the equilibrium equations.
Main Analysis
Since this post is about learning how to use the virtual work method, we will skip the stability analysis.
Reaction Analysis
Main Solution
This section will demonstrate solving the reaction load at roller support \(A\) using the virtual work method.
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.
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 \(\delta_A\)):
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.
Force Analysis
Shear
Main Solution
This section will demonstrate solving the shear at a point \(F(5,0)\) using the virtual work method.
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.
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 \(B\).
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.
\(\delta_C=3 d \theta\)
\(\delta_{F_1}=5 d \theta\)
\(\delta_{F_2}=3 d \theta\)
\(\delta_D=2 d \theta\)
\(\delta_E=3 d \theta\)
Substituting these values to the original equation, we get the following:
\(\left[V_F(5 d \theta)+V_F(3 d \theta)\right]+\left[40(3 d \theta)-80(2 d \theta)+16(3 d \theta)+4\left[\frac{1}{2}(3)(3 d \theta)\right]\right]=0\)
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.
Moment
Main Solution
This section will demonstrate solving the moment at a point \(F(5,0)\) using the virtual work method.
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.
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:
