1.3. Example

Consider the problem that we had solved using EBT in the previous lecture. A transverse force P is applied at the free end of a beam which is clamped at the other one end as shown in Figure 5.

In this example, we are looking at a beam that is fixed at one end (clamped) and has a force applied at the opposite end. This scenario represents a classic situation known as the cantilever beam problem. Understanding how to analyze this situation is key in engineering, as it helps predict how structures will behave under loads.

Imagine a cantilever as a tightrope walker who is anchored at one end. As they extend their arms and lean at the far end, we can see how forces and moments act upon their body—much like a beam under load.

The first step in solving is finding the shear force and bending moment profile in the beam. For that, we cut a section in the beam at a distance X from the clamped end and draw the free body diagram of the right portion of the beam as shown in Figure 6.

To solve the problem, we need to analyze how forces are distributed along the beam. By cutting the beam at a distance X from where it is clamped, we can create a free body diagram. This diagram allows us to clearly see the applied force and how it interacts with the shear forces and moments within the beam, which are essential for further calculations.

Think of this step like slicing into a piece of fruit to view its internal structure. Just as we need to see where the seeds and juice are located to understand the distribution within the fruit, we need to see the internal forces in the beam to understand how loads affect it.

The transverse load P acts on the free end while a shear force and moment act on its left end. Moment balance about the centroid of the left-end yields -M(X) + P(L-X) = 0 ⇒ M(X) = P(L-X). Whereas force balance yields V(X) = P.

Here, we perform calculations to find the moment (M) and shear force (V) acting on the beam. The equation derived from moment balance tells us how the moment changes along the length of the beam when a force is applied, while the force balance shows the relationship between the applied load and the shear force. This is fundamental in understanding how beams respond under load.

Consider balancing a seesaw with one person at each end. The moment is akin to the strength of the efforts at each end; if one side exerts too much force without a counteracting force, it tips over. In this scenario, keeping a balance of forces ensures the seesaw remains level, much like how forces and moments must be balanced in a beam.

Plugging them into equations of TBT (equations (7) and (11)), we get the equations to describe the deformation of the beam based on shear forces and moment.

In this segment, we take the shear force and moment equations we derived and substitute them into the Timoshenko Beam Theory equations. This step is crucial as it connects our previously calculated values to the broader theory, enabling us to analyze the beam's deflection considering both shear and bending effects.

Think of this process like cooking—once you have all of your ingredients (shear force and moment equations), you need to mix them properly (substituting into TBT equations) to create a finished dish (the final deformation equations). Each ingredient adds to the flavor (accuracy) of the result.

We have two boundary conditions both at the clamped end. The displacement y and the cross-section rotation θ must be zero here, i.e., y(0) = 0, θ(0) = 0.

Boundary conditions are essential as they define the state of the system at the points where loads act or supports exist. Setting these conditions allows us to solve the equations accurately and reflects the reality that at the clamped end, there is no displacement or rotation of the beam.

Consider a pole planted firmly in the ground; it does not move at its base (the clamped end). Similarly, by setting the displacement and rotation to zero, we mimic this fixed condition in our calculations.

Integrating equation (15), we get (19). Using equation (18) init, we get C = 0. Now, we can plug the above expression for θ in equation (16) to get (20) whose integration yields (using (17)) (21), this is the final expression for deflection y obtained by TBT.

Through integration of our derived equations, we can find the deflection (y) of the beam under load. This final outcome illustrates how beams react to loads and is crucial for design and safety assessments in engineering.

Similar to solving a puzzle: once we have all the pieces (integrated equations), we fit them together to see the complete picture (final deflection). Each integrated piece contributes to understanding how the beam will perform under load.

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