Axial Deformation: Statically Determinate and Indeterminate Bars

Hello! Welcome back to our course.

In the past few lessons, we've focused on how external forces create internal stresses within a material—specifically normal stress, shear stress, and bearing stress. Today, we'll connect stress back to a concept from an earlier lesson: strain. We will explore how axial stress causes a member to physically change its length. This change, known as deformation or elongation, is a critical design consideration. An aircraft wing strut must not only be strong enough to not break, but also stiff enough to not stretch or shrink excessively under load.

Our learning outcome for this lesson is to analyze deformation and elongation in bars under axial loads, including statically indeterminate cases. We will start with the fundamental formula for calculating elongation in simple cases and then move on to more complex situations where the basic equations of statics are not enough to solve the problem.

1. Axial Elongation: The "PL/AE" Formula

When an axial force $P$ is applied to a bar, it causes a normal stress $\sigma =P/A$. This stress, in turn, causes a normal strain $ϵ=\delta /L$, which is the change in length per unit length. Recalling Hooke's Law ($\sigma =Eϵ$), we can combine these simple relationships into one powerful formula that directly calculates the total change in length, $\delta$.

Where:

• $\delta$ (delta) is the total change in length (elongation or contraction).
• $P$ is the internal axial force in the member.
• $L$ is the original length of the member.
• $A$ is the cross-sectional area.
• $E$ is the Young's Modulus of the material.

This formula is a cornerstone of mechanics of materials. Let's watch a video that introduces this formula and shows how to apply it.

Principle of Superposition

As demonstrated in the video, if a bar consists of multiple segments with different loads, cross-sections, or materials, we can find the total deformation by calculating the deformation of each segment individually and adding them up. This is an application of the Principle of Superposition.

The total deformation ${\delta }_{total}$ is the algebraic sum of the deformations of the individual segments:

A critical first step is always to find the internal force $P_i$ in each segment. This is done using the method of sections, as shown in the video. You make an imaginary "cut" in the segment you're interested in and use the equations of static equilibrium ($\Sigma F_x=0$) on the free-body diagram to find the internal force. Remember the sign convention:

• Tension (pulling apart) is positive (+).
• Compression (pushing together) is negative (-).

2. Statically Indeterminate Problems

So far, we've looked at problems where you could find all the reaction forces using only the equations of static equilibrium ($\Sigma F=0,\Sigma M=0$). These are called statically determinate problems.

Now, let's consider a bar that is fixed at both ends and has a load applied somewhere in the middle.

In the bottom case, we have two unknown reaction forces ($R_A$ and $R_B$) but only one relevant equilibrium equation ($\Sigma F_x=0$). We have more unknowns than equations. This is a statically indeterminate problem.

How do we solve it? We need an additional equation. This equation comes from the geometry of the deformation. It's called a compatibility equation.

The Solution Strategy

As the video and supporting text below explain, solving statically indeterminate axial load problems involves a clear, four-step process.

Let's summarize the four-step method for solving these problems:

1. Equilibrium: Draw a free-body diagram of the entire structure and write down the relevant static equilibrium equations. You will have more unknowns than equations.

   • Example: $R_A+R_B-P=0$

2. Compatibility: Write an equation that describes the geometric constraints on the deformation.

   • For a bar fixed at both ends, the total change in length must be zero.
   • Example: ${\delta }_{total}={\delta }_{AC}+{\delta }_{CB}=0$, where AC is the segment from the left wall to the load, and CB is the segment from the load to the right wall.

3. Force-Displacement Relationship: Express the deformations in the compatibility equation in terms of the unknown forces using the elongation formula, $\delta =\frac{PL}{AE}$. Be careful to correctly identify the internal force $P$ in each segment in terms of the unknown reactions.

   • Example: The internal force in segment AC is $R_A$, and the internal force in segment CB is $-R_B$ (or $R_A-P$). So the compatibility equation becomes: $\frac{R_A{L}_{AC}}{A_{AC}{E}_{AC}}+\frac{-R_B{L}_{CB}}{A_{CB}{E}_{CB}}=0$

4. Solve: You now have a system of two equations (one from equilibrium, one from compatibility) and two unknowns ($R_A$ and $R_B$). Solve them simultaneously to find the reaction forces. Once the reactions are known, you can find the stresses and deformations in each segment.

This method can also be extended to problems involving thermal expansion, where a change in temperature adds another term to the deformation equation: ${\delta }_T=\alpha (\Delta T)L$. You will see this in future, more advanced problems.

Conclusion

In this lesson, you've learned how to quantify the change in length of components under axial load, a crucial step in ensuring a design is not just strong, but also sufficiently stiff.

Key Takeaways:

• The deformation of an axially loaded member is given by the formula $\delta =PL/AE$.
• For members with multiple segments, the total deformation is the algebraic sum of the individual segment deformations.
• Statically indeterminate problems have more unknown forces than can be solved by static equilibrium alone.
• To solve them, we introduce a compatibility equation that describes the geometric constraints of the system (e.g., total deformation is zero).
• The solution requires simultaneously solving the equilibrium equations and the compatibility equations.

Next Lesson Preview:

We've now analyzed 1D stress (axially loaded bars) quite thoroughly. However, many components in engineering, especially in aerospace, experience stresses in two directions simultaneously. A classic example is a pressurized aircraft fuselage or a fuel tank. In our next lesson, we will begin our look into 2D stress states by learning how to determine stresses in thin-walled pressure vessels using hoop and longitudinal stress formulas.