Mechanics of Materials: Lesson 20 -Statically Indeterminate Superposition Material Between Two Walls

Summary

In this video, the instructor tackles a statically indeterminate axial elongation problem involving two A36 steel bars of different diameters positioned between walls, with one wall initially separated by a small air gap. Using the method of superposition and compatibility equations, the video demonstrates step-by-step how to calculate the reaction forces at both walls. The process includes computing free elongation under a 200 kN force, determining the excess elongation beyond the gap, and calculating the forces needed to compress the bars back into place. The explanation emphasizes careful unit conversion, precise calculations, and common pitfalls, providing a clear approach to complex structural mechanics problems.

Takeaways

• 😀 The problem discussed is a statically indeterminate axial problem involving two steel bars fixed between walls.
• 😀 Compatibility equations are required to solve statically indeterminate problems, providing extra necessary equations beyond simple statics.
• 😀 The method of superposition is used to separate the effects of applied forces and boundary constraints for easier calculation.
• 😀 Step 1 of superposition: remove the wall at D and calculate the free elongation of the bar using δ = PL/AE.
• 😀 Only the section of the bar directly experiencing the applied force stretches; other parts move along without elongation until constrained.
• 😀 For the bar in the example, the free elongation under a 200 kN force was calculated to be 0.30558 mm.
• 😀 Step 2: reintroduce the wall at D and calculate the reaction force needed to compress the bar back by the difference between elongation and air gap (0.15558 mm).
• 😀 The reaction at wall D (RD) was calculated to be 20.36 kN, and the reaction at wall A (RA) was determined by equilibrium to be 179.64 kN.
• 😀 A common mistake is to subtract the applied force from RD during compression; the applied force is already accounted for in Step 1.
• 😀 Carrying extra decimal precision throughout calculations is important to avoid rounding errors in final results.
• 😀 This problem demonstrates how statically indeterminate systems require careful accounting of deformation and forces, similar to combining effects in shear and moment diagrams.

Q & A

What is meant by a 'statically indeterminate' problem?

A statically indeterminate problem is one where the number of unknown forces exceeds the number of available equations from statics alone. To solve it, you need additional equations, known as compatibility equations, which ensure that deformations are consistent with the system's constraints.

What are compatibility equations, and why are they needed?

Compatibility equations are extra equations used in statically indeterminate problems. They ensure that the displacements or deformations of the structure are compatible with the physical constraints, such as the closure of gaps or the position of walls.

How does the method of superposition work in solving this problem?

The method of superposition involves breaking the problem into simpler parts. First, you remove one constraint (like the wall) and calculate the deformation (elongation). Then, you add the constraint back and compute the effect of compression. These effects are added together to find the total deformation.

How is the elongation of the bar calculated in this example?

Elongation is calculated using the formula Δ = PL / AE, where P is the applied force, L is the length of the bar, A is the cross-sectional area, and E is the Young's modulus of the material (A36 steel in this case). The elongation accounts only for the section that is being stretched.

Why does only the section of the bar that is being stretched elongate?

Only the section being directly subjected to the force will elongate because, like a rubber band, if you stretch one part, the other parts remain unaffected. In this case, the part of the bar that is between the gap will stretch, while the other part does not.

How do we determine how much the bar will compress after the gap is closed?

The compression needed to close the gap is calculated by subtracting the elongation (calculated earlier) from the original gap size. The force causing this compression is the reaction force at point D, which is then used in the equation for deformation.

What is the significance of the reaction forces at points A and D?

The reaction forces at points A and D are critical to understanding how the bar responds to the applied load. Point A reacts by pulling away from the wall, while point D compresses to close the gap. These forces must satisfy the equilibrium equation RA + RD = 200 kN.

How is the force RD calculated in this problem?

RD is calculated by first finding the total elongation and then using the compatibility equation to determine how much the wall at point D needs to compress the bar to close the gap. The resulting force RD can be found by solving the deformation equation using the known material properties and geometry.

Why is it important to carry extra decimal places in calculations?

Carrying extra decimal places ensures precision in calculations, especially when dealing with small values. In this case, small differences in the results could significantly affect the final answer, particularly when calculating forces or deformations.

How do the forces in the two sections of the bar contribute to the total deformation?

The forces in the two sections of the bar, one with a 50 mm diameter and the other with a 25 mm diameter, cause different deformations due to their differing cross-sectional areas. These forces are applied in the method of superposition, where the elongation from the initial force is added to the compression caused by the reaction force.