The A 992 steel drill shaft of an oil well extends 12000 ft into the ground. Assuming that the p

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The A 992 steel drill shaft of an oil well extends 12000 ft...

The $A 992$ steel drill shaft of an oil well extends $12000 \mathrm{ft}$ into the ground. Assuming that the pipe used to drill the well is suspended frecly from the derrick at $A$, determine the maximum average normal stress in each pipe segment and the elongation of its end $D$ with respect to the fixed end at $A$. The shaft consists of three different sizes of pipe, $A B, B C,$ and $C D,$ each having the length, weight per unit length, and cross-sectional area indicated.

Key Concepts

Material Property (Young’s Modulus)

Young’s modulus is a fundamental material property that quantifies its stiffness under elastic deformation. It is used in the calculation of elongation and stress, linking the applied load to the resulting strain in the member. For materials like A992 steel, using the correct value of Young’s modulus ensures accurate predictions of both stress and deformation under the axial load.

Composite Member Analysis

A composite member is made up of different segments having varying cross-sectional areas or material properties. In analyzing such a structure, each segment must be evaluated individually for stress and deformation, while ensuring that continuity and equilibrium conditions are met at the interfaces. This approach is crucial for correctly determining the overall behavior of the structure under load.

Axial Normal Stress

Axial normal stress is the force per unit area experienced in a member when an axial force is applied. It is computed by dividing the axial force (which in this case includes the self-weight of the pipe segment) by the cross-sectional area. This concept is essential for assessing whether the material can safely handle the load without yielding or failing.

Axial Deformation (Elongation)

Axial deformation refers to the change in length of a member under an applied axial load. It is determined using the relationship ? = (Force × Length) / (Area × Young’s Modulus). This calculation is important in evaluating how much a structure like a drill shaft elongates under its own weight and ensuring that deformations do not compromise structural integrity.

Self Weight and Distributed Loading

When a structure supports its own weight, the load is distributed along its length rather than being applied at a single point. Each segment in the drill shaft effectively supports the weight of all material below it, resulting in varying axial forces along its length. Understanding this concept is key to accurately obtaining stress and deformation profiles in vertically loaded, long slender structures.

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