A vertical pipe column with an outside diameter of D = 114 mm and an inside diameter of d = 102 mm supports the loads shown. Determine the normal and shear stresses acting at point H, and show these stresses on a stress element.

Question

Plan the Solution
The cross-sectional properties will be computed for the pipe column. Each of the applied loads will be considered in turn. The normal and/or shear stresses created by each at point H will be computed.
Both the stress magnitude and its direction must be evaluated and shown on the proper face of a stress element. By the principle of superposition, the stresses will be combined appropriately so that the state of stress at point H is summarized succinctly by the stress element.

Accepted Answer

The outside diameter of the pipe is D = 114 mm, and the inside diameter is d = 102 mm.
The area, the moment of inertia, and the polar moment of inertia for the cross section are

[latex]A=\frac{\pi}{4}\left[D^{2}-d^{2} ight]=\frac{\pi}{4}\left[(114 mm )^{2}-(102 mm )^{2} ight]=2,035.752 mm ^{2}[/latex].

[latex]I=\frac{\pi}{64}\left[D^{4}-d^{4} ight]=\frac{\pi}{64}\left[(114 mm )^{4}-(102 mm )^{4} ight]=2,977,287 mm ^{4}[/latex].

[latex]J=\frac{\pi}{32}\left[D^{4}-d^{4} ight]=\frac{\pi}{32}\left[(114 mm )^{4}-(102 mm )^{4} ight]=5,954,575 mm ^{4}[/latex].

Stresses at H
The forces and moments acting at the section of interest will be evaluated sequentially to determine the type, magnitude, and direction of any stresses created at H.
The 17-kN axial force creates compression normal stress, which acts in the y direction:

[latex]\sigma_{y}=\frac{F_{y}}{A}=\frac{17,000 N }{2,035.752 mm ^{2}}=8.351 MPa ( C )[/latex].

The 3.2-kN force acting in the positive z direction creates transverse shear stress (i.e., τ = VQ/It) throughout the cross section of the pipe. However, the magnitude of the transverse shear stress is zero at point H.
The 3.2-kN force acting in the positive z direction also creates a bending moment at the section where H is located. The magnitude of the bending moment is

[latex]M_{x}=(3.2 kN )(0.8 m )=2.56 kN - m[/latex].

By inspection, we observe that this bending moment about the x axis creates compression normal stress on the horizontal faces of the stress element at H:

[latex]\sigma_{y}=\frac{M_{x} c}{I_{x}}[/latex].

[latex]=\frac{(2.56 kN - m )(57 mm )(1,000 mm / m )(1,000 N / kN )}{2,977,287 mm ^{4}}[/latex].

[latex]=49.011 MPa ( C )[/latex].

The 3.75 kN-m torque acting about the y axis creates shear stress at H. The magnitude of this shear stress can be calculated from the elastic torsion formula:

[latex] au=\frac{T c}{J}[/latex].

[latex]=\frac{(3.75 kN - m )(57 mm )(1,000 mm / m )(1,000 N / kN )}{5,954,575 mm ^{4}}[/latex].

[latex]=35.897 MPa[/latex].

Combined Stresses at H
The normal and shear stresses acting at point H can be summarized on a stress element. Note that at point H, the torsion shear stress acts in the –x direction on the +y face of the stress element. After the proper shear stress direction has been established on one face, the shear stress directions on the other three faces are known.

Question 12.1: A vertical pipe column with an outside diameter of D = 114 m...

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