Show that p must be directed through centroids of cross sections if axial stress ? is not to vary over a cross section.

1 answer:

7 0

Assuming P (usually written in upper case) represents a force normal to a given cross section.

If a point load is applied to any point of the section, stress concentration will cause axial stress to vary.

The context of the question considers the uniformity of axial stress at a certain distance away from the point of application (thus stress concentration can be neglected).

If a force P is applied through the centroid, sections will be stressed uniformly. However, if the force P is applied at a distance "e" from the centroid, the equivalent load on the section equals an axial force and a moment Pe. The latter causes bending of the member, causing non-uniform stress.

If we assume A=(uniform) cross sectional area, and I=moment of inertia of the section, then stress varies with the distance y from the centroid equal to
stress=sigma=P/A + My/I
where P=axial force, M=moment = Pe.
Therefore when e>0, the stress varies across the section.

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The comparison graph is also attached in 3rd figure. In the 3rd figure, the graph with vertex (0, 0) is representing $f(x) = x^2$ and $\:f\left(x\right)=\left(x-6\right)^2$ is represented as being shifted 6 units to the right as compare to the function $f(x) = x^2$ .

Step-by-step explanation:

When we Add or subtract a positive constant, let say c, to input x, it would be a horizontal shift.

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Type of change Effect on y = f(x)

$y = f(x - c)$ horizontal shift: c units to right

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The graph of $\:f\left(x-6\right)=\left(x-6\right)^2$ is shown below in second figure. It is clear that the graph of $\:f\left(x-6\right)=\left(x-6\right)^2$ is shifted 6 units to the right as compare to the function $f(x) = x^2$ .