# Curved Beam Analysis on a Hand-Forged, Non-Rectangular Cross Section

In a recent post I discussed curved beam analysis with a rectangular cross section, and as a follow-up a comparison of that type of analysis with finite element results. Many curved beams, such as the lifting hook shown above, do not have rectangular cross-sections. This is an interesting topic, not only because of the engineering involved, but also because it is an interesting specimen of late nineteenth and early twentieth century craftsmanship, as the piece was (as indicated on the drawing) hand forged. Thus we combine analysis with craftsmanship.

## Overview of the Application

The hook shown above was intended to be used with a pile pulling chain. The complete assembly is shown below, taken from the 1929 catalogue.

The hook shown in the drawing is the largest of the series. The eye on the back side of the hook–and the shackle to go with it–was for the tugger line. The chain (a curved beam problem in its own right) was wrapped around the pile (usually wood) and the hook connected back into the chain. The top of the chain was attached to the crane line and the pile pulled out of the ground (hopefully.)

## Loads and Moments on the Hook

As was the case before, the main axis of loading is the axis between the centre of the lifting eye of the hook and the centre of the radius of curvature of the inside of the hook. The moment is thus the product of the main force $P$ and the distance from the centre of the radius of curvature to the centroid of the section. It’s harder to see with this hook, but that’s just about where the lozenge shaped shaded section is on the hook, which suggests that Vulcan applied some kind of analysis in some way to the design of this hook.

The load on the hook is ostensibly the load given in the literature above, i.e. 61 kips. However, that’s not the way it was used, because the hook was wrapped around the pile in a “choker” arrangement and hooked back to the chain. Let’s say for analysis purposes that the basic load on the hook is 61/2 = 30.5 kips. We still haven’t determined the moment because we still haven’t determined the centroid of the section.

## Developing the Equations

The problem here is obvious: with rectangular cross-sections, the centroid is easy to find, and the integration is likewise simple. Here we have a composite, non-rectangular cross-section. Belyaev gives us values for several non-rectangular cross-sections but doesn’t give us a way to put them together. For that we will turn to Boresi et.al. (1993).

Let’s start by stating the basic equation of tangential stress, from the previous post:

$\sigma_{\theta} = \frac {P}{A}+\frac {M}{\rho}\frac{z}{S}$ (1)

where

• $P =$ basic load on hook
• $A =$ cross-sectional area of hook
• $M = PR_0 =$ moment on curved beam
• $R_0 =$ distance from centre of curvature to centroid of section
• $\rho$ = distance of a particular point of interest from the centre of curvature
• $z$ = distance from centroid of section to neutral axis of bending stress
• $S$ = result of integration from the inner to the outer radius of the section

From Boresi et.al. (1993), let us rewrite this as follows:

$\sigma_{{\theta}}=\frac{P}{A}+\frac{P{\it R_{0}}\,\left(A-\rho\,{\it A_{m}}\right)}{A\rho\,\left({\it R_{0}}\,{\it A_{m}}-A\right)}$ (2)

where

$A_{m}=\int_{A}\frac{dA}{\rho}$ (3)

This integral appears often in Belyaev and we will have recourse to it.

The advantage of using Equation (2) is the following:

• We can take several sections and compute their values of $A$, adding them up for the total area.
• We can take several sections and compute their values of $A_m$, likewise adding them up for the total integral
• We can compute the centroid $R_0$ of the entire section by computing the areas and centroids of each part and them compute $R_0$ by summing the products of the centroids and sections of each area and then dividing them by the total area.

Now let’s look at a CAD-drawn diagram of the critical area shown in the original drawing.

The geometry of this section is defined by the radii, which are a little tricky to pull together. This is not unusual in Vulcan designs. It’s also worth noting that, although the drawing calls for the section to be 2 3/4″ at its thickest, the section calls for a 3″ thickest point. For our analysis we’ll use the latter. That lack of precision–and some of the other vague dimensioning on the drawing–may be frustrating to the modern engineer, but it reflects a feature on Vulcan drawings of the era: many decisions were left to the craftsmen at the shop floor.

In any case the three regions we’ll use for analysis are obvious. Also obvious are the centroid location and cross-sectional area, both of which came from the CAD software. It’s possible to compute $A_m$ by a) dividing the section into thin vertical strips, b) computing the area for each strip by multiplying the width by the length, c) measuring the distance from the centroid of the strip to the centre of curvature, d) dividing the area by the centroidal distance for each strip, and e) summing up the results.

From the drawing the distance from the centre of curvature to the centroid is 3 3/4″ + 2.994 ~ 6 3/4″, and the area is 15.3 sq.in. Let’s see how this compares with using the method of Boresi.

There are three zones shown above:

1. The “ellipsoid” closest to the centre of curvature;
2. The trapezoid in the tapered section; and
3. The semicircle at the far side.

The variables and their values are as follows:

With all of that, the formulae for these three sections are as follows:

#### Section 1 (ellipsoid):

$A = 1/2\,\pi \,ab_4$ (4a)
$R = R_2-4/3\,{\frac {a}{\pi }}$ (4b)
$A_m = 2\,b_4+{\frac {\pi \,b_4\left (R_2-\sqrt {{R_2}^{2}-{a}^{2}}\right )}{a}}-2\,b_4\sqrt {{R_2}^{2}-{a}^{2}}\arcsin({\frac {a}{R_2}}){h}^{-1}$ (4c)

#### Section 2 (Trapezoid):

Note the diagram to the right, which will helpful to understand the variables.

$A = 1/2\,\left ({\it b_2}+{\it b_1}\right )\left (h\right )$ (5a)
$R = {\frac {R_2\left (2\,{\it b_2}+{\it b_1}\right )+R_1\left ({\it b_2}+2\,{\it b_1}\right )}{3\,{\it b_2}+3\,{\it b_1}}}$ (5b)
$A_{m}=\left(b_{1}+R_{1}\frac{b_{2}-b_{1}}{h}\right)-\left(b_{2}-b_{1}\right)$ (5c)

#### Section 3 (Semicircle)

These equations assume the semicircle is “pointing outward,” i.e., the semicircle is outside of the diameter that bisects the circle relative to the radius of curvature. Equation (6c) also assumes that $R_1 > b_3$.

$A = 1/2\,{b_3}^{2}\pi$ (6a)
$R = R_1+4/3\,{\frac {b_3}{\pi }}$ (6b)
$A_m = R_1\pi -2\,b_3-\pi \,\sqrt {{R_1}^{2}-{b_3}^{2}}+2\,\sqrt {{R_1}^{2}-{b_3}^{2}}\arcsin({\frac {b_3}{R_1}})$ (6c)

#### Putting Them Together

Making all of the appropriate substitutions yields the following:

The results for $A$ and $R_0$ are very close.

Substituting these values and the load into Equation (2) and plotting the results yields the following:

The stress at the inside radius is 22.8 ksi and at the outside radius -7.7 ksi.

## Some Notes About the “Blacksmith”

The “note to blacksmith” on the drawing reveals more than the draughtsman probably meant to, because it means that these were basically forged by hand. Vulcan was started as a foundry and had one until after World War II, but several parts on the hammers were consistently forged:

The last is on the same size scale as the hook. Although I haven’t investigated it, it’s likely that Vulcan had some kind of forging hammer to assist the blacksmith in making these parts. That said, the basic geometry of the part came into being through the interaction of the blacksmith and the forging hammer. Such shows the level of craftsmanship that existed in the manufacture of the Vulcan product line, a significant part of its success. It’s also an interesting glimpse into a manufacturing base long gone; its loss has and will prove expensive to our economy.

## Other Reference

• Boresi, A.P., Schmidt, R.J. and Sidebottom, O.M. (1993) Advanced Mechanics of Materials. Fifth Edition. New York: John Wiley and Sons, Inc.