Elasticity — Question 4

Question 4

The effective area of a tapered rod is not its arithmetic mean

A slender rod of length $L$ and constant modulus $E$ has slowly varying area $A(x)=A_0(1+x/L)$. It carries axial tension $F$. In a one-dimensional small-strain model, a slice extends by $d\delta=F\,dx/[EA(x)]$.

1. Integrate the extension and find the uniform area $A_{eff}$ that gives the same extension.

2. Would replacing the rod by its arithmetic-mean area $3A_0/2$ overestimate or underestimate the extension?

Question 4 - Solution

The effective area of a tapered rod is not its arithmetic mean

(a) Derivation and calculation

Integrate the local compliance: $$\delta=\frac{F}{EA_0}\int_0^L\frac{dx}{1+x/L} =\boxed{\frac{FL\ln2}{EA_0}}.$$ Equating this to $FL/(EA_{\mathrm{eff}})$ gives $$\boxed{A_{\mathrm{eff}}=A_0/\ln2=1.443A_0}.$$

(b) Result and interpretation

Using the arithmetic-mean area instead predicts $$\delta_{\mathrm{arith}}=\frac{FL}{E(3A_0/2)}=\frac{2FL}{3EA_0}.$$ Since $2/3<\ln2$, $$\boxed{\delta_{\mathrm{arith}}<\delta}.$$ The relative underestimate is $1-2/(3\ln2)=3.82\%$. The narrow portions contribute disproportionately because compliance depends on $1/A$.

Physical check

The effective area lies between $A_0$ and $2A_0$, but below their arithmetic mean.