Which of the following statements is/are correct in relation to the analysis of cables?

1. Under horizontal uniformly distributed load, the shape of the cable is parabolic.

2. If the maximum dip (d) of the cable is small as compared with the span (L), then the total length of the cable is equal to \(\rm \left[L+\frac{3d^2}{8L}\right]\)

Concept:

Cables subjected to uniformly distributed loads take specific geometric shapes based on how the load is applied.

When the load is uniformly distributed in the horizontal direction, the shape of the cable is parabolic. If the load is uniformly distributed along the cable’s length, the shape is a catenary.

Derivation for the length of cable under UDL:

Consider a cable of length ACB, supported at A and B, and length of cable before the dip occurs is l and length of cable after the dip occur is L at the same level and carrying the UDL as shown, below

Let C be the lowest point on the cable. Taking C as origin, let's consider a point P(x,y) on the cable and draw a tangent to the cable at point P meets the horizontal line CS at R. From the geometry of the curve, we have

CR = RS = \(\frac{X}{2}\)

Part CP is in equilibrium, under the horizontal force (H), Downward load (w/unit length), Tension in cable (T)

Triangle PSR represents to some scale the triangular forces that are in equilibrium. So, we can write

\(\frac{PS}{wx}=\frac{RS}{H}=\frac{RP}{T}\)

\(\frac{y}{wx}=\frac{x}{2H}\)

\(y =\frac{wx^{2}}{2H}\)

\(\frac{dy}{dx}=\frac{wx}{h}\)      .... (1)

Now, consider an elementary length of curve ds between two points P and Q. Taking the length of the arc PQ equal to the length of the chord PQ, We can write

\(ds = \sqrt{dx^{2}+dy^{2}}\)

= \(dx\sqrt{1+\left ( \frac{dy}{dx} \right )^{2}}\) ..... (2)

Putting dy/dx from equation (1) into equation 2, we have

\(ds =dx\sqrt{1+\left ( \frac{wx}{H} \right )^{2}}\)

If  \(\frac{Wx}{H}\) is a fraction, expanding it by binomial theorem we have 

\(\left ( 1+\left ( \frac{Wx}{H} \right ) ^{2}\right )^{1/2} = 1 + \frac{w^{2}x^{2}}{2H^{2}}+ .....\)

Neglecting higher powers of  \(\left ( \frac{Wx}{H} \right ) ^{2}\)

\(ds = \left ( 1+\frac{1}{2} \left [ \frac{wx}{H} \right ]^{2} \right )dx\)

Integrating this from x = 0 to x = l/2

\(\int_{0}^{l/2}ds =\int_{0}^{l/2}\left ( 1+\frac{w^{2}x^{2}}{2h^{2}} \right )dx\)

\(\frac{L}{2} =\left [ x+ \frac{w^{2}x^{3}}{6H^{2}} \right ]^{1/2}\)

\(\frac{L}{2} = \frac{l}{2}+ \frac{w^{2}l^{3}}{48H^{2}}\)

Total length = \(L = l+ \frac{w^{2}l^{3}}{24H^{2}}\)

Substituting \(H =\frac{wl^{2}}{8h}\) (Formula for horizontal thrust parabolic arc carrying UDL)

\(L = l + \frac{8h^{2}}{3l}\)

∴ The total length of the cable after deflection is \(l+\frac{8h^{2}}{3l}\)

Confusion Points

1.  In question, it is given that, there is a dip in the beam, but not given the loading under which this dip occurs, So we have to assume that dip occurs due to the self-weight of the beam and we know self-weight always occurs as UDL.
2. Once the beam is deflected and takes parabolic shape (given in question), we can apply the formula for arches on it, as applied to find horizontal thrust in this question.