Question 5 of 6: Froude-Krylov Wave-Exciting Force and Moment on a Box Barge

Question 5: Froude-Krylov Wave-Exciting Force and Moment on a Box Barge (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Box barge $L\times B\times T$, incident wave $\eta(x,t)=\eta_a\sin(kx-\omega t)$, $k=2\pi/\lambda$; midship at $x=0$, barge spanning $-L/2\le x\le L/2$.

Find. Derive the amplitude coefficients $F_1,F_2$ (heave force, in- and out-of-phase with $\sin\omega t$ at the midship) and $M_1,M_2$ (pitch moment, likewise) from Froude-Krylov theory.

Approach. In the thin-draft (long-wave) Froude-Krylov approximation, the exciting force per unit length along the barge is the surface dynamic pressure $\rho g\eta(x,t)$ times the beam $B$; integrating this (and its moment arm $x$, for pitch) over the length, and splitting $\sin(kx-\omega t)=\sin(kx)\cos\omega t-\cos(kx)\sin\omega t$, isolates the in-phase and quadrature force/moment components using only the odd/even symmetry of $\sin(kx)$ and $\cos(kx)$ about the midship.

1. Set up the Froude-Krylov force integral. The vertical force per unit length is $B\rho g\eta(x,t)=B\rho g\eta_a\sin(kx-\omega t)$. Expanding, $$\sin(kx-\omega t)=\sin(kx)\cos(\omega t)-\cos(kx)\sin(\omega t),$$ so the total heave force is $$F(t)=B\rho g\eta_a\left[\cos(\omega t)\!\!\int_{-L/2}^{L/2}\!\!\sin(kx)\,dx \;-\;\sin(\omega t)\!\!\int_{-L/2}^{L/2}\!\!\cos(kx)\,dx\right].$$
2. Apply symmetry. $\sin(kx)$ is an odd function of $x$, so its integral over the symmetric interval $[-L/2,L/2]$ vanishes identically — this is exactly $F_2/\rho g\eta_a$, the coefficient of $\cos\omega t$: $$\boxed{F_2/\rho g\eta_a = 0}\quad\text{(exact, by odd symmetry, for any barge symmetric about midship).}$$ $\cos(kx)$ is even, so $\int_{-L/2}^{L/2}\cos(kx)\,dx=2\sin(kL/2)/k$, giving the coefficient of $-\sin\omega t$: $$F_1=-\frac{2B\rho g\eta_a}{k}\sin\!\left(\frac{kL}{2}\right).$$
3. Convert to $\lambda$ and match the target form. With $k=2\pi/\lambda$, $kL/2=\pi L/\lambda$ and $1/k=\lambda/2\pi$: $$\frac{F_1}{\rho g\eta_a}=-\frac{2B}{k}\sin\!\left(\frac{\pi L}{\lambda}\right)=-\frac{B\lambda}{\pi}\sin\!\left(\frac{\pi L}{\lambda}\right)=\boxed{-\frac{BL}{\pi}\cdot\frac{\lambda}{L}\sin\!\left(\frac{\pi L}{\lambda}\right)},$$ which is exactly the printed target.
4. Pitch moment — set up and apply symmetry. The pitch moment (arm $x$, same force per length) is $$M(t)=B\rho g\eta_a\left[\cos(\omega t)\!\!\int_{-L/2}^{L/2}\!\! x\sin(kx)\,dx\;-\;\sin(\omega t)\!\!\int_{-L/2}^{L/2}\!\! x\cos(kx)\,dx\right].$$ $x\cos(kx)$ is odd (odd $\times$ even), so its integral vanishes — this is $M_1/\rho g\eta_a$, the coefficient of $-\sin\omega t$: $$\boxed{M_1/\rho g\eta_a = 0}\quad\text{(exact, by odd symmetry).}$$ $x\sin(kx)$ is even, so $\int_{-L/2}^{L/2}x\sin(kx)\,dx=2\int_0^{L/2}x\sin(kx)\,dx=\dfrac{2\sin(kL/2)}{k^2}-\dfrac{L\cos(kL/2)}{k}$ (by parts), giving the coefficient of $\cos\omega t$: $$M_2=B\rho g\eta_a\left[\frac{2\sin(kL/2)}{k^2}-\frac{L\cos(kL/2)}{k}\right].$$
5. Convert to $\lambda$. Substituting $k=2\pi/\lambda$ throughout, $$\frac{M_2}{\rho g\eta_a}=B\left[\frac{\lambda^2}{2\pi^2}\sin\!\left(\frac{\pi L}{\lambda}\right)-\frac{L\lambda}{2\pi}\cos\!\left(\frac{\pi L}{\lambda}\right)\right] = \boxed{\frac{B\lambda^2}{2\pi^2}\left[\sin\!\left(\frac{\pi L}{\lambda}\right)-\frac{\pi L}{\lambda}\cos\!\left(\frac{\pi L}{\lambda}\right)\right]}.$$