Question 3.SP.5: Figure S3.5 represents a gate, 2 ft wide perpendicular to th......

In addition to the reactive forces R_H at the hinge and R_E at end E, there are three forces acting on the gate: its weight W, the vertical hydrostatic force F_v upward on the rectangular bottom of the gate, and the slanting hydrostatic force F_s acting at right angles to the sloping rectangular portion of the gate. The magnitudes of the latter three forces are:

Given:              W=500 \mathrm{lb}

Eq. (3.16):      F_v=\gamma h_c A=\gamma(x)(4 \times 2)=8 \gamma x

Eq. (3.16):      F_s=\gamma h_c A=\gamma\left(\frac{x}{2}\right)\left(\frac{x}{\cos 30^{\circ}} \times 2\right)=1.155 \gamma x^2

A diagram showing these three forces is as follows:

The moment arms of W and F_v with respect to H are 1.2 \mathrm{ft} and 2.0 \mathrm{ft}, respectively. The moment arm of F_s gets larger as the water depth increases because the location of the center of pressure changes. We can find the location of the center of pressure of F_s from Eq. (3.18):

y_p=y_c+\frac{I_c}{y_c A}, \quad where, from Table A.7, I_c=\frac{b h^3}{12} with h=x / \cos 30^{\circ} and y_c=0.5 h. So

y_p=\frac{0.5 x}{\cos 30^{\circ}}+\frac{(1 / 12) 2\left(x / \cos 30^{\circ}\right)^3}{\left(0.5 x / \cos 30^{\circ}\right)\left[2\left(x / \cos 30^{\circ}\right)\right]}

i.e., for F_s: \quad O P=y_p=0.577 x+\frac{2 x}{12 \cos 30^{\circ}}=0.770 x

Hence the moment arm of F_s with respect to H is P H=x / \cos 30^{\circ}-0.770 x= 0.385 x. [Note: In this case we need not use Eq. (3.18) to find the lever arm of F_s because we know the line of action of F_s for the triangular distributed load on the rectangular area is at the one-third point between H and O, i.e., H P= (1 / 3)\left(x / \cos 30^{\circ}\right)=0.385 x ].

When the gate is about to open (incipient rotation), R_E=0 and the sum of the moments of all forces about H is zero, viz

\sum M=P_s(0.385 x)-F_v(2.0)+W(1.2)=0

i.e., \quad 1.155 \gamma x^2(0.385 x)-8 \gamma x(2)+500(1.2)=0

Substituting \gamma=62.4 \mathrm{lb} / \mathrm{ft}^3 gives

27.73 x^3-998.4 x+600=0

This is a cubic, polynomial equation (Appendix B). With a polynomial solver, available on some hand calculators (see Appendix B and Appendix C.1), we may find the three roots directly. With an equation solver, available on some scientific calculators (Appendixes C.1 and D.1) and in some spreadsheets and mathematics software (Appendixes C.2-3 and D.2-3), we may obtain the root closest to a guessed value we provide.

Without any of these aids, we can solve this equation by trials (“trial and error”), seeking an x value that makes the left side of the equation equal to zero. After two trials, we can use linear interpolation (or extrapolation) to estimate the next, better trial value. We then repeat this until x is sufficiently accurate, e.g., accurate to three significant figures after rounding:

We can find the other two roots by more, similar, trials. We could use a spreadsheet to facilitate such trials. But, more conveniently, dividing the cubic by (x-0.607) yields a quadratic (Eq. B.6) from which we can easily find that the other two roots (Eq. B.7) are x=5.67 and -6.28.

Thus x=0.607 \mathrm{ft} or 5.67 \mathrm{ft} or a negative (meaningless) root. Therefore, from inspection of the moment equation, the gate will remain closed when 0.607 \mathrm{ft}<x<5.67 \mathrm{ft} \quad

Note: Sections D.1-D.3 of Appendix D include complete example input and output for solutions to this problem using an HP 48G programmable calculator, and using Excel (spreadsheet) and Mathcad (mathematics software).

 F=\gamma h_c A                   (3.16)

 y_p=y_c+\frac{I_c}{y_c A}           (3.18)

 a x^2+(b+a r) x-\frac{d}{r}=0          (B.6)

x=-\left(\frac{b}{2 a}+\frac{r}{2}\right) \pm \sqrt{\left(\frac{b}{2 a}+\frac{r}{2}\right)^2+\frac{d}{a r}}              (B.7)