Civil Structural Girder Web Plate Centroid Analysis (Civil & Construction Engineering) Solution

A structural engineer and construction project manager at the UNLV Howard R. Hughes College of Engineering are detailing custom fabricated steel gusset connection plates for an elevated highway interchange bridge.

The planar gusset plate occupies a general region $R$ in the $xy$ -plane bounded below by the parabolic profile $y = \frac{1}{2}x^2$ and above by the horizontal top flange line $y = 2$ (with spatial coordinates measured in meters):

R = \left\{(x, y) \in \mathbb{R}^2 \;\middle|\; -2 \le x \le 2, \, \frac{1}{2}x^2 \le y \le 2\right\}

Step-by-Step Solution:

Locate Intersection Points (Outer Bounds): Equate the lower and upper boundary functions:
$\frac{1}{2}x^2 = 2 \implies x^2 = 4 \implies x = -2 \quad \text{and} \quad x = 2$ Because the plate is bounded below by $y = \frac{1}{2}x^2$ and above by $y = 2$ , this forms a symmetric Type I General Region (defined in Double Integrals over a General Region): $R = \left\{(x, y) \in \mathbb{R}^2 \;\middle|\; -2 \le x \le 2, \, \frac{1}{2}x^2 \le y \le 2\right\}$
Evaluate the Geometric Planar Area ( $A = \iint_R 1 \, dA$ ): State the Type I Area Integral Formula:
$A = \int_{-2}^{2} \int_{\frac{1}{2}x^2}^{2} 1 \, dy \, dx$ Evaluate the inner integral: $\int_{\frac{1}{2}x^2}^{2} 1 \, dy = [y]_{\frac{1}{2}x^2}^{2} = 2 - \frac{1}{2}x^2$ Evaluate the outer integral, exploiting symmetry over $[-2, 2]$ : $A = \int_{-2}^{2} \left( 2 - \frac{1}{2}x^2 \right) \, dx = 2 \int_{0}^{2} \left( 2 - \frac{1}{2}x^2 \right) \, dx$ $= 2 \left[ 2x - \frac{1}{6}x^3 \right]_{0}^{2} = 2 \left( 4 - \frac{8}{6} \right) = 2 \left( 4 - \frac{4}{3} \right) = 2 \left( \frac{8}{3} \right) = \frac{16}{3}\text{ m}^2$
Evaluate the First Moment of Area ( $M_x = \iint_R y \, dA$ ): State the First Moment of Area Integral Formula:
$M_x = \int_{-2}^{2} \int_{\frac{1}{2}x^2}^{2} y \, dy \, dx$ Evaluate the inner integral: $\int_{\frac{1}{2}x^2}^{2} y \, dy = \left[ \frac{1}{2}y^2 \right]_{\frac{1}{2}x^2}^{2} = \frac{1}{2}(2)^2 - \frac{1}{2}\left(\frac{1}{2}x^2\right)^2 = 2 - \frac{1}{8}x^4$ Evaluate the outer integral exploiting symmetry: $M_x = \int_{-2}^{2} \left( 2 - \frac{1}{8}x^4 \right) \, dx = 2 \int_{0}^{2} \left( 2 - \frac{1}{8}x^4 \right) \, dx$ $= 2 \left[ 2x - \frac{1}{40}x^5 \right]_{0}^{2} = 2 \left( 4 - \frac{32}{40} \right) = 2 \left( 4 - \frac{4}{5} \right) = 2 \left( \frac{16}{5} \right) = \frac{32}{5}\text{ m}^3$
Calculate the Vertical Centroid Coordinate ( $\bar{y}$ ): State the Centroid Formula:
$\bar{y} = \frac{M_x}{A} = \frac{\iint_R y \, dA}{\iint_R 1 \, dA}$ Substitute values: $\bar{y} = \frac{32 / 5}{16 / 3} = \frac{32}{5} \times \frac{3}{16} = \frac{2 \times 3}{5} = \frac{6}{5} = 1.20\text{ m}$

Meaning of the Answer & Real-Life Application:

Quantitative Interpretation: The vertical centroid sits at $\bar{y} = 1.20\text{ m}$ above the coordinate base $y = 0$ . Because the plate tapers to a narrow vertex at the bottom parabola $y = \frac{1}{2}x^2$ and has its widest flange width ( $4\text{ m}$ ) at the top $y = 2$ , the geometric mass center is shifted upward above the vertical geometric midpoint ( $y = 1.0\text{ m}$ ).
Engineering Application: In structural steel erection and construction crane rigging, bridge gusset plates must be picked and hoisted by spreader cables aligned with the plate's center of gravity. If rigging lines are hitched at the geometric midpoint ( $y = 1.0$ ) instead of the true centroid ( $\bar{y} = 1.2$ ), gravitational self-weight creates an eccentric overturning moment that tilts the plate in mid-air, creating severe pinch hazards for ironworkers during bolt alignment.