Computational Physics (Mechanical System)

Computer Science Level 5

Trusses are lightweight structures capable of carrying heavy loads. In bridge design, the individual members of the truss are connected with rotatable pin joints that permit forces to be transferred from one member of the truss to another. The accompanying figure shows a truss that is held stationary at the lower left endpoint $(1)$ , is permitted to move horizontally at the lower right endpoint $(4)$ , and has pin joints at $(1)$ , $(2)$ , $(3)$ , and $(4)$ . A load of $10,000\text{ N}$ is placed at joint $(3)$ , and the resulting forces on the joints are given by $f_1$ , $f_2$ , $f_3$ , $f_4$ , and $f_5$ , as shown. When positive, these forces indicate tension on the truss elements, and when negative, compression. The stationary support member could have both a horizontal force component $F_1$ and a vertical force component $F_2$ , but the movable support member has only a vertical force component $F_3$ .

If the truss is in static equilibrium, then based on Newton's $1^{\text{st}}$ law of motion the forces at each joint must add to the zero. Thus, the sum of the horizontal and vertical components at each joint must be $0$ . Calculate the value of $\lfloor\Phi\rceil$ where

$\Phi=|F_1|+|F_2|+|F_3|+|f_1|+|f_2|+|f_3|+|f_4|+|f_5|.$

The answer is 48966.

1 solution

Steven Chase
Jan 29, 2019

There are eight scalar unknowns and four joints, each with two equilibrium conditions (x and y). We therefore have eight linear equations to solve for eight unknowns.

$1X: -F_1 + f_2 + f_1 \, cos \frac{\pi}{4} = 0 \\ 1Y: f_1 \, sin \frac{\pi}{4} - F_2 = 0 \\ 2X: -f_1 \, sin \frac{\pi}{4} + f_4 \, sin \frac{\pi}{3} = 0 \\ 2Y: -f_1 \, cos \frac{\pi}{4} - f_4 \, cos \frac{\pi}{3} - f_3 = 0 \\ 3X: -f_2 + f_5 = 0 \\ 3Y: f_3 - 10000 = 0 \\ 4X: -f_5 - f_4 \, cos \frac{\pi}{6} = 0 \\ 4Y: f_4 \, sin \frac{\pi}{6} - F_3 = 0$

Solving and summing the force magnitudes gives 48966 (when taking the ceiling).

Computational Physics (Mechanical System)

The answer is 48966.

1 solution

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