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Bridge Truss Force Decomposition (Civil & Mechanical Engineering) Solution

In structural engineering, joint analysis relies on right-triangle trigonometry to decompose static force vectors into axial components along support members.

A bridge truss joint (Junction A) experiences a downward gravitational load of 150 kN150\text{ kN} from traffic. Two steel struts meet at this junction: Strut 1 is horizontal, and Strut 2 forms an angle of θ=30\theta = 30^\circ with the horizontal strut.

Determine the following under static equilibrium conditions (where the vertical component of the force in Strut 2 must balance the 150 kN150\text{ kN} downward load):

  1. Draw the force triangle representing the equilibrium state.
  2. Determine the exact force in Strut 2.
  3. Determine the exact force in Strut 1.

Step-by-Step Solution:

  1. Draw the Force Triangle: The downward gravitational load vector of 150 kN150\text{ kN} forms the vertical leg of a right-angled force triangle. The axial force in Strut 2 is along the hypotenuse, and the axial force in Strut 1 is along the horizontal leg.

  2. Determine the Force in Strut 2 (F2F_2): Using the sine trigonometric ratio:

    sin(30)=OppositeHypotenuse=150F2\sin(30^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{150}{F_2}
    F2=150sin(30)=1501/2=300 kNF_2 = \frac{150}{\sin(30^\circ)} = \frac{150}{1/2} = 300\text{ kN}

  3. Determine the Force in Strut 1 (F1F_1): Using the cosine trigonometric ratio:

    cos(30)=AdjacentHypotenuse=F1F2\cos(30^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{F_1}{F_2}
    F1=F2cos(30)=30032=1503 kNF_1 = F_2 \cdot \cos(30^\circ) = 300 \cdot \frac{\sqrt{3}}{2} = 150\sqrt{3}\text{ kN}

Meaning of the Answer & Real-Life Application:

Decomposing force vectors using trigonometry is essential for sizing steel beams and joints. Underestimating the axial forces in members can lead to structural buckling and catastrophic failure of the bridge under peak traffic loads.