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6.8: Method of Sections: Problem Solving I

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Mechanical Engineering

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Method of Sections: Problem Solving I

6.8: Method of Sections: Problem Solving I

Consider a symmetrical roof truss structure, composed of vertical, diagonal, and horizontal members. The length of each horizontal member is 4 m. The lengths of the vertical members FB and HD are 4 m, while the length of member GC is 6 m. The loads acting at joints F, G, and H are 2 kN, while those at joints A and E are 1 kN.

Figure 1

The method of sections is employed to calculate the forces acting on members DC and HC. The moment equilibrium condition is applied to point A, and the known values of forces and distances are substituted into the moment equation.

Equation 1

This results in an estimated reaction force of 4 kN at point E. Subsequently, the vertical force equilibrium condition at point A reveals that the reaction force at A is also 4 kN.

Equation 2

Due to the symmetry of the truss, the reaction forces at points A and E are equal.

The forces acting on members DC and HC can be obtained upon determining the reaction forces. A sectional cut is made along a plane intersecting members DC, HC, and HG, and a free-body diagram of the smaller section is considered.

Figure 2

By summing the moments about point H, the force along DC is calculated to be a positive 3 kN, indicating a tensile force.

Equation 3

The force along CH is resolved into its sine and cosine components. Trigonometry is used to find the angle between member CH and the horizontal axis to be 45°. Finally, applying the moment equilibrium condition at point E yields a force of -1.41 kN for FCH.

Equation 4

In this case, the negative result signifies a compressive force acting on member CH.

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Keywords: Method Of Sections Problem Solving Symmetrical Roof Truss Vertical Members Diagonal Members Horizontal Members Load Distribution Moment Equilibrium Force Equilibrium Reaction Forces Free-body Diagram Trigonometry Tensile Force Compressive Force

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