6.8:
Method of Sections: Problem Solving I
Consider a symmetrical roof truss structure comprising the vertical, diagonal and horizontal members.
A free-body diagram is drawn to analyze the forces on DC and HC members using the method of section.
Here, the loads and the lengths of the horizontal and vertical members are the known parameters.
First, summing the moments about point A, the reaction force at E is calculated.
Further, using the force equilibrium condition for the truss, the reaction force at A is determined.
The symmetry of the truss ensures that both the reaction forces are equal.
Now, a cut is made along a plane intersecting DC, HC and HG members, and a free-body diagram of the smaller section is drawn.
Taking the summation of the moments about H gives the force along DC. The positive sign indicates the tensile force.
The force along HC is resolved into its sine and cosine components, and trigonometry is used to estimate the angle theta.
The moment equilibrium condition at E yields the force on HC, with the negative sign indicating the compressive force.
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.
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.
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.
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.
By summing the moments about point H, the force along DC is calculated to be a positive 3 kN, indicating a tensile force.
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.
In this case, the negative result signifies a compressive force acting on member CH.
Suggested Reading
- R. C., Hibbeler Engineering Mechanics Statics, Pearson. Pp. 291-293
- F.P. Beer, E.R. Johnston, D.F. Mazurek, P.J. Cornwell, B.P. Self, Vector Mechanics For Engineers Statics and Dynamics Engineering Mechanics Statics, Mc Graw-Hill Education. Pp. 217