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Normal forces in members

This exercise is our introductory exercise in strength theory and deals with the following questions:

  • How to calculate the normal force in a member?
  • How to establish an equilibrium of forces with internal forces?

Task

A weight G2 is attached to a rod. The weight G1 is applied at connection point C. The members 1 and 2 are fastened with fixed bearings. The members are without mass.

The normal forces that occur in members 1 to 3 have to be calculated.

Members with loads G1 and G2
Members with loads G1 and G2

Solution

The normal forces are referred to as N and the index of their member number. The normal force in member 1 is the easiest to calculate.

Calculation of the normal force in member 1
Calculation of the normal force in member 1

The equilibrium of forces in the y-direction yields

\[ \tag{1} \sum F_y = 0 = N_1 -- G_2\]

\[ \tag{2} N_1 = G_2\]

Now, starting from point C, mentally cut through the members and apply their respective normal forces. A small trap lurks here in this task, because the force G1 must of course not be missing in this free cut.

Calculation of the normal forces in members 2 and 3
Calculation of the normal forces in members 2 and 3

The two normal forces in members 2 and 3 can now be calculated using the force equilibria in the x and y directions.

\[ \tag{3} \sum F_x = 0 = -- N_3 \cdot sin \alpha + N_2\]

\[ \tag{4} \sum F_y = 0 = N_3 \cdot cos \alpha -- G_1 -- N_1\]

or converted to N3 and the normal force N1 replaced by its value:

\[ \tag{5} N_3 = \frac{G_1 + G_2}{cos \alpha} \]

Finally, N2 follows from equation 3

\[ \tag{6} N_2 = \frac{G_1 + G_2}{cos \alpha} \cdot sin \alpha\]

or in other words:

\[ \tag{7} N_2 = (G_1+G_2) \cdot tan \alpha\]

The member lengths given in the sketch were not relevant for the calculation of the normal forces.