Solved Problem on Static Equilibrium

A structural beam with 3 m long and 120 kg of mass is supported at its ends A and B and supports two loads of 12 kg and 8 kg at 1 m and 2 m, respectively, from support A. Determine the reaction forces in the supports.

Problem data:

Length of the beam: L = 3 m;
Mass of the beam: m_V = 120 kg;
Mass of the load applied in C: m_C = 12 kg;
Distance AC: d_AC = 1 m;
Mass of the load applied in D: m_D = 8 kg;
Distance AD: d_AD = 2 m;
Acceleration due to gravity: g = 9.8 m/s².

Problem diagram:

We choose a reference frame in the center of the beam, where the gravitational force \( {\vec F}_{g} \), is applied, the clockwise direction is positive (Figure 1).

Figure 1

The reaction forces due to supports are applied to points A and B (\( {\vec F}_{A} \) and \( {\vec F}_{B} \)), the loads applied to the beam are represented by the masses at points C and D (\( m_{C} \) and \( m_{D} \)).

Solution

In order for the beam to remain in equilibrium, we must have the following conditions

\[ \begin{gather} \bbox[#99CCFF,10px] {\sum F_{i}=0} \tag{I-a} \end{gather} \]

\[ \begin{gather} \bbox[#99CCFF,10px] {\sum M_{i}=0} \tag{I-b} \end{gather} \]

The loads applied to points C and D are represented by the gravitational forces of the mass placed in these positions, the gravitational force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{II} \end{gather} \]

Applying the expression (II) to m_C and m_D masses

\[ \begin{gather} F_{C}=F_{gC}=m_{C}g\\[5pt] F_{C}=12\times 9.8\\[5pt] F_{C}=117.6\;\text{N} \tag{III} \end{gather} \]

\[ \begin{gather} F_{D}=F_{gD}=m_{D}g\\[5pt] F_{D}=8\times 9.8 \\[5pt] F_{D}=78.4\;\text{N} \tag{IV} \end{gather} \]

the gravitational force of the beam will be

\[ \begin{gather} F_{g}=m_{V}g \\[5pt] F_{g}=120\times 9.8\\[5pt] F_{g}=1176\;\text{N} \tag{V} \end{gather} \]

Drawing the forces that act on the beam on a coordinate system xy (Figure 2) and applying the condition (I-a)

\[ \begin{gather} F_{A}+F_{B}-F_{C}-F_{D}-F_{g}=0 \end{gather} \]

substituting the values (III), (IV), and (V)

\[ \begin{gather} F_{A}+F_{B}-117.6-78.4-1176.0=0\\[5pt] F_{A}+F_{B}-1372=0\\[5pt] F_{A}+F_{B}=1372 \tag{VI} \end{gather} \]

Figure 2

The torque of a force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {{\Large\tau}=Fd} \tag{VII} \end{gather} \]

Torque of the gravitational force of the beam:

Applying the expression (VII), the force F is represented by the gravitational force F_g of the beam, and the distance will be zero, d = 0, the gravitational force is applied on the same point taken as a reference

\[ \begin{gather} {\Large\tau}_{{F}_{g}}=0 \tag{VIII} \end{gather} \]

Torque of reaction force in support A:

The reference point G is in the center of the beam, the distance from point A to the center will be (Figure 3)

\[ \begin{gather} d_{AG}=\frac{L}{2}\\[5pt] d_{AG}=\frac{3}{2}\\[5pt] d_{AG}=1.5\;\text{m} \tag{IX} \end{gather} \]

Figure 3

Applying the expression (VII), the F force is represented by the support reaction force at point A, F_A, and the distance will be given by the value found in (IX). The reaction force tends to make the beam rotate in the same direction of the chosen orientation, the torque will be positive

\[ \begin{gather} {\Large\tau}_{{F}_{A}}=F_{A}d_{AG}\\[5pt] {\Large\tau}_{{F}_{A}}=1.5F_{A} \tag{X} \end{gather} \]

Torque of force due to load at point C:

The distance of point C to the G point will be (Figure 4)

\[ \begin{gather} d_{AG}=d_{AC}+d_{CG}\\[5pt] 1.5=1+d_{CG}\\[5pt] d_{CG}=1.5-1\\[5pt] d_{CG}=0.5\;\text{m}\tag{XI} \end{gather} \]

Figure 4

Applying the expression (VII), the force F is represented by the gravitational force of the load applied at point C, F_C, found in the expression (III), and the distance will be given by the value found in (XI). The gravitational force tends to make the beam rotate in the opposite direction of the chosen orientation, the moment will be negative

\[ \begin{gather} {\Large\tau}_{{F}_{C}}=-F_{C}d_{CG}\\[5pt] {\Large\tau}_{{F}_{C}}=-117.6\times 0.5\\[5pt] {\Large\tau}_{{F}_{C}}=-58.8\;\text{N.m} \tag{XII} \end{gather} \]

Torque of force due to load at point D:

The distance from point D to the point G will be (Figure 5)

\[ \begin{gather} d_{AG}=d_{AD}-d_{DG}\\[5pt] 1.5=2-d_{DG}\\[5pt] d_{DG}=2-1.5\\[5pt] d_{DG}=0.5\;\text{m} \tag{XIII} \end{gather} \]

Figure 5

Applying the expression (VII), the force F is represented by the gravitational force of the load applied at point D, F_D, found in the expression (IV), and the distance will be given by the value found in (XIII). The gravitational force tends to make the beam rotate in the same direction of the chosen orientation, the moment will be positive

\[ \begin{gather} {\Large\tau}_{{F}_{D}}=F_{D}d_{DG}\\[5pt] {\Large\tau}_{{F}_{D}}=78.4\times 0.5\\[5pt] {\Large\tau}_{{F}_{D}}=39.2\;\text{N.m} \tag{XIV} \end{gather} \]

Torque of reaction force in support B:

The reference point G is in the center of the beam (Figure 6), the distance from point B to the center will be the same value found in the expression (VIII)

\[ \begin{gather} d_{AG}=d_{BG} \end{gather} \]

Figure 6

Applying the expression (VII), the force F is represented by the support reaction force at point B, F_B, and the distance will be given by the value found in (IX). The reaction force tends to make the beam rotate in the opposite direction of the chosen orientation, the moment will be negative

\[ \begin{gather} {\Large\tau}_{{F}_{B}}=-F_{B}d_{BG}\\[5pt] {\Large\tau}_{{F}_{B}}=-1.5F_{B} \tag{XV} \end{gather} \]

Applying the condition (I-b)

\[ \begin{gather} {\Large\tau}_{F_{gV}}+{\Large\tau}_{F_{A}}+{\Large\tau}_{F_{C}}+{\Large\tau}_{F_{D}}+{\Large\tau}_{P_{B}}=0 \end{gather} \]

substituting the (VIII), (X), (XII), (XIV) and (XV)

\[ \begin{gather} 0+1.5F_{A}-58.8+39.2-1.5F_{B}=0\\[5pt] 1.5F_{A}-19.6-1.5F_{B}=0\\[5pt] 1.5F_{A}-1.5F_{B}=19.6 \tag{XVI} \end{gather} \]

Expressions (VI) and (XVI) can be written as a system of two equations to two variables, F_A and F_B

\[ \begin{gather} \left\{ \begin{array}{l} \,F_{A}+F_{B}=1372\\ \,1.5F_{A}-1.5F_{B}=19.6 \end{array} \right. \end{gather} \]

solving the first equation for F_A and substituting into the second equation

\[ \begin{gather} F_{A}=1372-F_{B} \tag{XVII} \end{gather} \]

\[ \begin{gather} 1.5\,\left(1372-F_{B}\right)-1.5F_{B}=19.6\\[5pt] 1.5\times 1372-1.5F_{B}-1.5F_{B}=19.6\\[5pt] 2058-3F_{B}=19.6\\[5pt] 3F_{B}=2058-19.6\\[5pt] 3F_{B}=2038.4\\[5pt] F_{B}=\frac{2038.4}{3} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {F_{B}\approx 679.5\;\text{N}} \end{gather} \]

substituting this value in the expression (XVII)

\[ \begin{gather} F_{A}=1372-679.5 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {F_{B}\approx 692.5\;\text{N}} \end{gather} \]

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