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3.4: Computation of Support Reactions for Planar Structures


The support reactions for statically determinate and stable structures on a plane are determined by using the equations of equilibrium. The procedure for computation is outlined below.

Procedure for Computation of Support Reactions

• Sketch a free-body diagram of the structure, identifying all the unknown reactions using an arrow diagram.
• Check the stability and determinacy of the structure using equation 3.3 or 3.4. If the structure is classified as determinate, proceed with the analysis.
• Determine the unknown reactions by applying the three equations of equilibrium. If a computed reaction results in a negative answer, the initially assumed direction of the unknown reaction, as indicated by the arrow head on the free-body diagram, is wrong and should be corrected to show the opposite direction. Once the correction is made, the magnitude of the force should be indicated as a positive number in the corrected arrow head on the free-body diagram

Example 3.3

A cantilever beam is subjected to a uniformly distributed load and an inclined concentrated load, as shown in figure 3.9a. Determine the reactions at support A.

$$Fig. 3.9$$. Beam

Solution

Free-body diagram. The free-body diagram of the entire beam is shown in Figure 3.9b. The support reactions, as indicated in the free-body diagram, are $$A_{y \prime}$$ $$A_{y \prime}$$ and $$M$$.

Computation of reactions. Prior to the computation of the support reactions, the distributed loading should be replaced by a single resultant force, and the inclined loading resolved to the vertical and horizontal components. The magnitude of the resultant force is equal to the area under the rectangular loading, and it acts through the centroid of the rectangle. As seen in Figure 3.9c, $$P=[(4 \mathrm{kN} / \mathrm{m})(2 \mathrm{m})]$$, and its location is at the centroid of the rectangle loading $$=\left[\left(\frac{1}{2}\right)(2 \mathrm{m})\right]$$. Applying the equations of static equilibrium provides the following:

$$\begin{array}{l} \curvearrowleft+\sum M_{A}=0 \\ -\left(16 \sin 75^{\circ}\right)(8)-(4 \times 2)(1)+M_{A}=0 \\ M_{A}=131.64 \mathrm{kN} \cdot \mathrm{m} \quad M_{A}=131.64 \mathrm{kN} \cdot \mathrm{m} \curvearrowleft \end{array}$$

$$\begin{array}{l} A_{y}-16 \sin 75^{\circ}-(4 \times 2)=0 \\ A_{y}=23.45 \mathrm{kN} \\ \rightarrow+\sum \mathrm{F}_{\mathrm{x}}=0 \\ \mathrm{A}_{\mathrm{x}}=0 \end{array} \quad \begin{array}{c} A_{y}=23.45 \mathrm{kN} \uparrow \\ \mathrm{A}_{\mathrm{x}}=0 \end{array}$$

Example 3.4

A 12ft-long simple beam carries a uniformly distributed load of 2 kips/ft over its entire span and a concentrated load of 8 kips at its midspan, as shown in Figure 3.10a. Determine the reactions at the supports $$A$$ and $$B$$ of the beam.

$$Fig. 3.10$$. Simple beam.

Solution

Free-body diagram. The free-body diagram of the entire beam is shown in Figure 3.10b.

Computation of reactions. The distributed loading is first replaced with a single resultant force, as seen in Figure 3.10c. The magnitude of the resultant force is equal to the area of the rectangular loading (distributed force). Thus, $$P=[(2 \mathrm{k} / \mathrm{ft})(12 \mathrm{ft})]$$, and its location is at the centroid of the rectangular loading $$=\left[\left(\frac{1}{2}\right)(12 \mathrm{ft})\right]$$ Since there is a symmetry in loading in this example, the reactions at both ends of the beam are equal, and they could be determined using the equations of static equilibrium and the principle of superposition, as follows:

$$\begin{array}{l} +\uparrow \sum F_{y}=0 \\ A_{y}=B_{y}=\left(\frac{2 \times 12}{2}\right)+\frac{2}{2}=13 \mathrm{kips} \\ +\rightarrow \sum \mathrm{F}_{\mathrm{x}}=0 \\ \mathrm{A}_{\mathrm{x}}=0 \end{array} \quad \begin{array}{l} A_{y}=B_{y}=13 \mathrm{kips} \uparrow \\ \mathrm{A}_{\mathrm{x}}=0 \end{array}$$

Example 3.5

A beam with an overhang is subjected to a varying load, as shown in Figure 3.11a. Determine the reactions at supports $$A$$ and $$B$$.

$$Fig. 3.11$$. Beam with an overhang.

Solution

Free-body diagram. The free-body diagram of the entire beam is shown in Figure 3.11b.

Computation of reactions. Observe that the distributed loading in the beam is triangular. The distributed load is first replaced with a single resultant force, as shown in Figure 3.11c. The magnitude of the single resultant force is equal to the area under the triangular loading. Thus, $$\rho=\left(\frac{1}{2}\right)(6 \mathrm{m})(10 \mathrm{kN} / \mathrm{m})$$, and its centroid is at the center of the loading (6m). Applying the equations of equilibrium provides the following:

$$\begin{array}{l} \curvearrowleft+\sum M_{A}=0 \\ -\left(\frac{1}{2}\right)(10)(6)(3)+3 B=0 \\ B_{y}=30 \mathrm{kN} \quad \quad B_{y}=30 \mathrm{kN} \\ \uparrow+\sum F_{y}=0 \\ 30+A_{y}-\left(\frac{1}{2}\right)(6)(10)=0 \quad \mathrm{A}_{\mathrm{y}}=0\\ \rightarrow+\sum \mathrm{F}_{\mathrm{x}}=0 \\ \mathrm{A}_{\mathrm{x}}=0 \quad \quad \mathrm{A}_{\mathrm{x}}=0 \end{array}$$

Example 3.6

A beam with overhanging ends supports three concentrated loads of 12 kips, 14 kips, and 16 kips and a moment of 100 kips.ft, as shown in Figure 3.12a. Determine the reactions at supports A and B.

$$Fig. 3.12$$. Beam with overhanging ends.

Solution

Free-body diagram. The free-body diagram of the beam is shown in Figure 3.12b.

Computation of reactions. Applying the equations of equilibrium provides the following:

$$\begin{array}{l} +\curvearrowleft M_{A}=0 \\ -100+12(2)-14(2)-16(8)+4 B_{y}=0 \\ B_{y}=58 \text { kips } \quad \quad B_{y}=58 \text { kips } \uparrow \\ +\uparrow \sum F_{y}=0 \\ 58+A_{y}-12-14-16=0 \quad A_{y} = 26 \text { kips } \uparrow\\ +\rightarrow \sum F_{x}=0 \\ A_{x}=0 \quad \quad A_{x} \end{array}$$

Example 3.7

A compound beam is subjected to the loads shown in Figure 3.13a. Find the support reactions at $$A$$ and $$B$$ of the beam.

$$Fig. 3.13$$. Compound beam.

Solution

Free-body diagram. The free-body diagram of the entire beam is shown in Figure 3.13b.

Identification of primary and complimentary structures. For correct analysis of a compound structure, the primary and the complimentary parts of the structure should be identified for proper understanding of their interaction. The interaction of these parts are shown in Figure 3.13c. The primary structure is the part of the compound structure that can sustain the applied external load without the assistance of the complimentary structure. On the other hand, the complimentary structure is the part of the compound structure that depends on the primary structure to support the applied external load. For the given structure, part $$AC$$ is the primary structure, while part $$CB$$ is the complimentary structure.

Computation of reactions. The analysis of a compound structure must always begin with the analysis of the complimentary structure, as the complimentary structure is supported by the primary structure. Using the equations of equilibrium, the support reactions of the beam are determined as follows:

Analysis of the complimentary structure $$CB$$.

Computation of support reaction. The isolated free-body diagram of the complimentary structure is shown in Figure 3.13c. First, the distributed loading is replaced by a single resultant force ($$P$$), which is equal to the area of the rectangular loading, as shown in Figure 3.13d and Figure 3.13e. Applying the equations of equilibrium, and noting that due to symmetry in loading, the support reactions at point C and B are equal in magnitude, provides the following:

$$\begin{array}{l} +\uparrow \sum F_{y}=0 \\ B_{y}=C_{y}=\frac{10(7)}{2}=35 \mathrm{kN} \quad B_{y}=\mathrm{C}_{y}=\frac{10(7)}{2}=35 \mathrm{k} \end{array}$$

Analysis of the primary structure $$AC$$.

Computation of support reaction. Note that prior to the computation of the reactions, the reaction at point $$C$$ in the complimentary structure is applied to the primary structure as a load. The magnitude of the applied load is the same as that of the complimentary structure, but it is opposite in direction. Applying the equations of equilibrium suggests the following: $$\begin{array}{l} +\curvearrowleft \sum M_{A}=0 \\ -25(2)-35(4)+M_{A}=0 \\ M_{A}=190 \mathrm{kN} \cdot \mathrm{m} \quad M_{A}=190 \mathrm{kN} \cdot \mathrm{m} \curvearrowleft \\ +\uparrow \sum F_{y}=0 \\ \mathrm{A}_{\mathrm{y}}-25-35=0 \\ \mathrm{A}_{\mathrm{y}}=60 \mathrm{kN} \quad \mathrm{A}_{\mathrm{y}}=60 \mathrm{kN} \uparrow\\ +\rightarrow \sum \mathrm{F}_{\mathrm{x}}=0 \\ \mathrm{A}_{\mathrm{x}}=0 \quad \mathrm{A}_{\mathrm{x}}=0 \end{array}$$

Example 3.8

Find the reactions at supports $$A$$, $$C$$, and $$E$$ of the compound beam carrying a uniformly distributed load of 10 kips/ft over its entire length as shown in figure 3.14a.

Fig. 3.14. Compound beam.

Solution

Free-body diagram. The free-body diagram of the entire beam is shown in Figure 3.14b.

Identification of primary and complimentary structures. The interaction diagram for the given structure is shown in Figure 3.14c. AB is the primary structure, while BD and DE are the complimentary structures.

Computation of reactions.

Analysis of complimentary structure $$DE$$.

Computation of support reaction. The isolated free-body diagram is shown in Figure 3.14c. First, the distributed loading is replaced by a single resultant force ($$P$$) equal the area of rectangular loading, as shown in Figure 3.14d. Applying the equations of equilibrium, and noting that due to symmetry in loading, the support reactions at point $$D$$ and $$E$$ are equal in magnitude, suggests the following:

$$\begin{array}{l} +\uparrow \sum F_{y}=0 \\ D_{y}=E_{y}=\frac{10(8)}{2}=40 \mathrm{kips} \quad E_{y}=40 \mathrm{kips} \uparrow \end{array}$$

Analysis of complimentary structure $$BD$$.

Computation of support reaction. The isolated free-body diagram is shown in Figure 3.14e. First, the distributed loading is replaced by a single resultant force ($$P$$) equal to the area of the rectangular loading, as shown in Figure 3.14f. The load from the complimentary structure is applied at point $$D$$. Applying the equations of equilibrium suggests the following:

$$\begin{array}{l} +\curvearrowleft \sum M_{B}=0 \\ -10(8)\left(\frac{8}{2}\right)-40(8)+4 C_{y}=0 \\ C_{y}=160 \text { kips } \quad C_{y}=160 \text { kips } \uparrow \\ +\uparrow \sum F_{y}=0 \\ 160-B_{y}-10(8)-40=0 \\ B_{y}=40 \text { kips } \end{array}$$

Analysis of primary structure $$AB$$.

Computation of support reaction. Note that prior to the computation of the reactions, the uniform load is replaced by a single resultant force, and the reaction at point $$B$$ in the complimentary structure is applied to the primary structure as a load. Applying the equilibrium requirement yields the following:

$$\begin{array}{l} \quad+\curvearrowleft \sum M_{A}=0 \\ \mathrm{M}-10(10)\left(\frac{10}{2}\right)+40(10)=0 \\ M_{A}=100 \mathrm{kips} . \mathrm{ft} \quad \quad M_{A}=100 \mathrm{kips} . \mathrm{ft} \curvearrowleft \\ +\uparrow \sum F_{y}=0 \\ A_{y}-10(10)+40=0 \\ A_{\mathrm{y}}=60 \mathrm{kips} \quad \quad A_{\mathrm{y}}=60 \mathrm{kips} \uparrow\\ +\rightarrow \sum \mathrm{F}_{\mathrm{x}}=0 \\ A_{x}=0 \quad \quad A_{x}=0 \end{array}$$

Example 3.9

Find the reactions at supports $$A$$, $$B$$, $$E$$, and $$F$$ of the loaded compound beam, as shown in Figure 3.15a.

$$Fig. 3.15$$. Compound beam.

Solution

Free-body diagram. The free-body diagram of the entire beam is shown in Figure 3.15b.

Identification of primary and complimentary structure. The interaction diagram for the given structure is shown in Figure 3.15c. $$CD$$ is the complimentary structure, while $$AC$$ and $$DF$$ are the primary structures.

Computation of reactions.

Analysis of complimentary structure $$CD$$.

Computation of support reaction. The isolated free-body diagram is shown in Figure 3.15c. First, the distributed loading is replaced by a single resultant force ($$P$$), which is equal to the area of the rectangular loading, as shown in Figure 3.15d. Applying the equations of equilibrium, and noting that due to symmetry in loading, the support reactions at point $$C$$ and $$D$$ are equal in magnitude, suggests the following:

$$\begin{array}{l} +\uparrow \sum F_{y}=0 \\ C_{y}=D_{y}=\frac{10(4)}{2}=20 \mathrm{kN} \end{array}$$

Analysis of primary structure $$AC$$.

Computation of support reaction. Note that the reaction at $$AC$$ of the complimentary structure is applied as a downward force of the same magnitude at the same point on the primary structure. Applying the equation of equilibrium suggests the following:

$$\begin{array}{l} +\curvearrowleft \sum M_{A}=0 \\ -24(2)-20(5.5)+4 B_{y}=0 \\ B_{y}=39.5 \mathrm{kN} \quad \quad B_{y}=39.5 \mathrm{kN} \uparrow\\ +\uparrow \sum F_{y}=0 \\ \mathrm{A}_{y}+39.5-24-20=0 \\ A_{y}=4.5 \mathrm{kN} \quad \quad A_{y}=4.5 \mathrm{kN} \uparrow\\ +\rightarrow \sum \mathrm{F}_{\mathrm{x}}=0 \\ A_{x}=0 \quad \quad A_{x} = 0 \end{array}$$

Analysis of primary structure $$DF$$.

Computation of support reaction. The isolated free-body diagram is shown in Figure 3.15f. First, the distributed loading is replaced by a single resultant force ($$P$$) equal to the area of the triangular loading, as shown in Figure 3.15g. Applying the equations of equilibrium, and noting that the support reaction at point $$D$$ of the complimentary structure is applied as a load on the primary structure, suggests the following:

$$\begin{array}{l} +\curvearrowleft \sum M_{F} =0 \\ -20+\left(\frac{1}{2} \times 8 \times 10\right)\left(\frac{2}{3} \times 8\right)+20(8)-4 E_{y}=0 \\ E_{y}=88.33 \mathrm{kN} \quad E_{y}=88.33 \mathrm{kN} \uparrow \\ +\uparrow \sum F_{y}=0 \\ F_{y}+88.33-\left(\frac{1}{2} \times 8 \times 10\right)-20=0 \\ F_{y}=28.33 \mathrm{kN} \quad F_{y}=28.33 \mathrm{kN}uparrow \end{array}$$

Example 3.10

Determine the reactions at supports A and D of the frame shown in Figure 3.16a.

$$Fig. 3.16$$. Frame.

Solution

Free-body diagram. The free-body diagram of the entire beam is shown in Figure 3.16b.

Computation of reactions. The distributed loads in column $$AB$$ and beam $$BC$$ are first replaced by single resultant forces determined as the area of their respective shade of loading, as shown in Figure 3.16c. Applying the conditions of equilibrium suggests the following:

$$\begin{array}{l} +\curvearrowleft \sum M_{A}=0 \\ D_{y}(6)-\left(\frac{1}{2}\right)(6)(3)(3)-(2)(8)(4)=0 \\ D_{y}=15.7 \text { kips } \quad D_{y}=15.7 \mathrm{kips} \uparrow \\ +\uparrow \sum F_{y}=0 \\ \mathrm{A}_{y}+15.17-3(6)=0 \\ A_{y}=2.830 \mathrm{kips} \quad A_{y}=2.830 \mathrm{kips} \uparrow \\ +\rightarrow \sum \mathrm{F}_{\mathrm{x}}=0 \\ -\mathrm{A}_{x}+(2 \times 8)=0 \\ A_{x}=16 \mathrm{kips} \quad A_{x}=16 \mathrm{kips} \leftarrow \end{array}$$

Example 3.11

A rigid frame is loaded as shown in Figure 3.17a. Determine the reactions at support $$D$$.

$$Fig. 3.17$$. Rigid frame.

Solution

Free-body diagram. The free-body diagram of the entire beam is shown in Figure 3.17b.

Computation of reactions. The distributed load in portion AB of the frame is first replaced with a single resultant force, as shown in Figure 3.17c. Applying the equations of equilibrium suggests the following:

\begin{aligned} &+\curvearrowleft \sum M_{D}=0\\ &-M_{D}-16(8)+(4 \times 14)\left(\frac{14}{2}\right)-10(10)=0\\ &M_{D}=164 \text { kips. } \mathrm{ft} \quad M_{D}=164 \text { kips. } \mathrm{ft} \curvearrowright \mathrm{A}\\ &+\uparrow \sum F_{y}=0\\ &D_{y}-4(14)-10=0\\ &D_{y}=66 \text { kips } \quad D_{y}=66 \text { kips } \uparrow\\ &+\rightarrow \sum \mathrm{F}_{\mathrm{x}}=0\\ &-D_{x}+16=0\\ &D_{x}=16 \text { kips } \quad D_{x}=16 \text { kips } \leftarrow \end{aligned}

Example 3.12

Find the reactions at supports E and F of the frame shown in Figure 3.18a.

$$Fig. 3.18$$. Frame.

Solution

Free-body diagram. The free-body diagram of the frame is shown in Figure 3.18b.

Computation of reactions. The distributed loads are first replaced with single resultant forces, as shown in Figure 3.18c. Applying the equations of static equilibrium suggests the following:

$$\begin{array}{l} +\curvearrowleft \sum M_{E}=0 \\ \left(\frac{1}{2} \times 4 \times 8\right)\left(\frac{1}{3} \times 4\right)-(4 \times 7)\left(\frac{7}{2}\right)-\left(\frac{1}{2} \times 4 \times 8\right)\left(\frac{7}{2}+\frac{1}{3} \times 4\right)+7 F_{y}=0 \\ F_{y}=22 \mathrm{kN} \quad F_{y}=22 \mathrm{kN} \uparrow \\ +\uparrow \sum F_{y}=0 \\ E_{y}+22-2\left(\frac{1}{2} \times 4 \times 8\right)-4(7)=0 \\ E_{y}=38 \mathrm{kN} \quad E_{y}=38 \mathrm{kN} \uparrow \\ +\rightarrow \sum \mathrm{F}_{\mathrm{x}}=0 \\ E_{x}=0 \quad E_{x}=0 \end{array}$$

Example 3.13

Determine the reactions at support A of the rigid frame shown in Figure 3.19a.

$$Fig. 3.19$$. Rigid frame.

Solution

Free-body diagram. The free-body diagram of the frame is shown in Figure 3.19b.

Computation of reactions. The distributed load in column $$AB$$ is first replaced with a single resultant force, as shown in Figure 3.19c. Applying the equations of static equilibrium suggests the following:

$$\begin{array}{l} +\curvearrowleft \sum M_{A}=0 \\ -M_{A}-20(3)-\left(\frac{1}{2} \times 10 \times 10\right)\left(\frac{1}{3} \times 10\right)=0 \\ M_{A}=-226.67 \mathrm{kN} \cdot \mathrm{m} \quad \quad M_{A}=226.67 \mathrm{kN} \cdot \mathrm{m} \curvearrowleft \\ +\uparrow \sum F_{y}=0 \\ A_{y}-20=0 \\ A_{y}=20 \mathrm{kN} \quad \quad A_{y}=20 \mathrm{kN} \uparrow\\ +\rightarrow \sum F_{x}=0 \\ -A_{x}+\left(\frac{1}{2} \times 10 \times 10\right)=0 \\ A_{x}=50 \mathrm{kN} \quad \quad A_{x}=50 \mathrm{kN} \leftarrow \end{array}$$

Example 3.14

Determine the reactions at supports $$A$$ and $$E$$ of the frame hinged at $$C$$, as shown in Figure 3.20a.

$$Fig. 3.20$$. Frame.

Solution

Free-body diagram. The free-body diagram of the frame is shown in Figure 3.20b.

Computation of reactions. The reactions in a compound frame are computed considering the freebody diagrams of both the entire frame and part of the frame. Prior to computation of the reactions, the distributed load in the column is replaced by a single resultant force. The vertical reactions at $$E$$ and $$A$$ and the horizontal reactions at $$A$$ are found by applying the equations of static equilibrium and considering the free-body diagram of the entire frame. The horizontal reaction at $$E$$ is found by considering part $$CDE$$ of the free-body diagram.

$$\begin{array}{l} +\curvearrowleft \sum M_{A}=0 \\ 8 E_{y}-(2 \times 10)\left(\frac{10}{2}\right)-10(4)=0 \\ E_{y}=17.5 \text { kips } \quad \quad E_{y}=17.5 \mathrm{kips} \uparrow \\ +\uparrow \sum F_{y}=0 \\ A_{y}+17.5-10=0 \\ A_{y}=-7.5 \text { kips } \quad \quad A_{y}=7.5 \text { kips } \downarrow \end{array}$$

The negative sign implies that the originally assumed direction of $$A_{y}$$ was not correct. Therefore, $$A_{y}$$ acts downward instead of upward as was initially assumed. This should be corrected in the subsequent analysis.

To determine $$E_{x}$$,consider the moment of forces in member $$CDE$$ about the hinge.

$$\begin{array}{l} \curvearrowleft +\sum M_{C}=0 \\ 17.5(4)-10 E_{x}=0 \\ E_{x}=7 \text { kips } \quad \quad E_{x}=7 \text { kips } \leftarrow \\ +\rightarrow \sum F_{x}=0 \\ -A_{x}-7+2 \times 10=0 \\ A_{x}=13 \text { kips } \quad \quad A_{x}=13 \text { kips } \leftarrow \end{array}$$

Example 3.15

Find the reactions at support $$A$$ and $$B$$ of the loaded frame in Figure 3.21a. The frame is hinged at $$D$$.

$$Fig. 3.21$$. Loaded frame.

Solution

Free-body diagram. The free-body diagram of the frame is shown in Figure 3.21b.

Computation of reactions. The distributed load in column $$AC$$ is first replaced with a single resultant force by finding the area of loading, as shown in Figure 3.21Figurec. The reaction at $$B$$ is computed by taking the moment of the forces in part $$DB$$ of the frame about the pin at $$D$$, and other reactions are determined by applying other conditions of equilibrium.

\begin{aligned} &+\curvearrowleft \sum M_{D}=0\\ &B_{y}(0)-15(4)=0\\ &B_{y}=0\\ &+n \sum M_{A}=0\\ &M_{A}+6 \times 0-\left(\frac{1}{2} \times 8 \times 20\right)\left(\frac{1}{3} \times 8\right)-12(3)+15(4)=0\\ &M_{A}=189.33 \mathrm{kN} \cdot \mathrm{m} \quad M_{A}=189.33 \mathrm{kN} \cdot \mathrm{m} \curvearrowright\\ &+\uparrow \sum F_{y}=0\\ &A_{y}+0-12=0\\ &A_{y}=-12 \mathrm{kN} \quad A_{y}=12 \downarrow \end{aligned}

The negative sign implies that the originally assumed direction of Ay was not correct. Therefore, Ay acts downward instead of upward as was initially assumed. This should be corrected in the subsequent analysis.

$$\begin{array}{l} +\rightarrow \sum F_{x}=0 \\ -A_{x}-15+\left(\frac{1}{2} \times 8 \times 20\right)=0 \\ A_{x}=65 \mathrm{kN} \quad \quad A_{x}=65 \mathrm{kN} \rightarrow \end{array}$$

Chapter Summary

Conditions of static equilibrium: A structure is in a state of static equilibrium if the resultant of all the forces and moments acting on it is equal to zero. Mathematically, this is expressed as follows:

$$\Sigma F=0$$ $$\Sigma M=0$$

For a body in a plane, there are the following three equations of equilibrium:

$$\Sigma F_{X}=0$$ $$\Sigma F_{Y}=0$$∑ $$\Sigma M_{0}=0$$

Types of support: Various symbolic representations are used to model different types of supports for structures. A roller is used to model a support that prevents a vertical movement of a structure but allows a horizontal translation and rotation. A pin is used to model a support that prevents horizontal and vertical movements but allows rotation. A fixed support models a support that prevents horizontal and vertical movements and rotation.

Determinacy, indeterminacy, and stability of structures: A structure is determinate if the number of unknown reactions is equal to the number of static equilibrium. Thus, the equations of static equilibrium are enough for the determination of the supports for such a structure. On the other hand, a statically indeterminate structure is a structure that has the number of the unknown reactions in excess of the equations of equilibrium. For the analysis of an indeterminate structure additional equations are needed, and these equations can be obtained by considering the compatibility of the structure. Indeterminate structures are sometimes necessary when there is a need to reduce the sizes of members or to increase the stiffness of members. A stable structure is one which has support reactions that are not parallel or concurrent to one another. The formulation of stability and determinacy of beams and frames are as follows:

 Beams and frames: $$3 m+r<3 j+C$$ Structure is unstable $$3 m+r=3 j+C$$ Structure is determinate $$3 m+r>3 j+C$$Structure is indeterminate

Practice Problems

3.1 Classify the structures shown in Figure P3.1a to Figure P3.1p as statically determinate or indeterminate, and statically stable or unstable. If indeterminate, state the degree of indeterminacy.

$$Fig. P3.1$$. Structure classification.

3.2. Determine the support reactions for the beams shown in Figure P3.2 through Figure P3.12.

$$Fig. P3.2$$. Beam.

$$Fig. P3.3$$. Beam.

$$Fig. P3.4$$. Beam.

$$Fig. P3.5$$. Beam.

$$Fig. P3.6$$. Beam.

$$Fig. P3.7$$. Beam.

$$Fig. P3.8$$. Beam.

$$Fig. P3.9$$. Beam.

$$Fig. P3.10$$. Beam.

$$Fig. P3.11$$. Beam.

$$Fig. P3.12$$. Beam.

3.3. Determine the support reactions for the frames shown in Figure P3.13 through Figure P3.20.

$$Fig. P3.13$$. Frame.

$$Fig. P3.14$$. Frame.

$$Fig. P3.15$$. Frame.

$$Fig. P3.16$$. Frame.

$$Fig. 3.17$$. Frame.

$$Fig. 3.18$$. Frame.

$$Fig. 3.19$$. Frame.

$$Fig. 3.20$$. Frame.

3.4 Determine the support reactions for the trusses shown in Figure P3.21 through Figure P3.27.

$$Fig. P3.21$$. Truss.

$$Fig. P3.22$$. Truss.

$$Fig. P3.23$$. Truss.

$$Fig. P3.24$$. Truss.

$$Fig. P3.25$$. Truss.

$$Fig. P3.26$$. Truss.

$$Fig. P3.27$$. Truss.

This page titled 3.4: Computation of Support Reactions for Planar Structures is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by René Alderliesten (TU Delft Open) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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