2019

Engineering Mechanics: Statics

Lecture Series

ST13: Truss Analysis Method of Joints

This document is a written version of video lecture ST13, which can be found

online at the web addresses listed below

Educative Technologies, LLC http://www.Lab101.Space

https://www.youtube.com/c/drstructure

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Statics – ST13 Truss Analysis: Method of Joints As a type of structural system, trusses can be found in buildings, bridges, towers, and other constructed facilities. Figure 1 shows three distinct uses of trusses.

Figure 1: Various uses of trusses in structures

We classify a structural system as a truss based on how it responds to the applied loads. For instance, a beam bends when subjected to a load, as depicted in Figure 2, resulting in the development of two internal forces: a shear force and a bending moment.

Figure 2: A beam with internal shear force and bending moment

A truss, however, responds differently to such a load. Its members displace and their lengths change, but they don’t bend the way a beam bends. Therefore, in theory, no shear force or bending moment develops in a truss member; it should carry an axial force only (see Figure 3).

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Figure 3: A two-dimensional truss and its deformed shape

In this lecture, we are going to examine the analysis of statically determinate trusses that are defined in the two-dimensional space. We shall employ the method of joints for this analysis.

As shown in Figure 4, a truss can be viewed as a network of interconnected slender members. A point at which two or more members are connected is called a joint. A truss can carry only loads that are placed at its joints.

Figure 4: A simple truss

If a load is placed directly on a member, as depicted in Figure 5, internal shear force and bending moment develop in the member. Such a member needs to be treated as a beam. Hence the structure can no longer be viewed as a truss.

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Figure 5: A hybrid structure with a beam member

We consider a truss joint to be a mechanism that allows the attached members to rotate relative to each other. However, the amount of rotation is restricted by the geometry and topology of the structure. For example, consider the simple truss shown in Figure 6. Under the applied load, the top joint is going to displace downward by a certain amount forcing the two inclined members to rotate inward.

Figure 6: Truss member rotation due to a displaced joint

Please keep in mind that although truss members change in length to accommodate a new configuration. They remain straight bars; they do not bend. That is, the deformation depicted in Figure 7 is not permissible in trusses.

Figure 7: An impermissible truss member deformation

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The purpose of truss analysis is to calculate the axial force in each member. For statically determinate trusses, we can use the method of joints to analyze the structure. This method involves first isolating the joints of the truss and then applying the equilibrium equations to each joint in order to determine the unknown member forces.

Let’s examine this method in the context of the truss shown in Figure 8.

Figure 8: A statically determinate truss

The truss consists of six joints and nine members. To simplify the process of calculating the member forces, we start by determining the support reactions. This is done by drawing the overall free-body diagram of the structure and then writing and solving the three static equilibrium equations for the unknown reactions.

Figure 9: The overall free-body diagram of a truss

The static equilibrium equations for the free-body diagram shown in Figure 9 are as follows:

x xF A 15 0= + =∑ [1]

y y yF A C 0= + =∑ [2]

@A yM (8.66 m)(15 kN) (20 m)C 0= − =∑ [3]

Solving the above equations for the unknowns, we get: xA 15 kN=− , yA 6.5 kN=− , and

yC 6.5 kN= .

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Having determined the support reactions, we are now in a position to isolate the truss joints and calculate the member forces.

How do we isolate the joints? Consider member AD. Imagine cutting and separating the member from the structure, as depicted in Figure 10.

Figure 10: The free-body diagram of a truss member

The member carries an axial force only. Since the force is not known yet, we can assume it puts the member in the tension mode. The tensile force is shown using an arrow pointing away from either end of the member.

Since member AD has been cut from the truss near joints A and D, the member’s tension force needs to appear at the other end of each cut as well. Therefore, we need to place an arrow at A pointing toward D, and an arrow at D pointing toward A, as shown in the figure above.

Since all four arrows refer to the same force, we denote their magnitude using the same symbol:ADF .

It is important to keep in mind that the arrow shown at the upper end of the truss member and the arrow shown at joint D must cancel each other out. The same is true for the force shown at the lower end of the member and the one placed at joint A. Their sum must be zero. Hence, we always draw each pair of forces in opposite directions, as shown in Figure 10.

Now consider member AB. Assuming it carries a tensile force, we can show the force acting at joints A and B, pointing away from the joint, as shown in Figure 11.

Note that both arrows are shown as having the same magnitude, ABF . This is necessary as they represent the same force: the tension force in member AB.

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Figure 11: An isolated truss joint

Using the same scheme, we can replace each remaining truss member with two arrows acting at its end joints. This results in the isolation of all the truss joints (see Figure 12).

Figure 12: Isolated joints of a truss structure

Given the above diagram, we can start calculating the unknown member forces. Since the truss is in the state of static equilibrium, the equilibrium equations must be satisfied for each truss joint. That is, for each joint we can write three equations:

xF 0=∑ [4]

yF 0=∑ [5]

M 0=∑ [6]

Note that Equation [6] is automatically satisfied. Since every force that acts at a joint passes through the joint, its moment about the joint is zero. Therefore, we are left with Equations [4] and [5] only at each truss joint. If the number of unknown forces at a joint does not exceed two, we can use these two equations to calculate the forces.

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For example, consider joint A. Four forces are acting on it, but only two of them are unknown: ADF and ABF (see Figure 12). Therefore, applying the equilibrium equations to the joint enables

us to determine the unknowns. The equilibrium equations for the joint are as follows:

x AB ADF F F cos(60) 15 0= + − =∑ [7]

y ADF F sin(60) 6.5 0= − =∑ [8]

Solving the equations for the unknowns, we get: ADF 7.5 kN= and ABF 11.25 kN= .

Let’s write these values on the force diagram (see Figure 13).

Figure 13: Results of applying the equilibrium equations to joint A of the example truss

There are seven additional unknown member forces to be determined. We can determine two of them by applying the equilibrium equations to another joint that has at most two unknown forces. Two joints qualify: D and F. Let’s use joint D. The unknowns at the joint are BDF and DEF . The resulting equilibrium equations are shown below.

x DE BDF F F cos(60) 7.5cos(60) 0= + − =∑ [9]

y BDF F sin(60) 7.5sin(60) 0=− − =∑ [10]

These equations yield: BDF 7.5 kN=− and DEF 7.5 kN= .

Note the negative value for the force magnitude in member BD. The negative sign means the direction of the force is opposite to what was assumed initially. We assumed the member was in tension. The sign tells us that BD is actually in compression. When showing the force in member BD on a free-body diagram, we can either keep the direction of the force arrow as is and write negative 7.5 kN next to it, or we can reverse the direction of the arrow and write positive 7.5 kN. Either way, we are stating that member BD is carrying a compressive force of 7.5 kN. These two alternative representations are depicted in Figure 14.

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Figure 14: The two possible ways of showing a compressive force in a truss member

For the sake of simplicity, let’s keep the direction of the force arrow as was initially assumed but place the negated force magnitude next to it. Consequently, the updated force diagram can be drawn as shown in Figure 15.

Figure 15: Results of applying the equilibrium equations to joint D of the example truss

Next, we can consider either joint B or F, since each is subjected to two unknown forces only. Let’s use joint F. The equilibrium equations for the joint are as follows:

x EF CFF 15 F F cos(60) 0= − − =∑ [11]

y CFF F sin(60) 0=− =∑ [12]

The above equations yield: CFF 0= and EFF 15 kN= . The updated force diagram is shown in Figure 16.

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Figure 16: Results of applying the equilibrium equations to joint F of the example truss

An examination of the figure above shows that only two unknown forces are present at each remaining joint. Therefore, we can use either joint B, C, or E to determine two more forces. Let’s use joint C. The unknowns at the joint are: CEF and BCF . The equilibrium equations for the joint can be written in the following manner.

x BC CEF F F cos(60) 0=− − =∑ [13]

y CEF F sin(60) 6.5 0= + =∑ [14]

Solving Equations [13] and [14] for the unknowns, we get: CEF 7.5 kN=− and BCF 3.75 kN= .

Figure 17 shows the updated force diagram.

Figure 17: Results of applying the equilibrium equations to joint F of the example truss

Now only one unknown force remains, BEF . We can use either joint B or E to determine the force. Let’s use joint E. The resulting equilibrium equations are shown below.

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x BEF 15 ( 7.5)cos(60) 7.5 F cos(60) 0= + − − − =∑ [15]

y BEF ( 7.5)sin(60) F sin(60) 0=− − − =∑ [16]

At this point, we can use either equation to determine the remaining unknown. Equation [16] yields: BEF 7.5 kN= .

This concludes our analysis of the truss as we have determined all the member forces. Let’s summarize the results by writing the force magnitudes on the members and indicate whether a member is in tension or compression.

Figure 18: The results of the truss analysis

As the above diagram shows the truss consists of two compression members (BD and CE), six tension members (AD, AB, DE, BE, BC, and EF), and one zero-force member (CF).

Exercise: Analyze the following statically determinate trusses using the method of joints.