Analysis of Structures - Simple Truss 

? Maplesoft, a division of Waterloo Maple Inc., 2008 

Introduction 

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The steps in the document can be repeated to solve similar problems. 

 

Problem Statement 

A truss pinned at point A and free at point J is displayed to the right in Figure 1. A vertical force of 5 kN is applied at F and a horizontal force is applied at I, which is midway between points G and J.     Using the method of joints, determine the internal forces in each member and the reaction forces at A and J.

Figure 1.0

Solution 

Method 1: Method of Joints - Classical Method 

Step	Result
To obtain numeric solutions for the forces in the members, start at a joint with at most two unknown forces and at least one known force. To obtain such a joint in this problem, it is first necessary to find the reactions at the supports.     Once the internal force in a member is determined, use its value and apply it to adjacent joints to find the forces in adjacent members.    Figure 1.1 shows a free body diagram of the entire structure as a rigid body. This will be used to determine the reaction forces at A and J.	Figure 1.1
There are two unknown vertical forces, namely, and . To eliminate one of these, moments must be taken either about A or J.     A moment is a "turning force," the product of a force and its perpendicular distance from a point. Since the system is in equilibrium, the law of moments must apply; i.e., the sum of all moments about any point must be equal to zero.
Sum the moments about leads and solve for .    Use the underscore ( _ ) to move the cursor to the subscript position, and the right arrow (→) to move back to the baseline. For example, to enter , type [A][_][y], then press the right arrow to move out of the subscript.    Right-click the expression and select Solve > Obtain Solutions for > .	(3.1.1)     (3.1.2)
Sum the forces in the -direction and solve for .    Reference the solution from the previous step with an equation label. Press [Ctrl][L], then enter the appropriate reference equation number.    Right-click the expression and select Solve > Obtain Solutions for > .	(3.1.3)     (3.1.4)
Sum the forces in the -direction and solve for .    Right-click the expression and select Solve > Obtain Solutions for > .	(3.1.5)     (3.1.6)
Now that the reaction forces at joint A have been obtained, the member forces, j and , can now be found since there are now only two unknowns and at least one known force acting on the joint. The forces acting on joint A are shown in Figure 1.2. (Known forces are shown in green, while unknown forces are shown in black.)    Member AB pushes on the joint and is therefore under compression. Member AC pulls on the joint and is therefore under tension.    To solve these new equilibrium equations, reference the reaction forces which were found in the previous step.	Figure 1.2
Sum the forces in the -direction on joint and solve for .    To solve the equation, repeat the instructions taken in previous steps to find a solution for the equation.    Obtain π from the Greek palette.	(3.1.7)     (3.1.8)
Sum the forces in the -direction on joint and solve for .	(3.1.9)     (3.1.10)
All the forces acting on joint B are shown to the right in Figure 1.3. Considering the forces acting in the x-direction only, one can see that .  Considering the forces acting in the y-direction only, one can see that .    Like member AB, member BD is under compression. Member BC is a zero force member and is not under compression or tension.	Figure 1.3
Sum the forces in the -direction and solve for .	(3.1.11)     (3.1.12)
All the forces acting on joint D are shown to the right in Figure 1.4.      Member DE is under compression while member DC is under tension.	Figure 1.4
Sum the forces in the -direction and solve for .	(3.1.13)     (3.1.14)
Sum the forces in the -direction and solve for .	(3.1.15)     (3.1.16)
All the forces acting on joint C are shown to the right in Figure 1.5.      Member CE is under compression while member CF is under tension.	Figure 1.5
Sum the forces in the -direction and solve for .	(3.1.17)     (3.1.18)
Sum the forces in the -direction and solve for .	(3.1.19)     (3.1.20)
All the forces acting on joint F are shown to the right in Figure 1.6.      Analyzing the horizontal and vertical components of the forces gives = 5 kN and = 6.50 kN.    Both members EF  and FH are under tension.	Figure 1.6
Sum the vertical forces and solve for .	(3.1.21)     (3.1.22)
Sum the horizontal forces and solve for .	(3.1.23)     (3.1.24)
All the forces acting on joint F are shown to the right in Figure 1.7.	Figure 1.7
Sum the vertical forces and solve for .    The negative value obtained for indicates that this force acts opposite to the direction shown in Figure 1.7.      Consequently, both members EH and EG are under compression.	(3.1.25)     (3.1.26)     (3.1.27)
Sum the horizontal forces and solve for .	(3.1.28)     (3.1.29)
As shown in Figure 1.8, there is one known force and two unknown forces acting on joint G.     The member GI is under compression, whereas the member GH is under tension.	Figure 1.8
Sum the horizontal forces and solve for .	(3.1.30)     (3.1.31)
Sum the vertical forces and solve for .	(3.1.32)     (3.1.33)
The join H contains three known forces and two unknown forces, displayed in Figure 1.9 to the right.    Both members, HI and HJ, are a tension force.	Figure 1.9
Sum the vertical forces and solve for .	(3.1.34)     (3.1.35)
Sum the horizontal forces and solve for .	(3.1.36)     (3.1.37)
Figure 1.10 shows joints I and J with their respective forces acting on them. Analyzing either the vertical or horizontal forces acting on either joint, J or I, will produce the internal force acting on member IJ.     The joint not used to find the force  can be used  as a check on the calculations.	Figure 1.10
Sum the horizontal forces acting on joint and solve for .	(3.1.38)     (3.1.39)
There are no unknown forces acting in the vertical direction at joint .  Hence, the sum of all vertical forces at this joint should be approximately zero.	(3.1.40)     (3.1.41)
Joint can be used as a check on these calculations. The sum of horizontal forces should be zero.	(3.1.42)     (3.1.43)
At joint , the sum of vertical forces should be zero.	(3.1.44)

 

Method 2: Method of Joints - Solving a Set of Equations Simultaneously 

Step	Result
The method of joints is essentially an iterative procedure whereby one starts at a solvable joint, solves the equilibrium equations at that joint and continuously substitutes the joint's solutions into the equilibrium equations of neighboring joints.	Figure 2.0
Figure 2.1 shows the free body diagram of the entire truss support. From this diagram we can obtain 3 equilibrium equations: one from the principle of moments (i.e., the sum of moments about any point is zero if the system is in equilibrium), one from the sum of horizontal forces and one from the sum of vertical forces.
Taking moments about J:	(3.2.1)
Sum of vertical forces = 0:	(3.2.2)
Sum of horizontal forces = 0:	(3.2.3)
Figure 2.2    Figure 2.2 shows each joint idealized as a particle. All the forces acting in the truss system are included. The applied loads are 5 kN and 2 kN. The reaction forces are , and . The internal forces are of the form, , where P and Q are two connected joints.    All the internal forces in Figure 2.2 are represented as tensile forces (they are all pulling away from the joint). Since it is difficult at this point to guess which members are under compression or tension, it will be assumed that all the members are under tension. If a negative answer is obtained for an internal member this will mean that that member is under compression. If a positive answer is obtained, this will mean that the assumption was true and the member is under tension.    From Figure 2.2, twenty equilibrium equations can be obtained, two from each joint. These equations are obtained below, and are color-coded to match Figure 2.2.
Joint :	(3.2.4)     (3.2.5)
Joint :	(3.2.6)       (3.2.7)
Joint :	(3.2.8)     (3.2.9)
Joint D:	(3.2.10)     (3.2.11)
Joint :	(3.2.12)     (3.2.13)
Joint :	(3.2.14)     (3.2.15)
Joint :	(3.2.16)     (3.2.17)
Joint :	(3.2.18)     (3.2.19)
Joint :	(3.2.20)     (3.2.21)
Joint :	(3.2.22)     (3.2.23)
There is now a set of 23 equations in 20 unknowns. Recall the 20 equations are from the two equilibrium equations of each joint. The additional three equations are from the moments about J, and summing the horizontal and vertical forces. If the equations are consistent, they can be solved.     Form a sequence of these 23 equations, then solve.    Use equation labels to reference each equation, and solve by right-clicking the set and selecting Solve > Solve.        A negative value for a force indicates that the direction assumed for it is incorrect; the actual direction of the force is in the opposite direction.     Solution to the truss system:    (to the left)  (up)  (up)    (tension)    ⇒ (compression)  (tension)  ⇒ (compression)  (tension)  ⇒ (compression)  ⇒ (compression)  ⇒ (compression)  (tension)  (tension)  ⇒ (compression)   (tension)  (tension)  ⇒   (compression)  (tension)  ⇒ (compression)	(3.2.24)     (3.2.25)

Method 3: Method of Joints - Using Matrices 

Step	Result
The method of joints is essentially an iterative procedure whereby one starts at a solvable joint and solves the equilibrium equations at that joint, propagating the solution to the equilibrium equations of neighboring joints.    If all the equilibrium equations of all the joints can be defined from the start, Maple can solve the system as a single entity. This approach is shown in Solution 2, above.    Alternatively, the set of 23 equations in 20 unknowns can be cast into a matrix-vector form. Matrix operations are computationally more efficient than solving large numbers of equations.
To use matrices, start by loading the Student - Linear Algebra package.    To do this click, Tools > Load Package > Student Linear Algebra.	Loading Student:-LinearAlgebra
Next, define the system as a list of equations.     A list must be enclosed in square brackets, [ and ].     Since all the equations were defined under common assumptions in Solution 2, there is no need to define the equations of equilibrium and the single principle of moments equation again. Simply reference all these equations using the correct reference labels.     Use an equation label to reference the correct equation.	(3.3.1)
Next define a list of variables. There are 20 unknowns in the system.    Recall, a list must be enclosed in a pair of square brackets.	(3.3.2)
Create an augmented matrix from the list of equations and variables.     To do this, use the GenerateMatrix command, as shown to the right. (Note that this command will not work unless the Student - Linear Algebra package is loaded.) This command automatically creates the augmented matrix of the system.     To view the matrix, double click on it to launch the Matrix Browser. Right-clicking the matrix and selecting Browse also works.    To solve the system, start by finding the row-reduced form of the matrix.    Right-click the matrix and select Standard Operations > Row-Reduced Form.    Since the solutions to the system are all in the rightmost column, the first 20 columns are no longer needed and could be removed. Additionally, the last three rows will simply be rows of zeroes and could be removed as well.    To restrict the columns, right-click and select Select Elements > Restrict Columns. Since only the 21st column is needed, type in [21] and click OK.    To remove the last three rows apply, right-click and select Select Elements > Restrict Rows. Type in [1 .. 20] and click OK.    To view the solutions without having to launch the matrix browser, convert the Matrix to a list.    Right-click the matrix and select Conversions > To List of Lists. A list of lists will be written to the screen.     Right-click again and select Flatten List.	(3.3.3)     (3.3.4)     (3.3.5)     (3.3.6)     (3.3.7)     (3.3.8)
The variable names must now be matched with their respective solutions.    Use the command, Equate. Type [var] as the first parameter, and the equation label of the solutions as the second parameter.	(3.3.9)

 

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