# In A Roof Supporting Truss The Load Is Transmitted When? • a) 566N b) 400N c) 773N d) 1090N

Answer: d Explanation: The direction of the unknown is not known to us. To know the direction of the unknown force we take assumption of it. That is we assume that the particular direction might be the direction of the force and then we do the calculations accordingly. And then we apply the equilibrium equations to the joints.12. Find the force in the member QS of the frame shown below.

1. a) 566N b) 400N c) 773N d) 1090N

Answer: a Explanation: The direction of the unknown is not known to us. To know the direction of the unknown force we take assumption of it. That is we assume that the particular direction might be the direction of the force and then we do the calculations accordingly.

And then we apply the equilibrium equations to the joints.13. A _ truss is in triangular section. a) Equilateral b) Simple c) Complex d) Lateral View Answer Answer: b Explanation: A simple truss in the shape of a triangle. It is the equilateral triangle. That is the angle between the legs are 60. This means the load is divided according and are equal in all the legs of the truss.

Hence the simple truss.14. Find the force in the member PQ of the frame shown below.

• a) 566N b) 546N c) 773N d) 1090N

Answer: b Explanation: The direction of the unknown is not known to us. To know the direction of the unknown force we take assumption of it. That is we assume that the particular direction might be the direction of the force and then we do the calculations accordingly.

1. And then we apply the equilibrium equations to the joints.15.
2. Which of the following is correct? a) To know the direction of the unknown force we take the assumption of it b) The direction of the unknown force is known to us already c) The direction of the unknown can’t be determined d) The direction of the unknown is of no use, it is not founded View Answer Answer: a Explanation: The direction of the unknown is not known to us.

To know the direction of the unknown force we take the assumption of it. That is we assume that the particular direction might be the direction of the force and then we do the calculations accordingly. Sanfoundry Global Education & Learning Series – Engineering Mechanics.

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#### How are forces distributed in a truss?

Lesson Background and Concepts for Teachers – This challenging lesson is originally designed as an application of right triangle trigonometry and is directed to pre-calculus or trigonometry students. This section is divided into three parts. Parts 1 and 3 are core of this lesson, and the see the associated activity Trust in the Truss: Design a Wooden Truss Bridge and have to be taught during class,

Teacher may or may not teach Part 2, depending on the students’ background. The Notetaking Sheet will help students follow along and easily annotate all of the explanations. There are also some video tutorials available (see Additional Multimedia Support below) that the teacher can assign to students to provide background or to reinforce the concepts in the lesson.1.

Truss Bridges We are going to focus on a specific class of bridges: truss bridges. In civil engineering, a truss is a regular structure built with straight members with end point connections. No member is continuous through a joint. The straight elements usually form triangular units, because this is the most stable structure in this type of bridge.

Truss bridges were widely used in the 19th century, because of their relative low cost and efficient use of materials. The truss design uses only tension and compression elements, which makes this structure strong and allows for simple analysis of forces on its structure. Engineers have designed different kinds of truss bridges while searching for the optimal combination of strength, weight, span, and cost.

(See Figure 5.) Figure 5. A pure truss can be represented as a pin-jointed structure, where the only forces on the truss members are tension or compression, and not bending. Engineers have created different kind of trusses that optimize span, weight, and strength copyright Copyright © University of North Carolina Charlotte, Public Domain, Learning Activity #1.

Build a Model of a Truss Bridge. https://webpages.uncc.edu/~jdbowen/1202/learning_activities_manual/Learning_Activity_1.pdf This lesson and its associated activity will focus on four truss bridges: the Warren, the Warren with verticals, the Pratt, and the Howe (Figure 6). Warren Truss. A design distinguished by equal-sized components and the ability of some of the diagonals to act in both tension and compression.

The type is generally characterized by thick, prominent, diagonal members, although verticals could be added for increased stiffness. This design was patented by the British engineer James Warren in 1848. Warren truss bridges gained popularity in the United States after 1900 as American engineers began to see the structural advantages of riveted or bolted connections over those that were pinned.

The design was well suited to a variety of highway bridge applications and was very popular from about 1900 to 1930. Pratt Truss. In this truss, its elements are arranged in right triangles. The United States railroad expansion in the 19th century required strong, dependable bridges to carry trains over ravines and rivers.

In 1844, Caleb and Thomas Pratt developed a bridge that was built initially with wood and diagonal iron rods. Later, the bridge was built entirely of iron. This bridge had the advantage of low-cost construction, and could also be quickly erected by semi-skilled labor.

This design became the standard American truss bridge for moderate spans from 7.62 meters (~25 feet) to 45.72 meters (~150 feet), well into the 20th century. Howe Truss. Similar to the Pratt truss, elements of the Howe truss are also arranged in right triangles, but with different orientation. Designed by William Howe in 1840, it used mostly wood in construction and was suitable for longer spans than the Pratt truss.

Therefore, it became very popular and was considered one of the best designs for railroad bridges back in the day. The diagonal structural beams slope toward the bridge center, while Pratt truss utilizes diagonal beams that slope outward from the center of the bridge. Figure 6. Diagrams of four types of truss engineering. copyright Copyright © 2019 Miguel R Ramirez, Independent Contributor 2. Analysis of Forces. Basic Concepts In physics, a force is any action that tends to maintain or alter the motion of a body or to distort it.

• When a bridge is loaded it is expected to be still, so all the forces applied on the bridge have to be absorbed by its structure.
• In other words, all forces are in equilibrium.
• In a truss, it is assumed that the forces along the elements converge at the nodes of the structure.
• This fact allows us to use a free body diagram to find the acting forces values.

By definition, a free body diagram (FBD) is a representation of an object with all the forces that act on it. The external environment, as well as the forces that the object exerts on other objects, are omitted in a FBD. This allow us to analyze an object in isolation.

A FBD can be constructed in three simple steps: first, sketch what is happening on the body; second, identify the forces that act on the object; and third, represent the object as a point with the forces as arrows pointing in their acting direction: with an origin at the point representing the object, with a size proportional to their magnitude and a label indicating the force type.

Figure 7 shows the FBD for a spherical object rolling on an incline. Figure 7. The forces acting on a spherical body freely rolling on an incline are the body’s weight, and the contact friction force. The FBD represents these forces. The body’s weight is split in two components, one parallel to the incline and the other perpendicular to this.

A coordinate system is also drawn centered at the point and aligned with the incline, and this make the forces analysis simple. copyright Copyright © 2019 Miguel R Ramirez, Independent Contributor When all the forces that act upon an object are balanced, then the object is said to be in state of equilibrium,

This does not mean that the forces are equal, but that the sum of all forces add up to zero. FBD’s are used to decompose forces into their vertical (x) and horizontal (y) components; then the state of equilibrium can be stated as the sum of all these vertical and horizontal components equal to zero.

1. Using the sigma notation to represent sum, the equilibrium conditions can be mathematically written as: Ʃ F x = 0 Ʃ F y = 0 (1) where F x represents all the forces’ components along the x -axis (horizontal), and F y all the components along the y -axis (vertical).
2. These components are determined using the magnitude of the force, and the value of the sine or cosine of the angle the force make with the horizontal.

These components will have a (+) sign when pointing in the positive direction of the axis, or (-) sign when pointing in the negative direction of the axis. Next is an example (included in the Notetaking Sheet ) you can use with students about how set up a FBD to obtain equilibrium equations. Figure 8. (a) A weight suspended from the ceiling by two ropes at different angles. (b) FBD representing the forces acting on a body, depicted as a dot. FBD also specifies the angle the forces make respect the x- axis. (c) The force T components along the coordinate axis can be found multiplying the T-force magnitude times the trigonometric ratios for the angle the T-force makes with the x-axis.

C) The force S components along the coordinate axis can be found multiplying the S-force magnitude times the trigonometric ratios for the angle the S-force makes with the x-axis. copyright Copyright © 2019 Miguel R Ramirez, Independent Contributor Forces T and S make horizontal angles of 60° and 40° respectively.

Force W works along the vertical axis. What should be the values for forces T and S to keep the systems static when W = 10 N? T x = T ·cos60° T y = T ·sin60° (2) The equilibrium conditions (1) requires that all the forces’ x -components and y -components add up to zero.

The T -force components ( T x, T y ) can be found using the trigonometric ratios sine and cosine of the 60° angle, because the force makes a right triangle with these axes (Figure 8(c)): S x = S ·cos40° S y = S ·sin40° (3) The S-force components ( S x, S y ) can be found in a similar way (Figure 8(d)): The W -force only has a y -component: W x = 0, W y = W,

Adding up the x -components and the y -components of the forces in equations (2) and (3), and making these additions equal to zero, we obtain the equilibrium equations for this problem: Ʃ F x = T x + S x = T ·cos60° – S· cos40° = 0 (4) Ʃ F y = T y + S y + W = T· sin60° + S· sin40° – W y = 0 (5) where the minus signs in equations (4) and (5) are assigned because these components are pointing to the negative direction of the axis.

Using now the given value for W = 10 N, and taking cos60° = 0.5, sin60° = 0.866, cos40° =0.766, and sin40° = 0.643, equations (4) and (5) can be used to determine the values for T and S to keep the system in equilibrium: 0.5 T – 0.766 S = 0 (6) 0.866 T + 0.643 S – 10 = 0 (7) The system of linear equations (6)-(7) can be easily solved.

Using the substitution method, we can solve for T in equation (6) T = 1.532 S and substituting this expression for T in equation (7), is possible to find a value for S : 0.866 (1.532 S ) + 0.643 S – 10 = 0 S = 5.077 N Using the found value for S, is now possible top find the value for T : T = 1.532 (5.77) = 7.778 N 3.

• The Joints Method Note to Teacher: In this section, a step-by-step example of the simplest method used in civil engineering to solve for the unknown forces acting on members of a truss is presented.
• The bridge structure will be the smallest and simplest possible: The Warren truss with three equilateral triangles (Figure 9).
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Larger structures, or different like Pratt or Howe, are solved in the same way. Use the Notetaking Sheet to help students to follow you during this long process. Figure 9. Warren truss made up with equilateral triangles 4 inches side, to be solved using the Joints Method. This method will produce a system of linear equations whose solutions will be the tensions and compressions on the truss elements. copyright Copyright © 2019 Miguel R Ramirez, Independent Contributor The joints method determine forces at the truss joints or nodes using FBD’s.

• The general assumptions to apply this method are: (a).
• All truss elements are considered rigid, they never bend. (b).
• A force applied to the truss structure will only produce compression or tension on the elements. (c).
• Tension – compression forces’ directions are parallel to the elements. (d).
• Any force on a truss element is transmitted to its ends.

(e). A truss structure in equilibrium means that every joint or node is at equilibrium. (f). Once determined the value of a tension or compression force at one of the ends of an element, the complementary force at the other end of the element will be equal but in opposite direction.

• Equilibrium condition).
• In this specific example (Figure 9) it is assumed that: (g).
• Vertical and equal downward forces of 10 lbf are applied on the top nodes (nodes 2 and 4), and at the central node (node 3). (h).
• The bridge stands only at its end bottom nodes (nodes 1 and 5). (i).
• The equilateral triangles are 4 inches side.

Step 1. Identify the corresponding forces acting on each of the truss elements. It is convenient to identify the forces on the truss elements making reference to the corresponding end nodes. For example, the force acting on the element joining nodes 1 and 2, is denoted F 12, (Figure 10), and the force acting on element between nodes 4 and 3 is denoted F 43, Figure 10. The force acting on the element joining the nodes i and j, will be denoted F_ij. So the force acting on the element between nodes 3 and 5 will be F_35, and the force on the element between nodes 3 and 2 will be F_32 copyright Copyright © 2019 Miguel R Ramirez, Independent Contributor To simplify a little the big number of variables generated, assumption ( f ) is applied. Figure 11. Because the element joining the nodes i and j, is in equilibrium, the forces acting at the ends of this element must be equal in magnitude but opposite. This means that the force F_ji can be replaced by the force F_ij. This assumption reduce the number of variables to work with in half.

2. Find the value of reaction forces on the end bottom of the truss (nodes 1 and 5) Reactions are forces developed at the supports of a structure, to keep the structure in equilibrium.
3. To find the reaction forces on the truss it is required to calculate the moments all the forces applied on the truss can produce, respect to every of the end bottom nodes.

By definition, the moment of a force is the product of the distance from the point to the point of application of the force and the component of the force perpendicular to the line of the distance: M = F·d The moment of a force M quantifies the turning effect or rotation the force can produce.

In the case of a structure in equilibrium, the moment of all the forces applied has to be zero. In this example (Figure 12), forces F 1, F 2 and F 3, tend to rotate the truss clockwise respect to node 1, but the reaction force R 5 on node 5 cancels this effect (Figure 12a), keeping the truss in equilibrium.

This equilibrium condition is expressed mathematically as the sum of the moments of all forces and reactions equal to zero: M 1 = Ʃ F 1 · x 1 = 0 (8) = R 1 · 0 – F 2 · 2 – F 3 · 4 + F 4 · 6 + R 5 · 8 = 0 where the minus signs indicate that the forces point in the negative direction. Figure 12. Forces applied on a rigid body certain distance away of a potential rotation point, tend to produce a rotation of the body around this point. copyright Copyright © 2019 Miguel R Ramirez, Independent Contributor These forces also tend to rotate the truss counter clockwise respect node 5, but the reaction force R 1 on node 1 cancels this effect (Figure 12b).

This equilibrium condition can be also expressed as: M 5 = Ʃ F 1 · x 1 = 0 (9) = R 1 · 8 – F 2 · 2 – F 3 · 4 + F 4 · 6 + R 5 · 0 = 0 For this example, given F 1 = F 2 = F 3 = 10 lbf, the value of reaction R 5 can be found solving equation (8): = R 1 · 0 – 10 · 2 – 10 · 4 + 10 · 6 + R 5 · 8 = 0 -120 + R 5 · 8 = 0 R 5 = 15 lbf and the value of reaction R 1 can be found solving equation (9): = R 1 · 8 – 10 · 6 – 10 · 4 + 10 · 2 + R 5 · 8 = 0 R 1 · 8 – 120 = 0 R 1 = 15 lbf Step 3.

Analysis of Forces on Nodes using FBD and the equilibrium conditions ƩF y = 0 and ƩF x = 0. Figure 13. Free Body Diagrams (FBD) for the analysis of forces at truss nodes. These FBD’s show the components along the vertical and horizontal axis necessary for the equilibrium conditions copyright Copyright © 2019 Miguel R Ramirez, Independent Contributor Putting together equations (10)-(19) we can see that we have obtained a system of linear equations: ten equations with seven variables. The solution of this system will give the values of the tension – compression forces on the truss elements because the 10 lbf loads. For those experienced solving systems of linear equations, is going to be odd to have an overdetermined system for a truss structure (more equations than variables). However, we know from Algebra that even though most of the times an overdetermined system of equations has no solution, there are overdetermined systems of equations that have a solution.

• This is the case for this example.
• In structural analysis, a truss is statically determinate when all the forces on its elements can be found by equations of statics alone.
• This is what we have in this example.
• In Step 4 we can verify that our system has a solution. Step 4.
• Solve the System of Equations.

Identify the obtained forces as Compressions (-) or Tensions (+) Note to teacher: Continue using the Notetaking Sheet to help students to follow you during this long process. Is recommended to model how to the first four or five equations and then ask students to work in small groups to obtain the solutions for the rest of the equations. It is simple to verify that the obtained values for F 34, F 35 and F 45 satisfy equations (17), (18), and (19): Step 5. Create the Matrix for the above System of Equations Note: This simple concept is important in the computer graphic interface explained in Step 6. Continue using the Notetaking Sheet, Guide students to obtain the associated matrix for the system of equations (10)-(19): Step 6. Spreadsheets to Calculate Trusses The procedure shown in Steps 1 – 5 is also used to solve other type of trusses, larger trusses, isosceles triangles or right triangles trusses, or when different loads are applied on the nodes. However, a larger truss will produce a larger system of equations, which it will be more difficult to solve by hand.

For example, a Warren truss with nine equilateral triangles will produce a 19 x 22 system of linear equations. Also, every change in the triangles angles or in the loads values will require to perform all the calculations again. Because the procedure is always the same, no matter the truss type or the number of elements, it is possible to use Excel or Google Sheets and create a graphical interface to perform all these calculations automatically, giving the loads and the elements’ angles as entry values.

Note to teacher: It is not in the scope of this lesson that students write themselves such interface. But for those advanced or curious, the procedure used to write this interface is detailed in the Annex, Figure 14. Computer Graphic Interface created in Google Sheets to find the tensions-compressions in a Warren truss. Calculations are performed automatically once the loads on the nodes are typed in, and the material and thickness of the truss elements are chosen copyright Copyright © 2019 Miguel R Ramirez, Independent Contributor This section summarizes how to use a very friendly computation interface developed by the author in Google Sheets.

• Calculates the tensions-compressions for the Warren truss, Warren with Vertical truss, Pratt truss, and Howe truss.
• Estimates maximum strength of the trusses considering the kind of wood used and element’s thickness.
• Gives solutions when the truss is supported only on its bottom end nodes, the truss’ diagonal elements are round or square, and the truss’ rails are square.

Figure 15. The graphic interface worksheet has specific cells to type values, and sections that perform the operations and display the results. copyright Copyright © 2019 Miguel R Ramirez, Independent Contributor Visit this Trusses Calculations webpage for more information.

 Truss Types Size (No. of triangles) Warren 3, 5, 7, 9, 11 (equilateral-isosceles) Warren w/ Verticals 3, 5, 7, 9, 11 (equilateral-isosceles) Pratt 6, 10, 14, 18 (right triangles) Howe 6, 10, 14, 18 (right triangles)

To activate an interface on a PC, laptop, or tablet, click on the gray square at the right top of the window. (Figure 14(a)); click on the window when using a cellphone. (Note: these sheets are open to the public for distribution and to copy; the master will always remain locked.) Once the spreadsheet is active, you can see its different sections (Figure 14(b)):

1. A truss diagram with entries for the loads on each node.
2. Two entries for the truss elements’ length and angle respect the horizontal.
3. Cells to select the truss elements’ thickness and kind of wood they are made of.
4. Section displaying the calculated truss elements’ tensions-compressions
5. The operational section: matrix associated to the system of linear equations obtained from the FBD’s, the inverse matrix, and the solutions of the system of equations.

Students only have to input values in the cells in sections (1) and (2), select round or square diagonals (3), and select the elements’ wood type and thickness in the cells in section (4) (Figure 15). Sections (5) and (6), or other cells do not have to be altered.

1. The tension-compression on the truss elements are automatically calculated once values in sections (1), (2), (3) or (4) are entered.
2. Next is suggested a guided practice to teach students the efficient use of these interfaces.
3. Important: Before this practice, reset the values in the Warren truss in the third sheet: 5 in the loads, 5 in the triangles’ base length, select round diagonals, and hardwood + 1/8 in the Diagonals entries, and also Hardwood + 1/8 in the Rails entries.

There is a hardcopy of these instructions in the Notetaking Sheet, make enough copies for students. You will also need to be familiar with how to access and work with Google Drive and Google Sheets with a cellphone or computer. Use projections of these notes to support the practice steps.

Tell students: Because it is required in the associated activity to estimate the strength of the bridge to build, and this bridge is going to be larger than the one solved here, the number of nodes is going to increase, so the resulting system of equations will be larger, and the time to solve will be longer.

For example, the analysis of a Warren truss with 11 equilateral triangles will produce a linear system with 26 equations and 23 variables! But there is more. Any change in the loads values, or truss angles, will require to repeat all the analysis again.

Repeating by hand the same process, even after the simplest change, is not really efficient. You are going to learn how to use a graphic interface to easily perform all this calculations. You should have a basic knowledge of Google Sheets or Excel, a Google Drive account, and your cellphone or a computer with internet access.

These interfaces are very easy to use. You have in your notes a hardcopy of these instructions in your notes for quick reference. First you have to open the webpage where these interfaces are located. Open Google Chrome in your cellphones or on the computer and type the next address: https://sites.google.com/gpapps.galenaparkisd.com/mramirez-math/courses-highlights/bridges/trusses-calculations In this page select the Warren Truss interface.

Click on this window to open the interface, and click on the Open button at the top of the document. (You will be required to use your Google account here.) Once the worksheet is open you con scroll this and identify the different parts of the graphic interface (Show to students Figure 15). You can easily see the cells where you can input information.

Do not type anything yet. Because this document is shared to every one of you, so every change made by one of you is going to be displayed in all other cellphones and computers. For this little practice it is necessary you work with your own copy of this document.

• So create your own copy in your Google Drive, using the commands Share & export + Make a copy, give a new name to the file and save it in your Drive.
• You will work now in your saved file.
• Note to Teacher.
• It is very important you inform students that even though the original file cells that do not have to be modified are protected, the copies they save possibly will not be protected.

So ask them to be very careful and not to modify or erase other cells but the indicated in the practices. Your first assignment is to use the interface to calculate the Warren truss with three equilateral triangles we have solved in class. You know the correct solutions for this problem, so the interface, if correctly developed, will give you the same answers. Figure 16. The graphic interface solution for the problem solved step-by-step in section 3. The Joints Method. Solutions are automatically updated after any change on the loads values or in the elements angles. copyright Copyright © 2019 Miguel R Ramirez, Independent Contributor Once you verified that the interface gave the same values, you can use the interface for more complex calculations.

• Change the 10 lbf load on the central bottom node (Node 3) by 65 lbf, and check the new tensions-compressions.
• What do you see now? The expected answer is: four cells changed color,
• Ask to students now: What does this mean? Check in the color code table.

The answer is: the four diagonal elements break under this new load, The next question to students is: Assuming you need a bridge able to support this load, what could be the solution to this problem, The expected answer is: use a stronger material,

1. Continue saying: Let’s first change the shape of the diagonal elements from round to square, what happened now? The expected answer is: Only two diagonals are broken now.
2. Ask students: Why is this possible? You only changed a round element of 1/8 inch dimeter by a square of 1.8 inch side? Draw on the board a square representing 1/8 inch and inside a circle with the same diameter to help students with their reasoning,

It is expected they see that the square’s area is greater than the circle’s area, and therefore it is a thicker, and consequently, a stronger element. Continue with the next: Change the diagonals to “round” again, and go now to the cells under the legend “Wood Compressive Strength – Truss Elements Strength.

• Here you have two options: Choose a different wood or choose a thicker element.
• Begin with the second option, click on the “Diameter” and “Side” cells and select thicker elements than 1/8.
• Select the next in size element.
• Students will find 3/16 for Diagonals, and Rails),
• What happened now? Students should answer something like: The thicker element of 3/16 inch resists this new load.

Ask students now to return Diagonals and Rails thicknesses to 1/8. Now ask them to change wood to Birch in both Diagonals and Rails, and ask students: What happened now? Students should answer: Birch is a stronger wood, 1/8 inch elements of this wood can resist the load 1/8 Hardwood elements canno t. Figure 17. Students have to find the Diagonals and Rails thicknesses such that a hardwood Warren Truss made up with nine equilateral triangles be able to resist 250 lbf at the top and 400 lbf at the bottom copyright Copyright © 2019 Miguel R Ramirez, Independent Contributor The next table contains the approximate solutions for the practice problems at the end of the Notetaking Sheet, A last remark about the graphic interfaces. The wood options in these worksheets include only the most common dowels can be found in stores: hardwood, pine, basswood, poplar, oak, and birch. During this assignment move around the room and check students’ work and provide the necessary help and support. It is very important they understand completely the use (and limitations) of these interfaces.

## What trusses lie on a plane?

A planar truss lies in a single plane. Planar trusses are typically used in parallel to form roofs and bridges.

## Which of the following forces is carried by truss members?

2. State whether the following statement is true or false.
3. Truss and column are different in physical appearance.

a) true b) false View Answer Answer: b Explanation: Truss and columns are similar looking, it depends on how they are used.4. State whether the following statement is true or false. DSI is difference between external degree of indeterminacy and internal degree of indeterminacy.

A) true b) false View Answer Answer: b Explanation: DSI is sum of external degree of indeterminacy and internal degree of indeterminacy.5. Internal degree of indeterminacy of a beam/frame member is :- a) always zero b) always non-zero c) can’t say d) depends upon internal hinge View Answer Answer: d Explanation: If internal hinge is there then it won’t be 0 otherwise it is always zero.

Check this: | 6. What is the general from of equation for DSI of a planar frame? a) R – 1 – c b) R – 2 – c c) R – 3 – c d) R – 4 – c View Answer Answer: c Explanation: 3 equations are there in case of planar frame until some extra equations c is not there.

1. Here, R is no.
2. Of external reactions. C is no.
3. Of extra equations.7.
4. What is the general from of equation for DSI of a space frame? a) R – 4 – c b) R – 5 – c c) R – 6 – c d) R – 7 – c View Answer Answer: c Explanation: 6 equations are there in case of planar frame until some extra equations c is not there.
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Here, R is no. of external reactions. C is no. of extra equations.8. How many extra equations are possible if 3 hinges are there in a planar frame in relation to DSI? a) 6 b) 9 c) 3 d) 1 View Answer Answer: c Explanation: Each hinge gives one extra equation in case of planar frame.9., a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry, He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at, Subscribe to his free Masterclasses at & technical discussions at, : Structural Analysis Questions and Answers – DKI and DSI – I

## Which axial force is determined while analyzing a truss?

There are several methods of truss analysis, but the two most common are the method of joint and the method of section (or moment).5.6.1 Sign Convention In truss analysis, a negative member axial force implies that the member or the joints at both ends of the member are in compression, while a positive member axial force indicates that the member or the joints at both ends of the member are in tension.5.6.2 Analysis of Trusses by Method of Joint This method is based on the principle that if a structural system constitutes a body in equilibrium, then any joint in that system is also in equilibrium and, thus, can be isolated from the entire system and analyzed using the conditions of equilibrium.

1. The method of joint involves successively isolating each joint in a truss system and determining the axial forces in the members meeting at the joint by applying the equations of equilibrium.
2. The detailed procedure for analysis by this method is stated below.
3. Procedure for Analysis •Verify the stability and determinacy of the structure.

If the truss is stable and determinate, then proceed to the next step. •Determine the support reactions in the truss. •Identify the zero-force members in the system. This will immeasurably reduce the computational efforts involved in the analysis. •Select a joint to analyze.

• At no instance should there be more than two unknown member forces in the analyzed joint.
• Draw the isolated free-body diagram of the selected joint, and indicate the axial forces in all members meeting at the joint as tensile (i.e.
• As pulling away from the joint).
• If this initial assumption is wrong, the determined member axial force will be negative in the analysis, meaning that the member is in compression and not in tension.

•Apply the two equations $$\Sigma F_ =0$$ and $$\Sigma F_ =0$$ to determine the member axial forces. •Continue the analysis by proceeding to the next joint with two or fewer unknown member forces. Example 5.2 Using the method of joint, determine the axial force in each member of the truss shown in Figure 5.10a. $$Fig.5.10$$. Truss. Solution Support reactions. By applying the equations of static equilibrium to the free-body diagram shown in Figure 5.10b, the support reactions can be determined as follows: $$\begin +\curvearrowleft \sum M_ =0 \\ 20(4)-12(3)+(8) C_ =0 \\ C_ =-5.5 \mathrm & C_ =5.5 \mathrm \downarrow \\ +\uparrow \sum F_ =0 \\ A_ -5.5+20=0 \\ A_ =-14.5 \mathrm & A_ =14.5 \mathrm \downarrow \\ +\rightarrow \sum F_ =0 \\ -A_ +12=0 \\ A_ =12 \mathrm & A_ =12 \mathrm \leftarrow \\ \end$$ Analysis of joints.

The analysis begins with selecting a joint that has two or fewer unknown member forces. The free-body diagram of the truss will show that joints $$A$$ and $$B$$ satisfy this requirement. To determine the axial forces in members meeting at joint $$A$$, first isolate the joint from the truss and indicate the axial forces of members as $$F_$$ and $$F_$$, as shown in Figure 5.10c.

The two unknown forces are initially assumed to be tensile (i.e. pulling away from the joint). If this initial assumption is incorrect, the computed values of the axial forces will be negative, signifying compression. Analysis of joint $$A$$. $$\begin +\uparrow \sum F_ =0 \\ F_ \sin 36.87^ -14.5=0 \\ F_ =24.17 \\ +\rightarrow \sum F_ =0 \\ -12+F_ +F_ \cos 36.87^ =0 \\ F_ =12-24.17 \cos 36.87^ =-7.34 \mathrm \end$$ After completing the analysis of joint $$A$$, joint $$B$$ or $$D$$ can be analyzed, as there are only two unknown forces. Analysis of joint $$D$$. $$\begin +\uparrow \sum F_ =0 \\ F_ =0 \\ +\rightarrow \sum F_ =0 \\ -F_ +F_ =0 \\ F_ =F_ =-7.34 \mathrm \end$$ Analysis of joint $$B$$. $$\begin +\rightarrow \sum F_ =0 \\ -F_ \sin 53.13+F_ \sin 53.13+15=0 \\ F_ \sin 53.13=-15+24.17 \sin 53.13= \\ F_ =5.42 \mathrm \end$$ 5.6.3 Zero Force Members Complex truss analysis can be greatly simplified by first identifying the “zero force members.” A zero force member is one that is not subjected to any axial load. Sometimes, such members are introduced into the truss system to prevent the buckling and vibration of other members. $$Fig.5.11$$. Zero force members.5.6.4 Analysis of Trusses by Method of Section Sometimes, determining the axial force in specific members of a truss system by the method of joint can be very involving and cumbersome, especially when the system consists of several members.

In such instances, using the method of section can be timesaving and, thus, preferable. This method involves passing an imaginary section through the truss so that it divides the system into two parts and cuts through members whose axial forces are desired. Member axial forces are then determined using the conditions of equilibrium.

The detailed procedure for analysis by this method is presented below. Procedure for Analysis of Trusses by Method of Section •Check the stability and determinacy of the structure. If the truss is stable and determinate, then proceed to the next step. •Determine the support reactions in the truss.

•Make an imaginary cut through the structure so that it includes the members whose axial forces are desired. The imaginary cut divides the truss into two parts. •Apply forces to each part of the truss to keep it in equilibrium. •Select either part of the truss for the determination of member forces. •Apply the conditions of equilibrium to determine the member axial forces.

Example 5.3 Using the method of section, determine the axial forces in members $$CD$$, $$CG$$, and $$HG$$ of the truss shown in Figure 5.12a. $$Fig.5.12$$. Truss. Solution Support reactions. By applying the equations of static equilibrium to the free-body diagram in Figure 5.12b, the support reactions can be determined as follows: $$\begin A_ =F_ =\frac =80 \mathrm \\ +\rightarrow \Sigma F_ =0 \quad A_ =0 \end$$ Analysis by method of section. First, an imaginary section is passed through the truss so that it cuts through members $$CD$$, $$CG$$, and $$HG$$ and divides the truss into two parts, as shown in Figure 5.12c and Figure 5.12d. Member forces are all indicated as tensile forces (i.e., pulling away from the joint). If this initial assumption is wrong, the calculated member forces will be negative, showing that they are in compression. Either of the two parts can be used for the analysis. The left-hand part will be used for determining the member forces in this example. By applying the equation of equilibrium to the left-hand segment of the truss, the axial forces in members can be determined as follows: Axial force in member $$CD$$. To determine the axial force in member $$CD$$, find a moment about a joint in the truss where only $$CD$$ will have a moment about that joint and all other cut members will have no moment. A close examination will show that the joint that meets this requirement is joint $$G$$. Thus, taking the moment about $$G$$ suggests the following: $$\begin +\curvearrowleft \sum M_ =0 \\ -80(6)+80(3)-F_ (3)=0 \\ F_ =-80 \mathrm & 80 \mathrm (C) \end$$ Axial force in member $$HG$$. $$\begin +\curvearrowleft \sum M_ =0 \\ -80(3)+F_ (3)=0 \\ F_ =80 \mathrm & 80 \mathrm (T) \end$$ Axial force in member $$CG$$. The axial force in member $$CG$$ is determined by considering the vertical equilibrium of the left-hand part. Thus, $$\begin +\uparrow \sum F_ =0 \\ 80-80-F_ \cos 45^ =0 \\ F_ =0 \end$$ Chapter Summary Internal forces in plane trusses: Trusses are structural systems that consist of straight and slender members connected at their ends. The assumptions in the analysis of plane trusses include the following: 1.Members of trusses are connected at their ends by frictionless pins.2.Members are straight and are subjected to axial forces.3.Members’ deformations are small and negligible.4.Loads in trusses are only applied at their joints. Members of a truss can be subjected to axial compression or axial tension. Axial compression of members is always considered negative, while axial tension is always considered positive. Trusses can be externally or internally determinate or indeterminate. Externally determinate trusses are those whose unknown external reactions can be determined using only the equation of static equilibrium. Externally indeterminate trusses are those whose external unknown reaction cannot be determined completely using the equations of equilibrium. To determine the number of unknown reactions in excess of the equation of equilibrium for the indeterminate trusses, additional equations must be formulated based on the compatibility of parts of the system. Internally determinate trusses are those whose members are so arranged that just enough triangular cells are formed to prevent geometrical instability of the system. The formulation of stability and determinacy in trusses is as follows: $$\begin m+r<2 j \quad \text \\ m+r=2 j \quad \text \\ m+r>2 j \quad \text \end$$ Methods of analysis of trusses: The two common methods of analysis of trusses are the method of joint and the method of section (or moment). Method of joint : This method involves isolating each joint of the truss and considering the equilibrium of the joint when determining the member axial force. Two equations used in determining the member axial forces are $$\Sigma F_ =0$$ and $$\Sigma F_ =0$$. Joints are isolated consecutively for analysis based on the principle that the number of the unknown member axial forces should never be more than two in the joint under consideration in a plane trust. Method of section: This method entails passing an imaginary section through the truss to divide it into two sections. The member forces are determined by considering the equilibrium of the part of the truss on either side of the section. This method is advantageous when the axial forces in specific members are required in a truss with several members. Practice Problems 5.1 Classify the trusses shown in Figure P5.1a through Figure P5.1r.   $$Fig. P5.1$$. Truss classification.5.2 Determine the force in each member of the trusses shown in Figure P5.2 through Figure P5.12 using the method of joint. $$Fig. P5.2$$. Truss. $$Fig. P5.3$$. Truss. $$Fig. P5.4$$. Truss. $$Fig. P5.5$$. Truss. $$Fig. P5.6$$. Truss. $$Fig. P5.7$$. Truss. $$Fig. P5.8$$. Truss. $$Fig. P5.9$$. Truss. $$Fig. P5.10$$. Truss. $$Fig. P5.11$$. Truss. $$Fig.5.12$$. Truss.5.3 Using the method of section, determine the forces in the members marked X of the trusses shown in Figure P5.13 through Figure P5.19. $$Fig. P5.13$$. Truss. $$Fig. P5.14$$. Truss. $$Fig. P5.15$$. Truss. $$Fig. P5.16$$. Truss. $$Fig. P5.17$$. Truss. $$Fig. P5.18$$. Truss. $$Fig. P5.19$$. Truss.

## How is load distributed in a roof truss?

What forces act on a roof truss? – As the loads supported by a truss are mainly applied to the joints, they only act along the axis of each individual piece, or member. This subjects the structure to two axial forces, compression and tension. As axial loads are carried equally by all parts of the member, weight bearing is as high as possible.

### What is a truss How does it transfer the loads?

Trusses consist of triangular units constructed with straight members. The ends of these members are connected at joints, known as nodes. They are able to carry significant loads, transferring them to supporting structures such as load-bearing beams, walls or the ground.

#### How does a truss roof work?

What is a truss? – A truss is a web-like roof design of wood or steel that uses tension and compression to create strong, light components that can span a long distance. The sides are in compression and the bottom is in tension to resist being pulled apart. Engineers design trusses to withstand the three types of loads associated with a building:

Live loads. Transient forces within the building include people, furniture, appliances, and cars. Dead loads. Permanent loads like beams, walls, and flooring comprise the structure of a building. Environmental loads. Forces like wind, rain, or snow act laterally against the building.

As an alternative to rafters, roof trusses are designed to hold more weight. Trusses are typically built in a factory rather than at the job site, which makes them less expensive—in fact, they can cut roof framing costs by as much as 50%.

#### How do I know if my roof Trusse is load-bearing?

Walls that run perpendicular to the ceiling/floor joists are typically load-bearing. In multi-story buildings, load-bearing walls typically will stack on top of each other. If there’s a wall directly above or below the wall on the main floor, you could assume that they’re all load-bearing.

### How are trusses connected?

3.9.1 General – Truss bridges are generally used for spans over 40 m. For spans between 40 and 70 m, parallel chord trusses are used, while for spans greater than 70 m, polygonal chord trusses are used. Trusses are, normally, designed to carry axial forces in its members, which are either tension or compression or reversible tension/compression depending on the worst cases of loading and load combinations.

Truss members are connected at joints using welds or bolts. Joints are designed as pins and the forces in truss members are in full equilibrium at the joints. In practice, gusset plates are used at the joints to collect the forces in the members meeting at the joints, where equilibrium takes place. Therefore, the size of the gusset plates should be as small as possible to simulate the behavior of pins.

If the maximum force in a truss is less than 3000 kN, single gusset plate trusses are used and truss members are designed as angles. On the other hand, if the maximum force in truss members is greater than 3000 kN, double gusset plate trusses are used and chord members are designed as box sections, while diagonals and verticals are designed as I-sections or box sections in case of long diagonals carrying compressive forces.

### Which source of loads are always acting on a truss?

Trusses::Fundamentals::Knowledgebase::SAFAS Trusses are linear structures made of members that resist applied loads mainly through axial tension or compression rather than bending, and therefore they are structurally very efficient. However, this is valid only if the truss members are pin-connected and the loads act at the,

 Deformation of a Truss Under Loads

Considering that the truss members are not subjected to any loads but at the joints (their ends), the of a truss member can be considered as shown below (note that there is no moment at the pins so the member can freely rotate):  Internal Forces in a Truss Member

where is the axial force and is the shear force. Satisfying the equilibrium equation by taking sum of the moments about point A equal to zero :

Since is the length of the member and cannot be equal to zero, therefore, This means that the truss members are subjected to axial forces(tension or compression) only. There are various truss configurations. Some are shown below:

 Pratt Truss Howe Truss Warren Truss Common Types of Truss

Importantly, trusses are stable only if they are triangulated. This means that their configuration is made of triangles. If any member is removed such that this condition is violated, the truss becomes unstable. There is, however, one exception (Vierendeel truss). : Trusses::Fundamentals::Knowledgebase::SAFAS

### Which of the following are true about roof trusses Mcq?

Design of Steel Structures Questions and Answers – Compression Members and Loads on Compression Members This set of Design of Steel Structures Question Bank focuses on “Compression Members and Loads on Compression Members”.1. What is compression member? a) structural member subjected to tensile force b) structural member subjected to compressive force c) structural member subjected to bending moment d) structural member subjected to torsion View Answer Answer: b Explanation: Structural member which is subjected to compressive forces along its axis is called compression member.

1. Compression members are subjected to loads that tend to decrease their lengths.2.
2. Which of the following is true about axially loaded column? a) member subjected to bending moment b) member subjected to axial force and bending moment c) net end moments are not zero d) net end moments are zero View Answer Answer: d Explanation: if the net end moments are zero, the compression member is required to resist load acting concentric to original longitudinal axis of member and is called axially loaded column or simply column.3.

Which of the following is true about beam column? a) member subjected to bending moment b) member subjected to axial force only c) member subjected to axial force and bending moment d) net end moments are zero View Answer Answer: c Explanation: If the net end moments are not zero, the member will be subjected to axial force and bending moments along its length.

• Such members are called beam-columns.4.
• What are columns? a) vertical compression members in a building supporting floors or girders b) vertical tension members in a building supporting floors or girders c) horizontal compression members in a building supporting floors or girders d) horizontal tension members in a building supporting floors or girders View Answer Answer: a Explanation: The vertical compression members in a building supporting floors or girders are normally called as columns.

They are sometimes called as stanchions. They are subjected to heavy loads. Vertical compression members are sometimes called posts.5. Which of the following are true about roof trusses? a) principal rafter are compression members used in buildings b) principal rafter is bottom chord member of roof truss c) struts are compression members used in roof trusses d) struts are tension members used in roof trusses View Answer Answer: c Explanation: The compression members used in roof trusses and bracings are called as struts.

They may be vertical or inclined and normally have small lengths. the top chord members of a roof truss are called principal rafter. Check this: | 6. Knee braces are _ a) long compression members b) short compression members c) long tension members d) short tension members View Answer Answer: b Explanation: Short compression members at junction of columns and roof trusses or beams are called knee braces.

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Which of the following is correct? a) moment due to wind loads is not considered in unbraced buildings b) wind load cause large moments in braced buildings c) wind loads in multi-storey buildings are not usually applied at respective floor levels d) wind loads in multi-storey buildings are usually applied at respective floor levels View Answer Answer: d Explanation: Wind loads in multi-storey buildings are usually applied at respective floor levels and are assumed to be resisted by bracings.

In such cases, wind load is applied to the column through sheeting rails and may be taken as uniformly distributed throughout the length of column.10. The strength of column does not depend on a) width of building b) material of column c) cross sectional configuration d) length of column View Answer Answer: a Explanation: The strength of column depends on material of column, cross sectional configuration, length of column, support conditions at the ends, residual stresses, imperfections.11.

Which of the following is not an imperfection in column? a) material not being isotropic b) geometric variations of columns c) material being homogenous d) eccentricity of load View Answer Answer: c Explanation: Imperfections in column include material not being isotropic and homogenous, geometric variations of columns and eccentricity of load., a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry, He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at, Subscribe to his free Masterclasses at & technical discussions at,

#### What is an axial load on a truss?

Axial loading is defined as applying a force on a structure directly along an axis of the structure. As an example, we start with a one-dimensional (1D) truss member formed by points P 1 and P 2, with an initial length of L (Fig.1.2) and a deformed length of L′, after axial loading is applied.

## Is axial force same as tension?

What Does Axial Tension Force Mean? – Axial tension force can be defined as the force acting on a body in its axial direction. It’s a pulling force that will cause the body to elongate linearly in the positive direction causing a change in its dimension.

The axial tension force is exactly opposite to the axial compression force where the body will experience a change in dimension due to compression in the negative direction. The axial tensile force or stretching forces acting on the body has two components, namely: tensile stress and tensile strain. This means that the material experiencing the force is under tension and the forces are trying to stretch it.

When a tensile force is applied to a material, it develops a stress corresponding to the applied force, contracting the cross-section and elongating the length.

## How are roof loads transferred?

In traditional sloped roof framing, loads are transmitted vertically downward through the roof framing to the exterior walls that transmit the load downward to the home’s foundation and, ultimately, the ground. Loads are transferred downward and outward through sloping rafters, the lower ends of which rest on the top plates of the exterior walls; the vertical load on the roof is transferred to the walls at this point. The upper ends of the rafters rest against a ridge beam which, in a typical gable roof, does provide support for the roof load.

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• Page ID 70258
• Key Questions

• What is a distributed load?
• Given a distributed load, how do we find the magnitude of the equivalent concentrated force?
• Given a distributed load, how do we find the location of the equivalent concentrated force?

Distributed loads are forces which are spread out over a length, area, or volume. Most real-world loads are distributed, including the weight of building materials and the force of wind, water, or earth pushing on a surface. Pressure, load, weight density and stress are all names commonly used for distributed loads.

Distributed load is a force per unit length or force per unit area depicted with a series of force vectors joined together at the top, and will be designated as $$w(x)$$ to indicate that the distributed loading is a function of $$x\text$$ For example, although a shelf of books could be treated as a collection of individual forces, it is more common and convenient to represent the weight of the books as as a uniformly distributed load,

A uniformly distributed load is a load which has the same value everywhere, i.e. $$w(x) = C\text$$ a constant (a) A shelf of books with various weights. (b) Each book represented as an individual weight (c) All the books represented as a distributed load. Figure 7.8.1. We can use the computational tools discussed in the previous chapters to handle distributed loads if we first convert them to equivalent point forces. This equivalent replacement must be the resultant of the distributed loading, as discussed in Section 4.7.

• Magnitude equal to the the area or volume under the distributed load function.
• Line of action that passes through the centroid of the distributed load distribution.

The next two sections will explore how to find the magnitude and location of the equivalent point force for a distributed load. The magnitude of the distributed load of the books is the total weight of the books divided by the length of the shelf \ It represents the average book weight per unit length.

1. Similarly, the total weight of the books is equal to the value of the distributed load times the length of the shelf or \begin W \amp = w(x) \ell\\ \text \amp = \frac } } \times\ \text \end This total load is simply the area under the curve $$w(x)\text$$ and has units of force.
2. If the loading function is not uniform, integration may be necessary to find the area.

Example 7.8.2, Bookshelf. A common paperback is about $$\cm$$ thick and weighs approximately $$\N \text$$ What is the loading function $$w(x)$$ for a shelf full of paperbacks and what is the total weight of paperback books on a $$\m$$ shelf? Answer \begin w(x) \amp = \Nperm \\ W \amp = \N \end Solution The weight of one paperback over its thickness is the load intensity $$w(x)\text$$ so \ The total weight is the the area under the load intensity diagram, which in this case is a rectangle.

So, a $$\m$$ bookshelf covered with paperbacks would have to support \ The line of action of this equivalent load passes through the centroid of the rectangular loading, so it acts at $$x = \m \text$$ To use a distributed load in an equilibrium problem, you must know the equivalent magnitude to sum the forces, and also know the position or line of action to sum the moments.

The line of action of the equivalent force acts through the centroid of area under the load intensity curve. For a rectangular loading, the centroid is in the center. We know the vertical and horizontal coordinates of this centroid, but since the equivalent point force’s line of action is vertical and we can slide a force along its line of action, the vertical coordinate of the centroid not important in this context.

1. Similarly, for a triangular distributed load — also called a uniformly varying load — the magnitude of the equivalent force is the area of the triangle, $$bh/2$$ and the line of action passes through the centroid of the triangle.
2. The horizontal distance from the larger end of the triangle to the centroid is $$\bar = b/3\text$$ Essentially, we’re finding the balance point so that the moment of the force to the left of the centroid is the same as the moment of the force to the right.

The examples below will illustrate how you can combine the computation of both the magnitude and location of the equivalent point force for a series of distributed loads. Example 7.8.3, Uniformly Varying Load. Find the equivalent point force and its point of application for the distributed load shown. Answer The equivalent load is $$\lb$$ downward force acting $$\ft$$ from the left end. Solution 1 The equivalent load is the ‘area’ under the triangular load intensity curve and it acts straight down at the centroid of the triangle. This triangular loading has a $$\ft$$ base and a$$\lbperft$$ height so \ and the centroid is located $$2/3$$ of the way from the left end so, \ Solution 2 Distributed loads may be any geometric shape or defined by a mathematical function.

1. If the load is a combination of common shapes, use the properties of the shapes to find the magnitude and location of the equivalent point force using the methods of Section 7.5.
2. If the distributed load is defined by a mathematical function, integrate to find their area using the methods of Section 7.7.

A few things to note:

• You can include the distributed load or the equivalent point force on your free body diagram, but not both !
• Since you’re calculating an area, you can divide the area up into any shapes you find convenient. So, if you don’t recall the area of a trapezoid off the top of your head, break it up into a rectangle and a triangle.

Once you convert distributed loads to the resultant point force, you can solve problem in the same manner that you have other problems in previous chapters of this book. Note that while the resultant forces are externally equivalent to the distributed loads, they are not internally equivalent, as will be shown Chapter 8. Answer \begin A_x\amp = 0\\ A_y \amp = \N(16)\\ M \amp = \Nm \end Solution Draw a free body diagram with the distributed load replaced with an equivalent concentrated load, then apply the equations of equilibrium. \begin \Sigma F_x \amp = 0 \amp \amp \rightarrow \amp A_x \amp = 0\\ \Sigma F_y \amp = 0 \amp \amp \rightarrow \amp A_y \amp = \N \\ \Sigma M_A \amp = 0 \amp \amp \rightarrow \amp M_A \amp = (\N )(\m ) \\ \amp \amp \amp \amp \amp = \Nm \end Example 7.8.5, Beam Reactions. Find the reactions at the supports for the beam shown. Answer \ Solution 1 \begin \sum M_B \amp = 0\\ +(\lbperin )(\inch ) (\inch ) -(\lb ) (\inch )\\ -(\lb )(\inch ) -(\lb ) ( \inch )\\ +(F_y) (\inch ) – (\lbperin ) (\inch ) (\inch )\amp = 0 \rightarrow \amp F_y \amp= \lb \\ \\ \sum F_y\amp = 0\\ -(\lbperin ) (\inch ) + B_y – \lb – \lb \\ – \lb +F_y – (\lbperin )( \inch )\amp = 0 \rightarrow \amp B_y\amp= \lb \\ \\ \sum F_x \amp = 0 \rightarrow \amp B_x \amp = 0 \end Solution 2 1.

### Can a truss have a distributed load?

In skeletal structures, the distributed loads act along the members that are defined as lines in the structural model. In truss analysis, distributed loads are transformed into equivalent nodal loads, and the effects of bending are neglected.

## How loads are transferred through a structure?

By shear wall/diaghragm function the load is transferred to the facade. The load from the roof area is transferred through the trusses (beam function) to the facades and then through beams, columns and masonry to the foundations (column function).

## How are structural loads transferred?

Walls, like columns, transmit loads by compressive force to the floor below, another wall, or earth through the foundation wall. The wall unit will react to force like a long slender column. A wall may also be required to act like a beam, resisting flexing force such as a wind load.

## Where are loads and supports applied in truss structures?

Use of trusses in buildings – Trusses are used in a broad range of buildings, mainly where there is a requirement for very long spans, such as in airport terminals, aircraft hangers, sports stadia roofs, auditoriums and other leisure buildings, Trusses are also used to carry heavy loads and are sometimes used as transfer structures.

• To carry the roof load
• To provide horizontal stability.

Two types of general arrangement of the structure of a typical single storey building are shown in the figure below.

Typical truss building arrangements

 Lateral stability provided by columns and truss connected to form a frame. Longitudinal stability provided by transverse wind girder and vertical cross bracing (blue) No longitudinal wind girder. Building braced in both directions. Lateral stability provided by longitudinal wind girder and vertical bracing in the gables (blue) Longitudinal stability provided by transverse wind girder and vertical bracing (green). Vertical bracing is generally provided on both elevations.

In the first case (left) the lateral stability of the structure is provided by a series of frames formed from a truss and columns; the connections between the truss and the columns provide resistance to a global bending moment. Loads are applied to the portal structure by purlins and side rails,

In the second case, (right) each truss and the two columns between which it spans, constitute a simple structure; the connection between the truss and a column does not resist the global bending moment, and the two column bases are pinned. Bracing in both directions is necessary at the top level of the simple structure; it is achieved by means of a longitudinal wind girder which carries the transverse forces due to wind on the side walls to the vertical bracing in the gable walls.

Longitudinal stability is provided by a wind girder in the roof and vertical bracing in the elevations.

## How are forces distributed?

A distributed force is a force that acts on a large part of a surface, not just on one place. The loading on the beam can be a distributed force or a force that acts at a single point. The intensity of a distributed force is the force per unit length, area, or volume.

### Can a truss have a distributed load?

In skeletal structures, the distributed loads act along the members that are defined as lines in the structural model. In truss analysis, distributed loads are transformed into equivalent nodal loads, and the effects of bending are neglected.

#### Where do the forces act in a truss?

A truss is a series of individual members, acting in tension or compression and performing together as a unit. On truss bridges, a tension member is subject to forces that pull outward at its ends. Even on a “wooden” truss bridge, these members are often individual metal pieces such as bars or rods. One bridge historian describes a truss bridge in this manner: “A truss is simply an interconnected framework of beams that holds something up. The beams are usually arranged in a repeated triangular pattern, since a triangle cannot be distorted by stress.

• In a truss bridge, two long – usually straight members known as chords – form the top and bottom; they are connected by a web of vertical posts and diagonals.
• The bridge is supported at the ends by abutments and sometimes in the middle by piers.
• A properly designed and built truss will distribute stresses throughout its structure, allowing the bridge to safely support its own weight, the weight of vehicles crossing it, and wind loads.

The truss does not support the roadway from above, like a suspension bridge, or from below, like an arch bridge; rather, it makes the roadway stiffer and stronger, helping it hold together against the various loads it encounters.” (Eric DeLony, The Golden Age, Invention and Technology, 1994).

1. The pattern formed by the members combined with the stress distribution (tension and compression) creates a specific truss type, such as a Warren or Pratt.
2. Most truss types bear the name of the person(s) who developed the pattern, such as the Pratt truss that is named for Caleb and Thomas Pratt who patented it in 1844.

For instance, the configuration or pattern of a Pratt and Howe truss appears identical (a series of rectangles with X’s), but a Howe’s diagonals are in compression and the verticals in tension. In a Pratt, the reverse is true. In theory, a truss bridge contained no redundant members.

• Builders considered each member or element essential to the functioning of the truss, although some were more important than others were.
• While most trusses could sustain considerable damage and lose the support of some members without collapsing, severe traffic damage to a member could result in the collapse of the bridge.

Tennessee’s four remaining historic covered bridges utilize one of these three truss types:

Kingpost (the Parks Covered Bridge Queenpost (the Harrisburg Covered Bridge and the Bible Covered Bridge Howe (Elizabethton Covered Bridge)

Kingpost Builders first developed the Kingpost as the most basic and earliest truss type. The outline consisted of two diagonals in compression and a bottom chord in tension that together formed a triangular shape. A vertical tension rod (called a Kingpost and thus the origin of the truss name) divided the triangle in half. Queenpost The Queenpost, another early and basic truss type, is a variation of the Kingpost truss. A Queenpost truss contains two vertical members (rather than the one in a Kingpost). These vertical members require the use of a top chord to connect them. Howe Truss William Howe patented the Howe truss in 1840. End diagonals connect the top and bottom chords, and all wood members act in compression. Each panel has a diagonal timber compression member and a vertical metal tension member, a material that conducts tensile forces better than wood.

#### How are loads distributed in structural system?

The load is transferred from the wall area by slab/beam function to vertical wind beams and then further to gable foundation and roof area. Through the purlins of the roof area (compression members) the load is transferred further to wind bracings and then by tension to the foundation of the facade.