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Calculating the “actual” internal force in truss bridge members

Calculating the “actual” internal force in truss bridge members. EGN1006. Learning Goals: The student WILL be able to:. Calculate the components of a force vector . Add two force vectors together . Draw a free body diagram.

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Calculating the “actual” internal force in truss bridge members

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  1. Calculating the “actual” internal force in truss bridge members EGN1006

  2. Learning Goals: The student WILL be able to: • Calculate the components of a force vector. • Add two force vectors together. • Draw a free body diagram. • Calculate whether a truss is statically determinate or indeterminate. • Write and solve a force vector equilibrium equation. • Use the Method of Joints to calculate the internal force in every member in a truss. • Evaluate a truss, to determine if it can carry a given load safely by calculating factors of safety for individual members.

  3. Static Determinacy and Stability A structure that cannot be analyzed using the equations of equilibrium alone is called statically indeterminate. A structure that can be analyzed using the equations of equilibrium alone is called statically determinate. Only statically determinate trusses can be analyzed with the Method of Joints. A statically determinate truss with two reactions must satisfy the mathematical equation Where j is the number of joints and m is the number of members.

  4. Static Determinacy and Stability 6 How many joints does this bridge have? 10 How many members does this bridge have? Is this bridge statically determinate? In other words does 2j = m+3? No , 2(6) = 12 and 10+3 = 13 Thus 2J does NOT equal m+3

  5. Static Determinacy and Stability 6 How many joints does this bridge have? 9 How many members does this bridge have? Is this bridge statically determinate? In other words does 2j = m+3? Yes , 2(6) = 12 and 9+3 = 12 Thus 2J does equal m+3

  6. Free Body Diagram A free body diagram is a pictorial representation of all of the forces which act on an object. Suppose we have a box being pushed by an applied force to the right. What forces act on the box? • Weight, aka, the Force due to gravity , - This force is ALWAYS drawn straight down. • Normal Force – The force that a SURFACE applies on an object. Always drawn PERPENDICULAR to the surface • Applied Force - Either a PUSH or a PULL • Friction force – the force that ALWAYS opposes the motion. Drawn at the surface.

  7. Free Body Diagrams Often times the free body diagram is drawn using what is called a POINT MODEL. The object is drawn as a single point with the forces labeled as “F”. A subscript is added according to the type of force it is.

  8. Free Body Diagrams FT1 FT2 Fg In the case of bridge members and ropes, we have a special type of force called TENSION. Since you can’t PUSH a rope, the tension is ALWAYS drawn as if your are pulling the object. In other words, it is always drawn AWAY from the object. If there are multiple ropes, subscripts must be used to classify them separately.

  9. Equations of Equilibrium When an object is at rest, the SUM of all of the FORCE vectors must be equal to zero. So when you write your equations they MUST equal ZERO.

  10. What is a vector? A vector is any quantity which has both MAGNITUDE (# and a unit) and DIRECTION. The vector is always represented as an ARROW. Suppose the vector below represents a displacement of 10m. 10 m, NORTH-EAST MAGNITUDE DIRECTION

  11. A vector’s direction Direction is best described by using a Cartesian Coordinate system. Forces on the negative x or negative y axis must have a negative sign. Using this idea allows us to write the equation of equilibrium. Assume the object is at rest. Fa2=? N Fa1=+10 N

  12. Let’s look at JOINT B on our truss FBI FAB FBC Assume all forces are TENSION!

  13. Joint B – Equations of Equilibrium FBI FAB FBC If you knew the FORCE in member AB, you would be able to solve for the FORCE in member BC. Isolating just ONE JOINT to analyze the force is called the METHOD OF JOINTS.

  14. Some Basic Concepts from Trigonometry A truss is a structure composed of members arranged in interconnected triangles. For this reason, the geometry of triangles is very important in structural analysis. This diagram shows a right triangle—a triangle with one of its three angles measuring exactly 90o. Sides a and b form the 90o angle. The other two angles, identified as θ1and θ2, are always less than 90o. Side c, the side opposite the 90o angle, is always the longest of the three sides. It is called the hypotenuse of the right triangle. Thanks to an ancient Greek mathematician named Pythagoras, we can easily calculate the length of the hypotenuse of a right triangle. The Pythagorean Theorem tells us that:

  15. Some Basic Concepts from Trigonometry The Pythagorean Theorem shows how the lengths of the sides of a right triangle are related to each other. But how are the lengths of the sides related to the angles? The sine of an angle (abbreviated “sin”) is defined as the length of the opposite side divided by the length of the hypotenuse. For example, the sine of the angle θ1would be calculated as: The cosine of an angle (abbreviated “cos”) is defined as the length of the adjacent side divided by the length of the hypotenuse. Applying this definition to our example, we have:

  16. Breaking a Vector into its Components Once the coordinate axis system is established, we can represent the direction of any vector as an angle measured from either the x-axis or the y-axis. For example, the force vector at right has a magnitude (F) of 20 Newtons and a direction (θ) of 50 degrees, measured counterclockwise from the x-axis. This force can also be represented as two equivalent forces, one in the x-direction and one in the y-direction. Each of these forces is called a component of the vector F.

  17. What if a member is at an angle?

  18. Calculating the Vector’s Components

  19. Let’s get started The required load our bridge must withstand is 49N or 5-kg. Since there are TWO trusses held together by lateral bracings, HOW much load does ONE truss bridge hold? 24.5 N

  20. Let’s get started The load acts downward at joints J, K, and L. How much force acts at each one of these locations? 8.17 N

  21. Let’s get started The two upward forces are both force normals. The “R” in this case stands for REACTION as they are a reaction to the load. How much force does each REACTION FORCE(force normal) support? 12.25 N

  22. Joint A RA FAI FAB

  23. Force AI must be broken into components FAI RA FAI q q FAB What is the VALUE of the angle THETA? LengthAI LengthBI q What other angle are also equal to theta? LengthAB

  24. Where is theta? q q q q q

  25. Force AI’s Components FAI RA FAI FAIsinq q q FAB FAIcosq Let’s now REDRAW the FBD! FAIsinq RA FAIcosq FAB

  26. Joint A’s Equations of Equilibrium FAIsinq RA FAIcosq FAB This force is COMPRESSION and NOT tension, thus it is a TUBE!

  27. Your task Use the Method of Joints to solve for the rest of the internal forces. Use the calculation guide for reference and to keep organized. Wait, there is ONE last thing…..

  28. Factor of Safety When an engineer designs a structure, he or she must consider many different forms of uncertainty. There are three major types of uncertainty that affect a structural design: • There is always substantial uncertainty in predicting the loads a structure might experience at some time in the future. • The strengths of the materials that are used to build actual bridges are also uncertain. • The mathematical models we use for structural analysis and design are never 100% accurate.

  29. Factor of Safety The engineer accounts for all forms of uncertainty by making the structure somewhat stronger than it really needs to be—by using a factor of safety in all analysis and design calculations. In general, when it is used in the analysis of an existing structure, the factor of safety is a defined as In a truss, the actual force in a member is called the internal member force, and the force at which failure occurs is called the strength. Thus we can rewrite the definition of the factor of safety as

  30. Factor of Safety For example, if a structural member has an internal force of 5000 pounds and a strength of 7500 pounds, then its factor of safety, FS, is If the factor of safety is less than 1, then the member or structure is clearly unsafe and will probably fail. If the factor of safety is 1 or only slightly greater than 1, then the member or structure is nominally safe but has very little margin for error—for variability in loads, unanticipated low member strengths, or inaccurate analysis results. Most structural design codes specify a factor of safety of 1.6 or larger (sometimes considerably larger) for structural members and connections.

  31. The next step • Calculate the internal member force • Use the previously found strengths to calculate the factors of safety for each bridge member

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