ENSC1002StaticsLectureNotes.pdf

# Just like we established a convention for external

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Just like we established a convention for external forces (positive x positive y and positive moment ), we must establish a convention for internal forces. In mechanics it is common to take tension forces as positive and compression forces as negative. This is the convention we will use.

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Week 10 Lecture 1 3 Later in this course we will look at "beams" which are members that can carry axial loads (tension or compression) as well as shear forces and bending moments (more on that later in the course). Members of a truss are commonly referred to as props, ties or bars. They are always "pinned" together so that one member of a truss is free to rotate with respect to the other. In a truss structure, loads are always applied at the pin joints. Support conditions will be either pin supports or roller supports. The above truss is statically determinate (3 unknown reaction forces) so we can easily solve for the reactions at the supports.
Week 10 Lecture 1 4

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Week 10 Lecture 1 5 we now know all the "external" forces acting on the truss. what about the internal axial forces in the members Mab, Mac and Mcb?? For the trust to be in equilibrium, each of the pin joints must be in equilibrium. i.e.
Week 10 Lecture 1 6 Consider the forces acting at Joint A by drawing a FBD Note the line of action of each member force acting on the joint is known from the geometry of the truss. However, we do not know if the member forces are in tension or compression. When we solved for external reactions, we always drew FBDs assuming the unknown forces and moments act in the positive external direction. When solving for internal forces, assume that unknowns are in tension (i.e positive internal convention). A tensile force will always act away from the joint, as indicated by the arrows on the members above. To solve, apply equilibrium by summing forces in x and y (note all forces act through the joint so ).

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Week 10 Lecture 1 7 Now we know Fac, substitute back into horizontal equilibrium and solve for Fab That is the internal axial forces are In solving for the member forces we assumed they were both in tension (positive by convention). The negative sign for Member AC indicates it is in compression. we can now draw the FBD with the known magnitude and direction of internal forces. T = tension; C=compression
Week 10 Lecture 1 8 Now consider Joint B to solve for Fbc In drawing the FBD of Joint B, we know Member AB is in tension ( ) so it will act away from Joint B (tensile forces always act away from the isolated joint). By convention we assume Fbc is tensile, so draw it acting away from the joint. Now sum forces This implies that Fbc is in compression. Re-draw as it acts

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Week 10 Lecture 1 9 we found that value of Fbc by considering horizontal equilibrium. Check that vertical equilibrium is satisfied..
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