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# PRESENTATION OF AXIAL FORCE DIAGRAM by 10.01.03.132

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Presentation of AXIAL FORCE DIAGRAM by SARAH SUBHA, I.D.: 10.01.03.132

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• @Kasvinda Kulathunga ,I hope you get your answer from the post, thank you.

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• Hi. Sorry I'm late. It can be explained in 2 different ways.Solution 1: 3KN has been given in question.If you add 3KN in the calculation for the segment T(CD), then you will get 0 which means in the end- diagram has to reach zero.So the last segment shows it is balanced 'IF' you add 3KN. And I've shown it in the force diagram,the diagram ends at zero line. Solution 2: If you want to find 'Why' the diagram comes to zero or where 3KN comes from then you can calculate like I've done.This implies that 'because' of tension +3KN,you'll be needing a -3KN to bring it to zero line of the diagram, since it is a balanced condition.Please look at the diagram,you'll see that till the end point 'D' ,the tension value is +3KN. So T(CD) is +3KN even at point D and after that the diagram moves to zero line to be balanced. I actually showed the second solution in the slide but explained the 1st solution orally in my class. Its difficult to explain deeply,I still tried,hope it helps you.Its a simple thing,think simply. Thank you.

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• Why 3kN is never taken to the account? Please explain .thank you

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### PRESENTATION OF AXIAL FORCE DIAGRAM by 10.01.03.132

1. 1. PRESENTATION ON AXIAL FORCE DIAGRAM By SARAH SUBHA 10.01.03.132 4th YEAR, 2nd SEMESTER DEPARTMENT OF CIVIL ENGINEERING AHSANULLAH UNIVERSITY OF SCIENCE AND TECHNOLOGY
2. 2. WHAT IS AXIAL FORCE? If the load on a structure is applied through the center of gravity of its cross section, it is called an axial load or that would be force applied to the lengthwise centerline of an object. Axial force is the compression or tension force of the member. Here, in the figure, the force, P is being applied at the centroid, C, which is the main condition of axial force. Axial load, torque, shear force are the Internal Forces of a structure.
3. 3. LOADING DEFINED BY AXIAL FORCE  If the axial force acts through the centroid of the member it is called concentric loading.  If the force is not acting through the centroid it's called eccentric loading. Eccentric loading produces a moment in a structure as a result of the load being a distance away from the centroid.
4. 4. WHAT IS AXIAL FORCE DIAGRAM (AFD)?  In statics, a graphical representation of the axial load acting at each section of a structural member, plotted to scale and with proper sign as an ordinate at each point of the member is known as AXIAL FORCE DIAGRAM. It acts along a reference line representing the length of the member.  The difference between the axial force and shear force and bending moment is that the axial forces are a pure compression or tension force that are applied through the member axes, where the shear force and bending moment, i.e. the radial forces act perpendicular to the member axes.
5. 5. CHARACTERISTICS OF AFD • Axial force diagrams change abruptly at the location of an axial force (or applied force) • It always acts along a reference point • The sign convention is simple, if one direction of the force is assumed positive, the other direction will be negative. • The forces indicate tension and compression only. Here, T(tension) and C(compression) are indicating the axial forces after cutting the Portal Frame.
6. 6. BASIC STEPS TO DRAW AFD Following some easy steps, the AFD can be developed from the internal forces at any point in a structure by: (1) Taking an imaginary cut at the point; (2) Drawing the free-body diagram; (3) Using the equations of equilibrium to calculate the internal forces and (4) Finally, plotting the values graphically according to their signs.
7. 7. MATHEMATICAL AND GRAPHICAL REPRESENTATION OF AFD Let, ABCD is a rectangular beam. Now the AFD of this beam is going to be drawn. 1. Take section 1-1, AB 2. If the segment is in equilibrium, internal force TAB must be equal to the externally applied force at A ∑Fx = TAB - 20 = 0 TAB = 20kN
8. 8. 1. Take section 2-2, BC 2. Considering the segment in equilibrium, internal force TBC must be equal to the externally applied force. ∑Fx = TBC + 22 - 20 = 0 TBC = - 2 kN
9. 9. 1. Now taking section 3-3, CD 2. Considering the segment in equilibrium, internal force TCD, ∑Fx = TCD + 22 - 20 – 5 = 0 TCD= 3 kN
10. 10. Finally, TAB = 20 kN TBC = -2 kN TCD = 3 kN The internal axial force diagram is drawn with tensile force above the zero line and compressive forces below:
11. 11. AFD OF INDETERMINATE STRUCTURES • First, to analyze the structures in a specific method. For analyzing frames under lateral load, the methods can be Portal method, Cantilever method, Factor method etc. Also, for analyzing frames under vertical load, the Approximate method can be taken • Calculate the values of the internal forces • Draw the AFD of each member along the reference line. Figures below are AFD of frame, beam and column respectively, based on the Portal Method.
12. 12. WHY AXIAL FORCE DIAGRAM IS IMPORTANT? The challenge with a member is that the internal forces can vary a greatly along the length. So, the diagram is drawn. • The diagram shows the internal forces along the length of the structure. This helps to visualize where the maximum stresses will occur due to axial force or axial loading. • Since the axial force variations affect the strength of the elements, the columns, with the greater axial compression caused by the vertical motion, tends to reach greater values of the moment in the plastic range. This increment of the maximum moment can be very high in the interior and in the upper columns which can be realized from diagram. • The greater tensile axial loads cause an approach to the shear failure in columns, so the diagram can help to modify design. • It helps to design the structure according to the stresses, occurring at specific points.
13. 13. THANK YOU