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# Ahsanullah university

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### Ahsanullah university

1. 1. WELCOME TO MY PRESENTATION
2. 2. AHSANULLAH UNIVERSITY OF SCIENCE AND TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING CE-416;PRESTRESS CONCRETE LAB COURSE TEACHER:MD.GALIB MUKTADIR PRESENTATION OF AXIAL FORCE PRESENTED BY H.M SURUZZAMAN ID NO:10.01.03.63
3. 3. BACKGROUND: Axial force is a force that tends to elongate or shorten a member & is normally measured in pounds. It is a system of internal forces whose resultant is force that is acting along the longitudinal axis of a structural member or assembly.
4. 4.  Axial force is the compression or tension force acting in a member. If the axial force acts through the censored of the member it is called concentric loading. If the force is not acting through the centroid it is called eccentric loading.  Eccentric loading produces a moment in the beam as a result of the load being a distaces away from the centroid.
5. 5. Learning objectives are: -Understanding the theory ; it’s limitations, and it’ supplications for design & analysis of axial members Axial members: Members with length significantly greater than the largest cross sectional dimension & with load applied along the longitudinal axis. -Developing the discipline to draw free body diagrams & approximate deformed shapes in the Design & analysis of structures.
6. 6. Nature of axial force: -These forces are typically stretching force or compression force, depending on direction. -Shear forces occupies a similar position to axial force, but operates perpendicular to the centre axis of the object. -When a force is acting directly on the central axis, it is an axial force. These force will often compress the axis from either end or stretch the axis in two opposing directions; as a result the object typically does not move.
7. 7. Figure of Axial Force
8. 8.  Example of axial force:A prime example of these forces can be seen on columns within buildings.  The column has an axis that runs through the entire from top to bottom. The column is constantly compressed as it supports the roof of the structure. In the column example, the axial force runs through the geometric centre of the form.
9. 9.  ColumnForce on different members: Axial   Beam
10. 10.  Axial Force on different members:  Shaft  Cylinder
11. 11.  Calculation Formula : Axial loading occurs when an object is loaded so that the force is normal to the axis that is fixed, as seen in the figure. Taking statics into consideration the force at the wall should be equal to the force that is applied to the part.
12. 12. APPROACH OF WORK:  Used to also solve statically indeterminate problems by using superposition of the forces acting on the freebody diagram  First, choose any one of the two supports as “redundant” and remove its effect on the bar  Thus, the bar becomes statically determinate  Apply principle of superposition and solve the equations simultaneously
13. 13. = +
14. 14. Compatibility Choose one of the supports as redundant and write the equation of compatibility. Known displacement at redundant support (usually zero), equated to displacement at support caused only by external loads acting on the member plus the displacement at the support caused only by the redundant reaction acting on the member.
15. 15. Equilibrium Draw a free-body diagram and write appropriate equations of equilibrium for member using calculated result for redundant force. Solve the equations for other reactions
16. 16. EXAMPLE A-36 steel rod shown has diameter of 5 mm. It’s attached to fixed wall at A, and before it is loaded, there’s a gap between wall at B’ and rod of 1 mm. Determine reactions at A and B’.
17. 17. Compatibility Consider support at B’ as redundant. Use principle of superposition, (+) 0.001 m = δP −δB Equation 1
18. 18. Compatibility Deflections δP and δB are determined from Eqn. 4-2 PLAC δP = = … = 0.002037 m AE FB LAB δB = = … = 0.3056(10-6)FB AE Substituting into Equation 1, we get 0.001 m = 0.002037 m − 0.3056(10-6)FB FB = 3.40(103) N = 3.40 kN
19. 19. Equilibrium From free-body diagram + Fx = 0; − FA + 20 kN − 3.40 kN = 0 FA = 16.6 kN
20. 20. THANK YOU