PRESENTATION ON AXIAL FORCE10.01.03.063

20
WELCOME TO MY PRESENTATION

description

 

Transcript of PRESENTATION ON AXIAL FORCE10.01.03.063

Page 1: PRESENTATION ON AXIAL FORCE10.01.03.063

WELCOME TO MY

PRESENTATION

Page 2: PRESENTATION ON AXIAL FORCE10.01.03.063

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

Page 3: PRESENTATION ON AXIAL FORCE10.01.03.063

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.

Page 4: PRESENTATION ON AXIAL FORCE10.01.03.063

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.

Page 5: PRESENTATION ON AXIAL FORCE10.01.03.063

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.

Page 6: PRESENTATION ON AXIAL FORCE10.01.03.063

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.

Page 7: PRESENTATION ON AXIAL FORCE10.01.03.063

Figure of Axial Force

Page 8: PRESENTATION ON AXIAL FORCE10.01.03.063

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.

Page 9: PRESENTATION ON AXIAL FORCE10.01.03.063

Axial Force on different members: Column Beam

Page 10: PRESENTATION ON AXIAL FORCE10.01.03.063

Axial Force on different members: Shaft Cylinder

Page 11: PRESENTATION ON AXIAL FORCE10.01.03.063

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.

Page 12: PRESENTATION ON AXIAL FORCE10.01.03.063

APPROACH OF WORK:Used to also solve statically indeterminate

problems by using superposition of the forces acting on the free-body diagram

First, choose any one of the two supports as “redundant” and remove its effect on the bar

Thus, the bar becomes statically determinateApply principle of superposition and solve the

equations simultaneously

Page 13: PRESENTATION ON AXIAL FORCE10.01.03.063

+=

Page 14: PRESENTATION ON AXIAL FORCE10.01.03.063

CompatibilityChoose 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.

Page 15: PRESENTATION ON AXIAL FORCE10.01.03.063

EquilibriumDraw a free-body diagram and write appropriate equations of equilibrium for member using calculated result for redundant force.Solve the equations for other reactions

Page 16: PRESENTATION ON AXIAL FORCE10.01.03.063

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’.

Page 17: PRESENTATION ON AXIAL FORCE10.01.03.063

CompatibilityConsider support at B’ as redundant. Use principle of superposition,

0.001 m = δP −δB

Equation 1( + )

Page 18: PRESENTATION ON AXIAL FORCE10.01.03.063

CompatibilityDeflections δP and δB are determined from Eqn. 4-2

δP =PLAC

AE= … = 0.002037 m

δB = FB LAB

AE= … = 0.3056(10-6)FB

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

Page 19: PRESENTATION ON AXIAL FORCE10.01.03.063

EquilibriumFrom free-body diagram

− FA + 20 kN − 3.40 kN = 0

FA = 16.6 kN

+ Fx = 0;

Page 20: PRESENTATION ON AXIAL FORCE10.01.03.063

THANK YOU