equilibrium and equation of equilibrium:2d

WELCOME TO MY WELCOME TO MY PRESENTATIONPRESENTATION

AHSANULLAH UNIVERSITY AHSANULLAH UNIVERSITY OF OF

SCIENCE AND TECHNOLOGYSCIENCE AND TECHNOLOGY

PRE-STRESSED CONCRETE PRE-STRESSED CONCRETE SESSIONALSESSIONAL

PRESENTED BYPRESENTED BY

MUNSHI MD. RASELMUNSHI MD. RASEL

ID: ID: 10.01.03.07510.01.03.075

SECTION BSECTION B

44thth YEAR AND 2 YEAR AND 2ndnd SEMESTER SEMESTER

DEPARTMENT OF CIVIL ENGINEERINGDEPARTMENT OF CIVIL ENGINEERING

PRESENTATION TOPICPRESENTATION TOPIC

EQUILIBRIUM AND EQUATION OF EQUILIBRIUM AND EQUATION OF EQUILIBRIUM:2DEQUILIBRIUM:2D

EquilibriumEquilibrium

A body is said to be in equilibrium if it is at rest or moving with uniform velocity.

Newton’s First Law of Motion: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line.

Factors that affect equilibrium

Area of the base: the bigger the area of the base, the more the stable the object is

Weight: the heavier the object is, the more stable it is

Types of EquilibriumTypes of Equilibrium

Static EquilibriumStatic EquilibriumDynamic EquilibriumDynamic Equilibrium

Static EquilibriumStatic Equilibrium

If some forces are If some forces are acting on a body acting on a body horizontally or horizontally or vertically, and the vertically, and the body remains it states body remains it states of rest is called Static of rest is called Static Equilibrium. Example: Equilibrium. Example: A book lying on a A book lying on a table.table.

Dynamic Equilibrium Dynamic Equilibrium

If some forces are If some forces are acting on a body acting on a body horizontally or horizontally or vertically, and the vertically, and the body remains it states body remains it states of motion is called of motion is called Dynamic Equilibrium. Dynamic Equilibrium. Example: A train is Example: A train is moving with uniform moving with uniform velocity.velocity.

Equations of equilibriumEquations of equilibrium

Consider an object Consider an object moving along the x-moving along the x-axis. If no net force is axis. If no net force is applied to the object applied to the object along the x-axis, it will along the x-axis, it will continue to move continue to move along the x-axis at a along the x-axis at a constant velocity with constant velocity with no acceleration. We no acceleration. We can extend this to the can extend this to the y- and z- axes. y- and z- axes.

In static systems, where motion does not In static systems, where motion does not occur, the sum of the forces in all occur, the sum of the forces in all directions must always equal zero directions must always equal zero (otherwise, it's a dynamics problem). This (otherwise, it's a dynamics problem). This concept can be represented concept can be represented mathematically with the following mathematically with the following equations:equations:

∑∑Fx=0Fx=0

∑∑Fy=0Fy=0

∑∑Fz=0Fz=0

The concept also applies to rotational The concept also applies to rotational motion.motion.

If the resultant moment about an axis is If the resultant moment about an axis is zero, the object will have no rotational zero, the object will have no rotational acceleration about the axis. Again, we can acceleration about the axis. Again, we can extend this to moments about the y-axis extend this to moments about the y-axis and the z-axis. This is represented and the z-axis. This is represented mathematically with the following.mathematically with the following.

∑∑Mx=0Mx=0

∑∑My=0My=0

∑∑Mz=0Mz=0

∑∑F=0 : The algebraic F=0 : The algebraic sum of all the sum of all the horizontal or vertical horizontal or vertical forces acting on a forces acting on a body which is in body which is in equilibrium must equilibrium must equal zero.equal zero.

∑∑M=0: The algebraic M=0: The algebraic sum of the moments sum of the moments of all the forces acting of all the forces acting on a body which is in on a body which is in equilibrium, about any equilibrium, about any point in the plane of point in the plane of those forces, must those forces, must equal zero.equal zero.

There are six equations expressing the There are six equations expressing the equilibrium of a rigid body in 3 dimensions. equilibrium of a rigid body in 3 dimensions. ∑Fx=0∑Fx=0 ∑Fy=0 ∑Fy=0 ∑Fz=0 ∑Fz=0

∑∑Mx=0Mx=0 ∑My=0 ∑My=0 ∑Mz=0 ∑Mz=0

In two dimensions one direction of force In two dimensions one direction of force and two directions of moments can be and two directions of moments can be ignored. When forces exist only in the x ignored. When forces exist only in the x and y directions, there cannot be a and y directions, there cannot be a moment in any direction except z. The moment in any direction except z. The equations of concern when forces only equations of concern when forces only exist in the x and y directions are shown exist in the x and y directions are shown below:below:

∑∑Fx=0Fx=0

∑ ∑Fy=0 Fy=0

∑ ∑Mz=0Mz=0

How to apply equations of How to apply equations of equilibrium?equilibrium?

First draw a free body diagram of the First draw a free body diagram of the structure or its member.structure or its member.

If a member is selected, it must be isolated If a member is selected, it must be isolated from its supports and surroundings.from its supports and surroundings.

All the forces and couple moments must All the forces and couple moments must be shown acting on the member.be shown acting on the member.

Then apply the equations of equilibrium.Then apply the equations of equilibrium.

A

FB

FD

30˚

FBD at A

A

y

x

Area to be cut or isolated

ExampleExample

Find the reactions at support of the Find the reactions at support of the following beam:following beam:

Applying eqn of equilibrium:Applying eqn of equilibrium:

∑∑Fx=0; Fx=0;

Ax=0Ax=0

∑ ∑Fy=0; Fy=0;

Ay+By-10-20*4=0Ay+By-10-20*4=0

Ay+By=90kn………(1) Ay+By=90kn………(1)

Considering Z axis passing through A and Considering Z axis passing through A and taking moment of all the forces about Z-taking moment of all the forces about Z-axis (taking clockwise –ve and axis (taking clockwise –ve and anticlockwise +ve)anticlockwise +ve)

∑∑Mz=0; By*10-10*8-20*4*2=0Mz=0; By*10-10*8-20*4*2=0

By=24knBy=24kn

Putting this value in eqn (1) we get,Putting this value in eqn (1) we get,

Ay=66kn.Ay=66kn.

2D Equilibrium - Applications

Since the forces involved in supporting the spool lie in a plane, this is essentially a 2D equilibrium problem. How would you find the forces in cables AB and AC?

For a given force exerted on the boat’s towing pendant, what are the forces in the bridle cables? What size of cable must you use?

This is again a 2D problem since the forces in cables AB, BC, and BD all lie in the same plane.

2D Equilibrium - Applications2D Equilibrium - Applications

SummarySummary

• In order for an object to be in equilibrium, there must be no net force on it along any coordinate, and there must be no net torque around any axis.

• 2D equations of equilibrium:

∑∑Fx=0Fx=0

∑ ∑Fy=0 Fy=0

∑ ∑Mz=0Mz=0

Thanks for your kind attentionThanks for your kind attention

Any Questions?Any Questions?