CE – 416 Pre-stress Concrete Design Sessional

Presented byFariya Rahman Moho

Student No. : 10.01.03.014

Course TeachersMunshi Galib MuktadirMs Sabreena Nasrin

Department of Civil EngineeringAhsanullah University of Science & Technology

Equilibrium & Equation of Equilibrium : 2D

The Concept Of Equilibrium

1. The concept of equilibrium is introduced to describe a body which is stationary or which is moving with a constant velocity. 

2. A body under such a state is acted upon by balanced forces and balanced couples only. 

3. There is no unbalanced force or unbalanced couple acting on it. 

Particles and Rigid Bodies

Particles : A particle is a body whose size does not have any effect on the results of mechanical analyses on it and, therefore, its dimensions can be neglected. 

Rigid body : A body is formed by a group of particles.  The size of a body affects the results of any mechanical analysis on it.  A body is said to be rigid when the relative positions of its particles are always fixed and do not change when the body is acted upon by any load (whether a force or a couple).   

Conditions for Equilibrium of a Rigid

Body For a rigid body which is not moving at all we have the following conditions:

1. The (vector) sum of the external forces on the rigid body must equal zero:

∑F = 0 When this condition is satisfied we say that the body is in translational

equilibrium.

2. The sum of the external torques on the rigid body must equal zero.

∑τ = 0 When this condition is satisfied we say that the body is in rotational

equilibrium. When these conditions are satisfied we say that the body is in static

equilibrium.

Equations of Equilibrium for Two Dimensions (2D) Rigid Body

y

xCouple moment

1F

2F

3F

1M

2M0M0F 0

0M

0F

0F

0

y

x

Here: Fx

Fy

0M

algebraic sum of x components of all force on the body.

algebraic sum of y components of all force on the body.

algebraic sum of couple moments and moments of allthe force components about an axis xy plane and⊥passing 0.

(1) Equilibrium Equation

Alternate Equations for 2D Equilibrium

O)point any (for 0 M 0 F 0 F Oyx

Most common equations for 2D equilibrium

line) verticalaon not B and(A 0 M 0 M 0 F BAx

Alternate equations of 2D equilibrium

line) horizontal aon not B and(A 0 M 0 M 0 F BAy

line) aon not C and B, (A, 0 M 0 M 0 M CBA

Note that with any of these sets of three equations, we will typically have three unknowns on our FBD.

BAR FFF 0

0Mo

Two-and Three-Force Members

1. Two-Force member

A member subject to no couple moments and forces applied at only two points on the member.

A

B

A

B

FA

FBEquations of Equilibrium

2. Three-Force member

0F 0F 0Mo 0Mo

A member subject to only three forces, which are either concurrent or parallel if the member is in equilibrium.

(1)Concurrent (2)parallel

o

F1F2

F3

F1 F2

F3

The Process of Solving Rigid-Body Equilibrium

Problems

Step 1: For analyzing an actual physical system, first we need to create an idealized model & identify any 2-force members .

Step 2: Then we need to draw a free-body diagram showing all the external (active and reactive) forces.

Step 3: Finally, we need to apply the equations of equilibrium to solve for any unknowns

The Process of Solving Rigid-Body Equilibrium Problems

(continued)

Free Body Diagram (FBD)

(1) F.B.D. • What ? - A sketch of the outlined shape of the body

represents it as being isolated or “free” from its surrounding , i.e ., a free body”.

It is a drawing that shows all external forces acting on the particle.

• Why ? - It helps to write the equations of equilibrium used

to solve for the unknowns . (usually forces or angles).

(a) roller or cylinder support

AyF

ByF

xF

Fy

xF

Fy

(b) pin support

Free Body Diagram (continued) (2) Support Reactions : A . Type of support :

(c) Fixed support

M

FAy

FAx

Free Body Diagram (continued)

B . General rules for support reaction:

If a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction . Likewise, if rotation is prevented, a couple moment is exerted on the body.

Free Body Diagram (continued)

(3) External and Internal forces

A. Internal force

Not represented on the F.B.D. became their net effect on the body is zero.

B. External force

Must be shown on the F.B.D.

(a) “Applied” loadings

(b) Reaction forces

(c) Body weights

(4) Weight and the center of gravity

Procedure for Drawing a Free-Body Diagram

1. Draw an outlined shape.

Imagine the body to be

isolated or cut “free” from its constraints and

draw its outlined shape.

2.Show all the external forces

and couple moments. These

typically include: a) applied loads,

b) support reactions, and, c) the weight of the

body.

3.Label loads and dimensions: All

known forces and couple moments should be labeled

with their magnitudes and

directions. For the unknown forces

and couple moments, use

letters like Ax, Ay, MA, etc.. Indicate

any necessary dimensions.

Procedure for Drawing a Free-Body Diagram

(continued)

1. Put the x and y axes in the horizontal and vertical directions, respectively.

2. Determine if there are any two-force members.

3. Draw a complete FBD of the beam.

4. Apply the Equation of Equilibrium to solve for the unknowns.

Given: The 4kN load at point B of the beam is supported by pins at A and C .

Find: The support reactions at A and C.

Plan:

Example

Note: Upon recognizing CD as a two-force member, the number of unknowns at C are reduced from two to one. Now, using the three equations of equilibrium:

FX = AX + 11.31 cos 45 = 0; AX = – 8.00 kN

FY = AY + 11.31 sin 45 – 4 = 0; AY = – 4.00 kN

MA = FC sin 45 1.5 – 4 3 = 0

Fc = 11.31 kN or 11.3 kN

FBD of the beam:

AX

AY

A

1.5 m

C B

4 kN

FC

45°

1.5 m

Note that the negative signs means that the reactions have the opposite direction to that shown on FBD.

Example (continued)

IMPORTANT NOTES

1. If there are more unknowns than the number of independent

equations, then we have a statically indeterminate situation.

We cannot solve these problems using just statics.

2. The order in which we apply equations may affect the

simplicity of the solution. For example, if we have two

unknown vertical forces and one unknown horizontal force,

then solving FX = 0 first allows us to find the horizontal unknown quickly.

3. If the answer for an unknown comes out as a negative number, then the

sense (direction) of the unknown force is opposite to that assumed when

starting the problem.

THANK YOU