Statics - Philadelphia University · 2 Equilibrium condition ... Coplanar Forces. 11/7/2015 10...

Post on 26-Aug-2020

5 views 0 download

Transcript of Statics - Philadelphia University · 2 Equilibrium condition ... Coplanar Forces. 11/7/2015 10...

11/7/2015

1

Statics

Equilibrium

Moves at

constant

velocity

V =C

At rest

V = 0

Constant Velocity

11/7/2015

2

Equilibrium condition

Satisfying Newtown’s 1st Law of Motion: ∑ F = 0

This condition implies that ma = 0. here m could not

equal zero and so a = 0 (i.e. not accelerating or

decelerating)

Free Body Diagram (FBD)

Definition:

Is a diagram for a particle or a body isolated

from its surrounding.

It shows all the forces acting on the particle

11/7/2015

3

Procedures to draw FBD

Drawing outlined shapes

Show all the forces

Identify each force

Example[1]

F1

F2

Mg

M

11/7/2015

4

Example[2]

AB C

D

T1

M1g

B

T2

C

M2g

T2

T3

M1

M2

Connection types: linear springs

LoLo

L

F

δ

Applying Load

deflection (δ) = L – Lo

F = K δ where K is thespring stiffness

increasing K makesthe spring stiffer.

Stiffer springs needsmore force to deflect it

11/7/2015

5

Forces in Linear Springs

As the load affects the spring, an internal resistance load creates.

The relation between the external load and the internal force of thespring is proportional and the proportional factor is the stiffness (K)

Linear springs could be:

tension spring compression springs

Equivalent of Linear Springs (parallel)

K1

K2

K3

Kn

Force Force

Keq = K1 + K2 + …. + Kn

11/7/2015

6

Equivalent of Linear Springs (series)

Force

Force

K1 K2 Kn

n21eq K

1...

K

1

K

1

K

1

Connection types: cables and pulleys

it has always tension

force in the direction of

the cable.

In most of engineering

problems, cables are

assumed massless and

unable to stretch.

T

T

11/7/2015

7

Example[3]

T

11/7/2015

8

Equilibrium conditions:

∑F = 0

∑Fx i + ∑Fy j = 0

∑Fx = 0

∑Fy = 0

11/7/2015

9

Procedures for analysis FBD

Coordinate system (x and y axes)

Force Labeling (magnitude and direction)

Assuming the sense of the unknown forces

Equilibrium equations

Positive and negative assigning

Define the direction of the solution

∑ Fx = 0 and ∑ Fy = 0

Coplanar Forces

11/7/2015

10

Coplanar Forces

11/7/2015

11

Coplanar Forces

11/7/2015

12

Equilibrium conditions:

∑F = 0

∑Fx i + ∑Fy j + ∑Fz k= 0

∑Fx = 0

∑Fy = 0

∑Fz = 0

S

t

a

t

i

c

s

.

11/7/2015

13

Procedures for analysis FBD

Coordinate system (x, y and z axes)

Force Labeling (magnitude and direction)

Assuming the sense of the unknown forces

Equilibrium equations

Positive and negative assigning

Define the direction of the solution

∑ Fx = 0, ∑ Fy = 0 and ∑ Fz = 0

S

t

a

t

i

c

s

.

Three dimensional Forces

11/7/2015

14

Three dimensional Forces

11/7/2015

15

Three dimensional Forces