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Amity - Forces - Why So Dumb

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Free body diagram, Equilibrium equations and applications. Plane truss, perfect and imperfect truss, assumption in the truss analysis, analysis of perfect plane trusses by the method of joints, method of section.

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Contents

Force 2

Force System Classification 2

Resolution of Forces 2

Laws of Forces 2

Triangular Law 3

Parallelogram Law 3

Polygon Law of Forces 4

Moment of Force 6

Varignon’s Theorem 6

Lami’s Theorem 7

Trusses - Formed Structures 8

Support Reactions 8

Analysis of a Frame 9

Force

Force System Classification :

i) Coplanar and Non coplanar

ii) Concurrent and Non Concurrent

iii) Parallel and Non Parallel

Resolution of Forces:

Laws of Forces

i) Triangular law

ii) Parallelogram law

iii) Polygon law (more than two forces)

θ

θ

(only for two forces

coplanar and concurrent

)

Triangular Law

F1 = 10 N F2 = 5 N = 30o

Scale consideration 1 N = 1cm

F1 = 10 cm F2 = 5 cm

We can find the value of R (resultant) & directly proportional to

𝛼 (direction) by measuring.

Parallelogram Law:

𝜃 𝛼

𝜃 𝛼

𝜃

𝜃

In ACE

(AC)2 = (AE)

2 + (CE)

2

R2 = (F1 + F2 cos )

2 (F2 sin )

2

R2 =

F1

2 + F2

2cos

2 + 2F1F2 cos + F22sin

2 )

R2 = F1

2 + F2

2 (cos

2 + sin2 ) + 2F1F2cos )

R2

= F12 + F2

2 + 2F1F2cos

R = √

=

Polygon Law of Forces:

i) Graphical method

In graphical method direction of forces should be same inward

or outward

Let

F1 = 10 N

F2 = 5N

F3 = 8N

F4 = 7N

Scale consideration 1 N = 1cm

F1 = 10cm F2 = 5cm F3 = 8cm F4 = 7cm

ii) Analytical method

R2 = √ ∑ ∑

Moment of Force

Moment = force x perpendicular distance

Varignon’s Theorem

MRO = MF1O + MF2O

OE perpendicular to AD, OF to AC, OA to AB

R x OF =

F2 x OE +

F1 x OA

R x OF = F2 x OE + F1 x OA

MRO = MF2O + MF1O

MRO = MF1O + MF2O

𝜃 𝛼

θ

θ

Lami’s Theorem:

Lami’s theorem is applicable only when there are three

coplanar concurrent forces equilibrium state.

P

Q

R

=

𝛼 𝛽

𝛾

Trusses - Formed Structures

M = 2j – 3 (m = members j = joints)

i) Perfect m = 2j – 3

ii) Deficient m < 2j – 3

iii) Reductant m = 2j – 3

Support Reactions

There are three types of supports:

i) Simply supported / fixed R = R =

ii) Roller supported R = perpendicular to the contact

surface

iii) Hinged supported R = Rh

Analysis of a Frame

i) The reaction at the supporters

ii) The internal stresses induced in the member due to external

loads.

Sign convention for the member under a tensile force

If the forces are pulling at the ends

Tensile force

If the forces are pushing at each other

Compressive

Principles that help to identify the member not subjected

to any force when the truss is loaded

i) A single force cannot form a system in equilibrium this implies that if

there is only one force acting on a point then for the equilibrium of this

joint this force equal zero.

ii) When two members meeting at a joint are not collinear and there is no

external force acting at that point then the value of both forces will

become zero.

iii) When three members are meeting at a unloaded joint and out of them

two are collinear then the force in the third member will be zero.

1.

2.

3.