Multipiece Structures - Engineering Mechanics Statics - Lecture Slides, Slides of Mechanical Engineering

These are the Lecture Slides of Engineering Mechanics Statics which includes Free Body Diagrams, Magnitude and Direction of Forces, Coordinate System, Newton's Third Law, Structural Supports, Sliding and Free Vectors, Center of Gravity, Center of Gravity, Plane of Symmetry etc.Key important points are: Multipiece Structures, Forces of Action and Reaction, Point of Appliction, Force Transmissibility, Force of Interest, Generation of Simultaneous Eqns, Method of Sections, Static Equilibrium, Free

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2012/2013

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Introduction: Multipiece Structures
For the equilibrium of structures made of several
connected parts, the internal forces as well the
external forces are considered.
In the interaction between connected parts, Newton’s
3rd Law states that the forces of action and reaction
between bodies in contact have the same magnitude,
same line of action, and opposite sense.
The Major Categories of Engineering Structures
Frames: contain at least one multi-force member, i.e.,
a member acted upon by 3 or more forces
Trusses: formed from two-force members, i.e.,
straight members with end point connections
Machines: structures containing moving parts designed
to transmit and modify forces
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Introduction: Multipiece Structures

 For the equilibrium of structures made of several

connected parts , the internal forces as well the

external forces are considered.

 In the interaction between connected parts, Newton’s

3 rd^ Law states that the forces of action and reaction

between bodies in contact have the same magnitude ,

same line of action , and opposite sense.

 The Major Categories of Engineering Structures

  • Frames: contain at least one multi-force member, i.e., a member acted upon by 3 or more forces
  • Trusses : formed from two-force members , i.e., straight members with end point connections
  • Machines: structures containing moving parts designed to transmit and modify forces

Definition of a Truss

 A truss consists of straight members connected at joints. No member is continuous through a joint.

 Bolted or welded connections are assumed to be pinned together. Forces acting at the member ends reduce to a single force and NO couple. Only two-force members are considered  LoA CoIncident with Geometry

 A truss carries ONLY those loads which act in its plane , allowing the truss to be treated as a two-dimensional structure.

 When forces tend to pull the member apart, it is in tension. When the forces tend to push together the member, it is in compression.

Trusses Made of Simple Trusses

  • Compound trusses are statically determinant, rigid, and completely constrained. m  2 n  3
  • Truss contains a redundant member and is statically indeterminate. m  2 n  3
  • Necessary but INsufficient condition for a compound truss to be statically determinant, rigid, and completely constrained,

mr  2 n

non-rigid

m  2 n  3

  • Additional reaction forces may be necessary for a nonrigid truss.

rigid m  2 n  4

Method of Sections

 To determine the force in member BD, pass a section through the truss as shown and create a free body diagram for the left side.  With only three members cut by the section, the equations for static equilibrium may be applied to determine the unknown member forces, including FBD

 When the force in only one member or the forces in a very few members are desired, the method of sections works well.

Example  Method of Sections

• Take Section to Expose FFD

 Now Take Σ M E = 0

0  15 kip 10 '  15 kip 20 '  FFD  10 '

45 kip

10ft

150 300 kip ft

FFD  FFD ^45 kip Compression

Example  Method of Sections

 Determine the force in

members just right of Center:

  • FH
  • GH
  • GI

 SOLUTION PLAN

  • Take the entire truss as a free body. Apply the conditions for static equilibrium to solve for the reactions at A and L.
  • Pass a section through members FH , GH , and GI and take the right-hand section as a free body.
  • Apply the conditions for static equilibrium to determine the desired member forces.

Example  Method of Sections

  • Pass a section (n-n) through members FH, GH, and GI and take the right -hand section as a free body

 Apply the conditions for static equilibrium to determine the desired member forces.

       

  1. 13 kN

7.50kN 10 m 1 kN 5 m 5. 33 m 0

 

GI

GI

H

F
F
M

FGI  13. 13 kN T

Example  Method of Sections

           

  1. 82 kN

cos 8 m 0

7.5kN 15 m 1 kN 10 m 1 kN 5 m

15 m

8 m tan

FH

FH

G

F
F
M
GL
FG

FFH  13. 82 kN C

 

        

  1. 371 kN

1 kN 10 m 1 kN 5 m cos 15 m 0

8 m

5 m tan 3 2

GH

GH

L

F
F
M
HI
GI

FGH  1. 371 kN C

Pick: Pivot & PoA

• When doing Sections Recall that the LoA for

Truss Members are defined by the Member

Geometry

• Use Force Transmissibility → Forces are

SLIDING Vectors

– Pick a Pivot Point, on or Off the Body, where the

LoA’s of many Force LoA’s Cross

– Apply the Force of interest so that ONE of its X-

Y Components passes Thru the Pivot

Pick: Pivot & PoA Example

• After Finding support RCNs find force in

Member ED → Use Section a-a

• Pick Pt-B as Pivot to Eliminate from

Moment Calc FAB, FFB, FEB, 1000N