ENGR 221 February 17, 2003February 17, 2003...Sep 06, 2010 · ENGR 221 February 17, 2003February 17, 2003. Lecture Goals • 6-4 Equilibrium in Three Dimensions4 Equilibrium in Three

Equilibrium and TrussesEquilibrium and TrussesEquilibrium and TrussesEquilibrium and Trusses

ENGR 221February 17, 2003February 17, 2003

Lecture GoalsLecture Goals

• 6-4 Equilibrium in Three Dimensions6 4 Equilibrium in Three Dimensions• 7-1 Introduction to Trusses

7 2 Pl T• 7-2 Plane Trusses• 7-3 Space Trusses• 7-4 Frames and Machines

Equilibrium Equilibrium ––ProblemProblem

Determine the reactions at A and the force in bar CD due to the loading.


Draw the free-body diagram of the main RAybody.

6 i⎛ ⎞

RAx

1 o6 in.tan 26.56512 in.

θ − −⎛ ⎞= = −⎜ ⎟⎝ ⎠ θ

TCD


Look at equilibrium

( )o26 565 125 lb 0F R T∑ ( )( )( )

ox Ax CD

oAx CD

cos 26.565 125 lb 0

cos 26.565 125 lb

F R T

R T

= + − + =

= − − +

∑( )( )

( )Ax CD

oy Ay CD sin 26.565 40 lb 60 lb 80 lb 0F R T= + − − − + =∑

( )( )oAy CD sin 26.565 20 lb R T= − − −

EquilibriumEquilibrium –– RAyEquilibrium Equilibrium ProblemProblem RAx

Ay

Take the moment about A

θ

TCD

θ

( )( ) ( )( ) ( )

oA CD0 cos 26.565 6 in. 40 lb 4 in.

60 lb 8 in 80 lb 12 in

M T= = − −

− +

∑( ) ( )

( )( )oCD

60 lb 8 in. 80 lb 12 in.

cos 26.565 6 in. 320 lb-inT

+

− = −

CD 59.628 lbT = −

EquilibriumEquilibrium –– RAyEquilibrium Equilibrium ProblemProblem RAx

Ay

Take the moment about A

θ

TCD

θ

( ) ( )( )( ) ( )( )oAx 59.628 lb cos 26.565 125 lb

71.667 lb

R = − − − +

= −

( ) ( )( )oAy 59.628 lb sin 26.565 20 lb R = − − − −

6.667 lb=

Equilibrium in 3Equilibrium in 3--DimensionsDimensionsI t di i th ti l dIn two dimensions, the equations are solved using the summation of forces in the x, y and z directions and the moment equilibrium includesdirections and the moment equilibrium includes moment components in the x, y and z directions.

x y z0 0 0F F F= = =∑ ∑ ∑

x y z0 0 0M M M= = =∑ ∑ ∑

Trusses Trusses --DefinitionDefinitionT t tTrusses are structures composed entirely of two force members Theyforce members . They consists generally of triangular sub-element and gare constructed and supported so as to prevent

iany motion.

FramesFrames DefinitionDefinitionFrames Frames --DefinitionDefinition

F t t th tFrames are structures that always contain at least one member acted on by forcesmember acted on by forces at three or more points. Frames are constructed and supported so as to prevent any motion. Frame like

h f llstructures that are not fully constrained are called machines or mechanismsmachines or mechanisms.

TrussTrussPlanar Trusses lie in aPlanar Trusses - lie in a single plane and all applied loads must lie in ppthe same plane.

TrussTrussSpace Trusses - are structures that are not contained in a single plane and/or are loaded out of the plane of the structureof the plane of the structure.

TrussTrussThere are four main assumptions made in theThere are four main assumptions made in the analysis of truss

Truss members are connected together at their1 Truss members are connected together at their ends only.

T t d t th b f i ti l

1

2 Truss are connected together by frictionless pins.

2

The truss structure is loaded only at the joints.

The weights of the members may be neglected.

3

4 g y g

Simple TrussSimple Truss

The basic building block of a truss is a triangle. Large truss are constructed by attaching several triangles together A ne triangle can be addednew triangle can be added truss by adding two members and a joint A trussand a joint. A truss constructed in this fashion is known as a simple truss.


It has been observed that the analysis of truss can be done by counting the number member and joints on the truss to determine the truss is determinate, unstable or indeterminate.


A truss is analysis by using m=2*j-3, where m is number of members, j represents the number of joints and 3 represents the external support reactions.


If m< 2j-3, then the truss is unstable and will collapse under load.

If m> 2j-3, then the truss has more unknowns th k ti d i i d t i tthan know equations and is an indeterminate structure.

If 2j 3 h i l l iIf m= 2j-3, ensures that a simple plane truss is rigid and solvable, it is neither sufficient nor necessary to ensure that a non-simple plane trussnecessary to ensure that a non-simple plane truss is rigid and solvable.

Simple TrussSimple Truss-- IdentifyIdentifyDetermine type of simple truss is it determinate, indeterminate or unstable.

Method of Joints Method of Joints --TrussTrussThe truss is made up of single bars, which areThe truss is made up of single bars, which are either in compression, tension or no-load. Themeans of solving force inside of the truss use equilibrium equations at a joint. This

th d i k thmethod is known as the method of joints.

Method of Joints Method of Joints --TrussTrussThe method of joints uses the summation ofThe method of joints uses the summation of forces at a joint to solve the force in themembers. It does not use the moment equilibrium equation to solve the problem. In a two di i l t f tidimensional set of equations,

x y0 0 F F= =∑ ∑In three dimensions,

0F =∑ z 0 F =∑

Method of Joints Method of Joints ––Example Example

Using the method of joints, determine the jforce in each member of the truss.

Method of Joints Method of Joints ––Example Example

Draw the free body diagram of the truss and gsolve for the equations

0F C∑ x x

x

00 lb

F CC

= =

=∑

y y0 2000 lb 1000 lbF E C= = − − + +∑ y y

y 3000 lbE C⇒ + =∑

Method of Method of Joints Joints ––ExampleExampleExample Example

Solve the moment about C

( ) ( ) ( )C 0 2000 lb 24 ft 1000 lb 12 ft 6 ft10000 lb

M EE

= = + −

⇒ =∑

y

10000 lbC 3000 lb 10000 lb 7000 lbE⇒ =

⇒ = − = −


Look at joint Aj

AD40 2000 lbF F= = − −∑

( )y AD

AD AD

0 2000 lb5

2500 lb 2500 lb C

F F

F F= − ⇒ =

∑

( )x AD AB AB3 30 2500 lb5 5

F F F F= = + = − +∑( )AB AB1500 lb 1500 lb TF F= ⇒ =


Look at joint Dj

( )y AD DB DB4 4 4 40 2500 lb5 5 5 5

F F F F= = + = − +∑( )DB DB

A

2500 lb 2500 lb T3 30

F F

F F F F

= ⇒ =

= = − + +∑

( ) ( )

x AD DB DE

DE

05 5

3 32500 lb 2500 lb5 5

F F F F

F

+ +

−= − + +

∑

( ) ( )

( )DE DE

5 53000 lb 3000 lb CF F= − ⇒ =


Look at joint B 4 4∑j

( )

y BD BE

DE

4 40 1000 lb5 5

4 42500 lb 1000 lb

F F F

F

= = − − −

= − − −

∑

( )

( )DE

DE DE

500 b 000 b5 53750 lb 3750 lb C

3 3F F= − ⇒ =

∑

( ) ( )

x BD BA BE BC

BC

3 305 5

3 32500 lb 1500 lb 3750 lb

F F F F F

F

= = − − + +

−= − − + − +

∑

( ) ( )

( )BC

BC DE

5 55250 lb 5250 lb TF F= ⇒ =


Look at joint E 4 4∑j

( )

y EB EC

DE

4 40 10000 lb5 5

4 43750 lb 10000 lb

F F F

F

= = − + +

= − − + +

∑

( )

( )DE

EC EC

3750 b 0000 b5 58750 lb 8750 lb C

3 3F F= − ⇒ =

∑

( ) ( )

x EB ED EC

EC

3 305 5

3 33750 lb 3000 lb

F F F F

F

= = − − +

= − − − − +

∑

( ) ( )

( )EC

EC EC

5 58750 lb 8750 lb CF F= − ⇒ =


Look at joint C to check jthe solution

y CE40 7000 lb5

F F= = − −∑

( )4 8750 lb 7000 lb 0 OK!5

3

= − − − =

∑

( ) ( )

x CE CB x305

3 8750 lb 5250 lb 0 0

F F F C= = − − +

= − − − + =

∑

( ) ( )5

Method of Joints Method of Joints ––Class Problem Class Problem

Determine the forces BC, ,DF and GE. Using the method of Joints.

Method of Sections Method of Sections --TrussTrussThe method of joints is most effective when the forces in all the members of a truss are to be determined If however the force is onlybe determined. If however, the force is only one or a few members are needed, then the method of sections is more efficient.method of sections is more efficient.

Method of Sections Method of Sections --TrussTrussIf we were interested in the force of member CE. We can use a cutting line orcan use a cutting line or section to breakup the truss and solve by taking theand solve by taking the moment about B.

Method of Sections Method of Sections –– Example Example Determine the forces in members FH, GH and GI of the roof truss.

Method of Sections Method of Sections –– Example Example Draw a free body diagram and solve for the reactions.

x Ax0F R= =∑

RAx LAx 0 kNR =

RAyy

Ay

0

20 kN

F

L R

=

⇒ + =∑

Ay

Method of Method of Sections Sections ––ExampleExampleExample Example

Solve for the A

RAx Lmoment at A. RAy

( ) ( ) ( )6 kN 5 6 kN 10 6 kN 15M∑ ( ) ( ) ( )( ) ( ) ( )

A 6 kN 5 m 6 kN 10 m 6 kN 15 m

1 kN 20 m 1 kN 25 m 30 m

M

L

= − − −

− − +∑

Ay

7.5 kN12.5 kN

LR

⇒ =⇒ =Ay

Method of Sections Method of Sections –– Example Example

Solve for the member GI. Take a cut between the hi d d f h i d d h f b dthird and fourth section and draw the free-body

diagram.

( )HIHI

8 m 10 m 8 m15 m 10 m 15 m

l l= ⇒ =

HI

1 o

5.333 m8 mtan 28.1

15

l

α −

=

⎛ ⎞= =⎜ ⎟⎝ ⎠15 m⎜ ⎟⎝ ⎠


The free-body diagram of the cut on the right side.

( ) ( ) ( )H GI1 kN 5 m 7.5 kN 10 m 5.333 mM F= − + −∑

( )GI GI13.13 kN 13.13 kN TF F= ⇒ =


Use the line of action of the forces and take the moment about G it will remove the FGI and FGH and shift FFH to the perpendicular of G.


Take the moment at G

( ) ( ) ( )G 1 kN 5 m 1 kN 10 m 7.5 kN 15 mM = − − +∑ ( ) ( ) ( )( )( )

G

oFH cos 28.1 8 mF+

∑

( )FH FH13.82 kN 13.82 kN CF F= − ⇒ =


Use the line of action of the forces and take the moment about L it will remove the FGI and FFH and shift FGH to point G.

1 o5 mtan 133.25.333 m

β − −⎛ ⎞= = −⎜ ⎟−⎝ ⎠


Take the moment at L

( ) ( ) ( )( )oL GH1 kN 5 m 1 kN 10 m cos 43.2 15 mM F= + +∑

( )GH GH1.372 kN 1.372 kN CF F= − ⇒ =

Method of SectionsMethod of Sections –– ClassClassMethod of Sections Method of Sections Class Class Problem Problem

Determine the forces in members CD and CE using method of sectionsof sections.

Homework (Due 2/24/03)Homework (Due 2/24/03)

Problems:

6 34 6 37 6 38 6 40 6 45 6 636-34, 6-37, 6-38, 6-40, 6-45, 6-63

Truss Truss ––Bonus ProblemBonus Problem

Determine whether the members are unstable, determinate or indeterminate.


Determine the loads in each of the members.





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ENGR 221 February 17, 2003February 17, 2003...Sep 06, 2010 · ENGR 221 February 17, 2003February 17, 2003. Lecture Goals • 6-4 Equilibrium in Three Dimensions4 Equilibrium in Three