3. structures - chulapioneer.netserv.chula.ac.th/~mkuntine/42-111/files/ch3... · 2012-03-08 · 3...

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3. Structures

2142111 S i 2011/22142111 Statics, 2011/2

© Department of Mechanical Engineering Chulalongkorn UniversityEngineering, Chulalongkorn University

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Objectives Students must be able to #1Objectives Students must be able to #1

Course ObjectiveAnalyze structures (trusses and frames/machines) in equilibrium Analyze structures (trusses and frames/machines) in equilibrium

Chapter Objectives Differentiate connecting bodies in equilibrium into

frames/machines and trusses For 2D/3D frames/machines State appropriate action/reactions between joints/members and State appropriate action/reactions between joints/members and

disassemble the structure by drawing FBDs of members and important jointsId tif t t ith t diti l i th ff t f Identify structures with symmetry condition, explain the effects of symmetry and identify the equal-values and zero-components due to symmetry

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Analyze structures for unknown loads/reactions by appropriate FBDs

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Objectives Students must be able to #2Objectives Students must be able to #2

Chapter ObjectivesF 2D/3D tFor 2D/3D trusses Describe characteristics and approximation justification for

members as 2-force members Analyze for forces in members by the method of joints and

method of sections with appropriate FBDs Explain the 2 causes of zero-force members and determine Explain the 2 causes of zero-force members and determine

zero-force members in trusses Identify trusses with the symmetry condition and explain the

ff t f t th bleffects of symmetry on the problem

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ContentsContents Analyses of Frames and Machines

Analyses of Trusses Truss Method of Joints Method of Sections

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Equilibrium ObjectsEquilibrium Objects

So far, you analyze simple rigid bodies Draw FBD Draw FBD

Apply equations of equilibrium

What’s you are going to do with complex systems Draw FBD of each members

I l t d b Isolated member Identify two and three-force members Add external and support loads Add external and support loads

Apply equations of equilibrium For two force members, also determine the internal

f

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forces.

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Equilibrium StructuresEquilibrium Structures

Members are interconnected parts in structures.

Structures are any connecting system of members that is built to support or transfer forces and safely withstand the applied loads.

We will study statically determinate We will study statically determinate Frames & Machines Trusses

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Frames and MachinesF&M

Frames and Machines Consider structures of

interconnected members that do not satisfy the definition of a truss.

Frames remain stationary and support l dloads.

Machines are designedMachines are designed to move and apply loads.

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Frames & Machines Procedures

F&M

Frames & Machines Procedures

Analyze the entire structure and find as many reaction forces at the supports as you canforces at the supports as you can.

Analyze individual members Disassemble structure and isolate members Identify two-force members

Identify important joints which connect three or more Identify important joints which connect three or more members/supports

Apply action/reaction forces on any two contacting members

Reassemble members to check for errors. All internal and action/reaction forces must cancel out.

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Example F & M 1 #1

F&M

Example F & M 1 #1

Determine the magnitudes of horizontal and vertical components pof force which the pin at B exerts on member CB.

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Example F & M 1 #2

F&M

p3 equations, 4 unknowns

Do you really unable to find reactions?

= − = 0 0x x xF A C = + − = = + °

0 2000 N 0

0 (2 m)(2000 N) (3sin60 m)

x x x

y y y

C

F A C

M A

10

+ − + ° =

0 (2 m)(2000 N) (3sin60 m)

(4 m 3cos60 m) 0C x

y

M A

A

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Example F & M 1 #3

F&MApproach 1

Example F & M 1 #3

= Consider FBD of

0C

BC

M

− =

=

(2 m)(2000 N) 4 0

1000 N Ansy

y

B

B

= − + =

0

(2000 N) 0y

y y

F

B C

=

= 1000 N

0y

x

C

F

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− = 0 (1)x xB C

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Example F & M 1 #4

F&MApproach 1

Example F & M 1 #4

Consider FBD of AB = − =

=

Consider FBD of

0 0

1000 Ny y y

y

AB

F A B

A

= ° − ° =

=

000

0 (3sin60 m) (3sin60 m) 0

1000 / 3 N

y

A x yM B B

B ==

= − =

1000 / 3 N577 N Ans

0 0

x

x

BB

F A B

=

=

0 0

1000 / 3 N

From (1) 1000 / 3 N

x x x

x

F A B

A

C

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=From (1), 1000 / 3 NxC

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Example F & M 1 #5

F&MApproach 2

Example F & M 1 #5

is a two-force member.AB

= − ° = Consider FBD of

0 (2 m)(2000 N) (4 m) sin60 0C

BC

M AB

=

°

2000 / 3 N is the force that pin exerts on ,

Horizontal component cos60 577 N

ABAB B BC

AB

13

= ° == ° =

Horizontal component cos60 577 NVertical component sin60 1000 N Ans

ABAB

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Example F & M 2 #1

F&M

Example F & M 2 #1

Disassembling the structure

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Example F & M 2 #2

F&M

Example F & M 2 #2


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Example F & M 2 #3

F&M

Example F & M 2 #3

Isolate the the member ACED

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Example F & M 2 #4

F&M

Example F & M 2 #4


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Example F & M 2 #5

F&M

Example F & M 2 #5


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Example Hibbeler Ex 6-10 #1

F&M

Example Hibbeler Ex 6 10 #1

A constant tension in theA constant tension in the conveyor belt is maintained by using the device shown.Draw the free-bodyDraw the free-body diagrams of the frame and the cylinder which supports the belt The suspendedthe belt. The suspended block has a weight of W.

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F&M


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Example F & M 4 #1

F&M

Example F & M 4 #1

The mechanism is used to weigh mail A package placed at A causesmail. A package placed at A causes the weighted pointer to rotate through an angle a. Neglect the weights of the members except for themembers except for the counterweight at B, which has a mass of 4 kg. If α = 20°, what is the mass of the package at A?of the package at A?

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Example F & M 4 #2

F&M

Example F & M 4 #2

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Example F & M 4 #3

F&M

Example F & M 4 #3

Given 20α = °Consider FBD of

0 0E EC x

AEC

M r C = = 0

0 cos20 0x

x x

C

F EF C

=

= − ° − = 0

0 0y y

EF

F W C

=

= − − =

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(1)yC W= −

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Example F & M 4 #4

F&M

Example F & M 4 #4

Consider FBD of BDC =

° + ° =

Consider FBD of

0

(0.1 m)sin20 (4 N) (0.1 m)cos10 0D

y

BDC

M

g C

° ⋅= −°

( ) ( ) ( )

0.4sin20 N m0.1cos10 m

y

y

g

gC

° ⋅= =°

Substitute (1)0.4sin20 N m

0 1cos10 mgW mg

°= =°

0.1cos10 m0.4sin20 1.3892 kg0.1cos10

m

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Mass of is 1.39 kg Ansm A

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F&M


Determine the horizontal and vertical components of force which the pin at C exerts on member ABCD of theforce which the pin at C exerts on member ABCD of the frame shown.

FBD

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FBD

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F&M

p

Whole frame in equilibrium = +

− + = 0

( 981 N)(2 m) (2.8 m) 0A

x

M

D

=

= → + 700.71 N

0 x

x

D

F

− ==

↑

0700.71 N

x x

x

A DAFBD

= ↑ + − =

0

981 N 0y

y

F

A

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= 981 NyA

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F&M


Apply equilibrium equations for member CEF = +

−

for member

0

( 981 N)(2 m)C

CEF

M

− ° == −

( 981 N)(2 m)( sin45 )(1.6 m) 0

1734.2 NB

B

FF

= → + − − ° = 0

cos 45 0x

x B

F

C F= − = −

= ↑ + 1226.3 N 1.23 kN Ans

0 x

y

C

F

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− ° − =

= − = −

sin45 981 N 0

245.26 N 245y B

y

C F

C N Ans

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F&M


The smooth disk shown is pinned at D and has a weight of 20 lbThe smooth disk shown is pinned at D and has a weight of 20 lb. Neglecting the weights of the other member, determine the horizontal and vertical components of reaction of pins B and D.

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FBD

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F&M

p

Equilibrium of entire frame =

− + =q

0

( 20 lb)(3 ft) (3.5 ft) 0A

x

M

C=

= 17.143 lb

0x

x

C

F

− ==

0 17.143 lb

x x

x

A CA

= − =

0

20 lb 0 y

y

F

A

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= 20 lbyA

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F&M

p

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F&M


Equilibrium of member AB =

− + =

Equilibrium of member

0

( )(6 ft) (3 ft) 0B

D

AB

M

A N+

=

=

( )(6 ft) (3 ft) 0

40 lb

0

y D

D

x

A N

N

F − == =

0

17.143 17.1 lb Ans

x

x x

x

A BB

= − + =

0

0y

y d y

F

A N B

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= 20 lb Ansy y

yB

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F&M

p

E ilib i f di k D = → + Equilibrium of disk

0

0 Ax

D

F

D =

= ↑ + 0 Ans

0

20 lb 0

x

y

D

F

N D− − =

− − =

=

20 lb 0

40 lb 20 lb 0

20 lb Ans

D y

y

N D

D

D = 20 lb AnsyD

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Example Hibbeler Ex 6-17 #6F&M

p

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F&M


Draw the free-body diagrams of the bucket and the vertical boom f th b k h h i th h t Th b k t d it t tof the back hoe shown in the photo. The bucket and its contents

have a weight W. Neglect the weight of the members.

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F&M


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Example F & M 3 #1

F&M

Example F & M 3 #1

For P = 150 N squeeze on the handles of the pliers, determine the force F applied by each jawdetermine the force F applied by each jaw.

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F & M Example3 #2

F&M

& p

Due to horizontal symmetry of foces acting on

37=

Due to horizontal symmetry of foces acting onon jaws and handles: 0xC Unit in mm

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Example F & M 3 #3

F&M

Example F & M 3 #3

= FBD of the upper handle

0M = − =

=

0

(0.03 m) (0.18 m) 0

6 (1)

C

y

M

B P

B P=

=

6 (1)

FBD of the upper jaw

0

y

A

B P

M

Unit in mm

− =

= =

0

(0.06 m) (0.02 m) 0

3 18

A

y

y

M

B F

F B P

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=

3 8

2.7 kN Ansy

F

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Example Hibbeler SI Ex 6-13 #1

F&M

Example Hibbeler SI Ex 6 13 #1

Disassembling the structure.

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Example Hibbeler SI Ex 6-13 #2

F&M

Example Hibbeler SI Ex 6 13 #2

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Example Meriam Ex 47 #1

F&M



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F&M


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F&M


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F&M



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F&M


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F&M

p

= + − = → = FBD of lever

0 (1 in) (8 lb)(4 in) 0 32 lbB EA EA

ABG

M F F

= ↑ + ° − ° = → = FBD of pin

0 sin60 sin60 0 y ED EG ED EG

E

F F F F F

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= → + ° + ° − = → = 0 cos60 cos60 0 32 lbx ED EG EA EDF F F F F

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F&M

p

FBD of arm DC = +

− ° =

FBD of arm

0

(6 in) + cos30 (3 in) 0C

DC

M

F F

==(6 in) + cos30 (3 in) 013.9 lb

S ED

S

F FF

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Example Bedford 6.114 #1

F&M

Example Bedford 6.114 #1


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Truss Definition

Truss

Truss Definition

A framework composed of bars joined at their ends bybars joined at their ends by smooth pins to form a rigid structure.

Trusses are supported and loaded at their joints.

If we neglect the weights of the bars each bar is a two-force

River Kwai Bridgebars, each bar is a two-force member.

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Truss Joints

Truss

Truss Joints

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Truss Type #1

Truss

Truss Type #1

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Truss Type #2

Truss

Truss Type #2

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Truss Type #3

Truss

Truss Type #3

Howe Bridge Truss

K Bridge Truss

53Subdivided Warren Bridge Trusswww.nasa.gov

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Truss Categories

Truss

Truss Categories

Plane Truss – 2D

Space Truss – 3D

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Plane Truss Rigid Frames

Truss

Plane Truss Rigid Frames

Rigid frames do not collapse and have negligible deformationdeformation.

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Plane Truss Simple Trusses

Truss

Plane Truss Simple Trusses

The basic element is a triangle.Rigid simple trusses are built by adding units of two end Rigid simple trusses are built by adding units of two end-connected bars.

56= −SD plane truss: 2 3m j

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Plane Truss Stability

Truss

Plane Truss Stability

> −2 3m j SI truss with redundant member(s).

= −2 3m j SD plane truss

< −2 3m j Truss collapses under load.

SI with redundant supports

SI with improper supports

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p p pp

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Plane Truss Solving Procedures

Truss

Plane Truss Solving Procedures

Main AnalysesMethod of joints disassembling all joints and Method of joints – disassembling all joints and members

Methods of sections – sectioning truss as needed

Helping HandsZero force members identifying members that Zero-force members – identifying members that do not support loads

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Method of Joints Procedure

Truss

For SD trussesIsolate the truss draw FBD and find the support Isolate the truss, draw FBD and find the support reactions

Isolate each joint, draw FBD and find forces of members on joints

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Method of Joints Demo #1

Truss

Draw FBD of the entire truss


Determine the axial forces in each member Draw FBD of the entire truss

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Truss


Determine the reactions at its supports.

= + = 0 500 N 0x xF A = −

= − + = 500 N

0 (2 m)(500 N) 2 0x

A y

A

M B =

= + = 500 N

0 0y

y y y

B

F A B = −

500 N

y y y

yA

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Truss

FBDs of members and joints Select order of joint analyses

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Truss


Draw the FBD of a joint having at least one known force and at most two unknown forcesforce and at most two unknown forces.

Orient x and y axes such that the forces on the FBD can be easily resolved into their x and y components. S l f thSolve for them.

= − = At joint

0 500 N 0x

A

F AB =

= − =

500 N

0 500 N 0

x

y

AB

F AC

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=

500 N

y

AC

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Truss


Continue with the rest of joints using the same procedureprocedure.

At joint B = ° + =

= −t jo t

0 sin45 500 N 0

707.1 NyF BC

BC

Axial forces in members are==

member : 500 N : 500 N : 707 N Ans

AB AB TAC AC TBC BC C

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= : 707 N AnsBC BC C

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Method of Joints Tips & Hints

Truss

p

Procedural tips to avoid confusionp Assign tension on joints and members in the original FBD

Check for their correct senses later + = tension, pulling on joints and members + tension, pulling on joints and members − = compression, pushing on the joints & members

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Method of Joints WWW Interactive

Truss

Bridge Designer @ Johns Hopkins UniversitySmall package Small package

http://www.jhu.edu/~virtlab/bridge/truss.htm FRAME3DD @ Duke University @ y

http://frame3dd.sourceforge.net/

D t F Doctor Frame http://www.drsoftware-home.com/index.html

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Example Method of Joints 1 #1

Truss


Determine maximum axial forces in each member when 0 ≤ θ ≤ 90°member when 0 ≤ θ ≤ 90 .

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Truss

p

θ

= + =

0

(10 kN)sin 0xF

A θθ

+ == −

=

(10 kN)sin 010sin kN

0

x

x

A

AA

M

θθ

− − =

=

(4 m) (8 m)(10 kN)cos 0

20cos kN

A

y

y

C

C

θ

= + − =

0

(10 kN)cos 0

y

y

y y

F

A CWhich joints should be chosen?

And in what order?

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θ= −

( )10cos kN

y y

yAAnd in what order?

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Truss

p

α −= 1At joint , tan (3 / 4)A

α θ

= − =

j , ( )

0

sin (10 kN)cos (10 kN) 0yF

ABθ=

=

( ) ( )(50 cos ) / 3 kN

0x

AB

F

α θθ θ

+ − == −

cos (10 kN) sin 010 sin (40 cos ) / 3 kN

AC ABAC

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Truss

p

At joint C

θθ

= + =

0

20cos kN 020 kN

yF

BCBC θ= −

= 20cos kN

0

0x

BC

F

CD ACθ θ

− == −

010 sin (40 cos ) / 3 kN

CD ACCD

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Truss


−1At joint tan (3 / 4)D α

α θ

=

=

1At joint , tan (3 / 4)

0

sin (10 kN)cos 0y

D

F

BD α θθ

− ==

sin (10 kN)cos 0(50 cos ) / 3 kN

BDBD

When you get used to the procedure

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When you get used to the procedure, you can draw only FBDs of joints.

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Truss


θ= =Magnitude of forces in members are

50 cos 3 kNAB BD θθ θ

θ

= == = −= −

50 cos 3 kN10 sin 40 cos 3 kN

20cos kN

AB BDAC CDBC

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Truss


θ =Maximum magnitude of when 0.F dF dθ≤ ≤ °

= == =

max max

Thus, in the range 0 90 ,16.7 kN 13 3 kN

AB BD TAC CD C

θ

= === °

max max

max

13.3 kN 20 kN

all at 0 Ans

AC CD CBC C

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Zero-Force Member identification #1

Truss

Zero Force Member identification #1

They support no load.They are used to increase stability of the truss during They are used to increase stability of the truss during construction.

They provide support if the applied load changes.

= → = 30 0yF F

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= → = 1 20xF F F

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Zero-Force Member identification #2

Truss

Zero Force Member identification #2

= → =

10 0

0 0xF F

F F′ = → = 20 0xF F

= → = 3 40xF F F

′

→ = → =

3 4

1 2

0

0x

x

F F F

F F F

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Zero-Force Member Removing

Truss

Zero Force Member Removing

= → = →

10 0

0 0yF F

F F

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= → = 20 0xF F

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Example Zero-Force Member 1 #1

Truss

p

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Truss

p

78

79


Truss

p

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80


Truss

Example Zero Force Member 2 #1

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Truss

Example Zero Force Member 2 #2

Howe Bridge TrussZero-force members in standard trussesstandard trusses

K Bridge Truss

81Subdivided Warren Bridge Truss

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Method of Sections Basics #1Imaginary Sections

Truss

Method of Sections Basics #1

Imaginary cut through the considered region If the whole body is in equilibrium, its parts must also be

in equilibrium.

Built-in Support

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TrussImaginary Sections


Find resultant internal load by method of section Internal loads are normal force N, shear force V, torsion T

and bending moment M

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Method of Sections Concept

Truss

Method of Sections Concept

If a body is in equilibrium, any partIf a body is in equilibrium, any part of the body is also in equilibrium.

We can draw an imaginary section th h th b dthrough the body.

If bodies are two-force members, there are only normal internal forces.

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Method of Sections Internal Forces

Truss

Method of Sections Internal Forces

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Method of Sections Procedure #1

Truss


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87


Truss


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88

Method of Sections Demo #1

Truss


Determine the axial forces in members BD, CD and CE.

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89


Truss


Draw FBD of entire structurestructure.

Choose an imaginary li t “ t” tiline to “cut” or section through the members where forces are to be d t i ddetermined.

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90


Truss


Consider FBDs of sections

90

91


Truss


Draw FBD of that part of the sectioned truss with the least number of forces on itleast number of forces on it.

A l ilib i ti t th h FBD

91

Apply equilibrium equations to the chosen FBD.

92


Truss


= + ° − =

0

( )(1 m)sin45 (10 kN)(1 m) 0DM

CE

′

=

= 10 2 kN

0y

CE

F

° − ° ==

sin45 (10 kN)sin45 010 kN

CDCD

′ = − − ° − − ° = 0

cos 45 (10 kN)cos 45 0

20 2 kN

xF

BD CD CE

BD= 10 2 kN CE T

92

= −20 2 kNBD ==

20 2 kN 10 kN Ans

BD CCD T

93

Space TrussesTruss

Space Trusses

Members are connected by joints that do not resist moments.

The basic element is a tetrahedron. Rigid simple space trusses can be built by Rigid simple space trusses can be built by

adding units of three end-connected bars. Analyze in the same way as 2D trusses.

93= −SD space truss: 3 6m j

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Example Space Trusses 1 #1

Truss


A space truss is placed on a smooth floor. Joint A is supported ppby the corner where the smooth walls meet, and joint C rests against the smooth back wall.against the smooth back wall. Determine axial forces in member AC.

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95


Truss

p p

Applied loadˆ ˆ ˆ2 6 kNF i j k= − − −

All smooth supportsJoint is nested in a corner

ˆ ˆ ˆ kN

A

A A i A j A k+ +

kNJoint is placed on a floor

ˆ kN

x y zA A i A j A k

B

B B j

= + +

=

kN

Joint rests against a wallˆ ˆ kN

yB B j

C

C C j C k

=

= +

95

kNy zC C j C k= +

96


Truss


= − + + + − + + − = ˆ ˆ ˆ0 ( 2) 6) ( 1) 0 (1)x y y y z zF A i A B C j A C k

= × + × + × =

0 ( ) ( ) ( ) 0 (2)A AB AC ADM r B r C r F

Solving (1) and (2)ˆ ˆ ˆ2 4 1 kNA i j k= + +

ˆ1 kN ˆ1 kN

B j

C j

=

=

96

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Truss


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98


Truss


At joint C

jˆˆ

ˆ ˆ ˆ ˆˆ ( 2 3 ) ( 0 55470 0 83205 )

CA CA CA CA

CB

F F n F iFF F n i k F i k

= = −

= = + = +

( 2 3 ) ( 0.55470 0.83205 )13

ˆ ˆ ˆˆ ( 2 3 1 )14

CB CB CB CB

CDCD CD CD

F F n i k F i k

FF F n i j k

= = − + = − +

= = − + +

98

14ˆ ˆ ˆ( 0.53452 0.80178 0.26726 )

CD CD CD

CD CDF F i j k= − + +

99


Truss


=

0F

+ + + =

− − − +

0

ˆ( 0.55470 0.53452 )CA CB CD y

CA CB CD

F F F C

F F F i

+ +

+ =

ˆ(0.80178 1)ˆ(0.83205 0.26726 ) 0

CD

CB CD

F j

F F k

= −Thus, 1.24722 kN 0 40062 kN

CDFF =

=0.40062 kN0.44444 kN

CB

CA

FF

99

= 444 N AnsCAF T

100

Trusses Tips & HintsTruss

p

Members All member are of two-force member types. Look for zero-force members to simplify problems

Internal forces Use tension as +ve and compression as −ve

Visualize the ‘flow’ of forces/loads through the trusses Visualize the ‘flow’ of forces/loads through the trusses Check the results

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SummarySummary Frame: A structure which is designed to remain stationary

and support loadsand support loads. Machine: A structure which is designed to move and exert

loads. Truss: A structure which is comprised entirely of two-force

members.

Tension: axial forces at the ends are directed away from each othereach other.

Compression: axial forces at the ends are directed toward each other

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ConceptsReview

Concepts Structures are systems of interconnecting rigid bodies and

fastening (pins joints etc ) Structured can be analyzes byfastening (pins, joints, etc.). Structured can be analyzes by disassembling the system into inter-related FBDs and considering the equilibrium of individual members.

Frames and machines are structures that are not trusses. Frames stay stationary while machines can move.

Trusses are structures which comprise of only Trusses are structures which comprise of only interconnecting 2-force members.

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