6.5 space trusses a space truss consists of members joined together at their ends to form a stable...

6.5 Space Trusses6.5 Space Trusses

A space truss consists of members joined together at their ends to form a stable 3D structure

The simplest space truss is a tetrahedron, formed by joined 6 members as shown

Any additional members added would be redundant in supporting force P


Assumptions for DesignThe members of a space truss may

be treated as two force members provided the external loading is applied at the joints and the joints consist of ball and socket connections

If the weight of the member is to be considered, apply it as a vertical force, half of its magnitude applied at each end of the member


Procedure for AnalysisMethod of Joints To determine the forces in all the members

of the truss Solve the three scalar equilibrium ∑Fx = 0,

∑Fy = 0, ∑Fz = 0 at each joint The force analysis begins at a point having

at least one unknown force and at most three unknown forces

Cartesian vector analysis used for 3D


Procedure for AnalysisMethod of Sections Used to determine a few member forces When an imaginary section is passes

through a truss and the truss is separated into two parts, the below equations of equilibrium must be satisfied

∑Fx = 0, ∑Fy = 0, ∑Fz = 0∑Mx = 0, ∑My = 0, ∑Mz = 0

By proper selection, the unknown forces can be determined using a single equilibrium equation


Example 6.8Determine the forces acting in the

members of the space truss. Indicate whether the members are in tension or compression.

SolutionJoint A

0577.0577.0577.04

0

;0

)577.0577.0577.0(

,,}4{

kFjFiFkFjFj

FFFP

F

kjiF

rr

FF

kFFjFFkNjP

AEAEAEACAB

AEACAB

AE

AE

AEAEAE

ACACABAB

6.5 Space Trusses6.5 Space TrussesView Free Body Diagram

SolutionJoint A

To show0

)(2

)(66.5

0707.02;0

045sin4;0

0707.045cos;0

CEDCDE

BD

BEB

BEBDz

By

BEBx

FFF

CkNF

TkNFR

FFF

RF

FRF


6.6 Frames and Machines6.6 Frames and Machines

Composed of pin-connected multi-force members (subjected to more than two forces)

Frames are stationary and are used to support the loads while machines contain moving parts, designated to transmit and alter the effects of forces

Apply equations of equilibrium to each member to determine the unknown forces


Free-Body Diagram Isolate each part by drawing its

outlined shape- show all the forces and the couple moments that act on the part- label or identify each known and unknown force and couple moment with reference to the established x, y and z coordinate system


Free-Body Diagram- indicate any dimension used for taking moments- equations of equilibrium are easier to apply when the forces are represented in their rectangular coordinates- sense of any unknown force or moment can be assumed


Free-Body Diagram Identify all the two force members in the

structure and represent their FBD as having two equal but opposite collinear forces acting at their points of application

Forces common to any contracting member act with equal magnitudes but opposite sense on the respective members


Free-Body Diagram- treat two members as a system of connected members - these forces are internal and are not shown on the FBD- if the FBD of each member is drawn, the forces are external and must be shown on the FBD


Example 6.9For the frame, draw the free-body

diagram of (a) each member, (b) the pin at B and (c) the two members connected together.

SolutionPart (a) members BA and BC are not two-force

members BC is subjected to 3 forces, the resultant

force from pins B and C and the external P

AB is subjected to the resultant forces from the pins at A and B and the external moment M


SolutionPart (b) Pin at B is subjected to two forces, force

of the member BC on the pin and the force of member AB on the pin

For equilibrium, theseforces and respective components must be equal but opposite


SolutionPart (b) But Bx and By shown equal and opposite

on members AB ad BC results from the equilibrium analysis of the pin rather from Newton’s third law


SolutionPart (c) FBD of both connected members without

the supporting pins at A and C Bx and By are not shown since

they form equal but opposite collinear pairs of internal forces


SolutionPart (c) To be consistent when applying the

equilibrium equations, the unknown force components at A and C must act in the same sense

Couple moment M can be applied at any point on the frame to determine reactions at A and C



Example 6.10A constant tension in the conveyor belt is maintained by using the device. Draw the FBD of the frame and the cylinder which supports the belt. The suspended black has a weight of W.


Solution Idealized model of the device Angle θ assumed known Tension in the belt is the same

on each side of the cylinder since it is free to turn


Solution FBD of the cylinder and

the frame Bx and By provide equal

but opposite couple moments on the cylinder

Half of the pin reactions at A act on each side of the frame since pin connections occur on each side


Example 6.11Draw the free-body diagrams of each part of the smooth piston and link mechanism used to crush recycled cans.


Solution Member AB is a two force

member FBD of the parts


Solution Since the pins at B and D connect only

two parts together, the forces are equal but opposite on the separate FBD of their connected members

Four components of the force act on the piston: Dx and Dy represent the effects of the pin and Nw is the resultant force of the floor and P is the resultant compressive force caused by can C


Example 6.12For the frame, draw the free-body diagrams of (a) the entire frame including the pulleys and cords, (b) the frame without the pulleys and cords, and (c) each of the pulley.

SolutionPart (a) Consider the entire frame, interactions at

the points where the pulleys and cords are connected to the frame become pairs of internal forces which cancel each other and not shown on the FBD



SolutionPart (b) and (c) When cords and pulleys are

removed, their effect on the frame must be shown


Example 6.13Draw the free-body diagrams of the bucket and the vertical boom of the back hoe. The bucket and its content has a weight W. Neglect the weight of the members.


Solution Idealized model of the assembly Members AB, BC, BE and HI are

two force members


Solution FBD of the bucket and boom Pin C subjected to 2 forces,

force of the link BC and force of the boom

Pin at B subjected to three forces, force by the hydraulic cylinder and the forces caused by the link

These forces are related by equation of force equilibrium


Equations of Equilibrium Provided the structure is properly

supported and contains no more supports and members than necessary to prevent collapse, the unknown forces at the supports and connections can be determined from the equations of equilibrium

The selection of the FBD for analysis are completely arbitrary and may represent each of the members of the structure, a portion or its entirety.


Equations of Equilibrium Consider the frame in fig (a) Dismembering the frame in fig (b),

equations of equilibrium can be used FBD of the entire frame in fig (c)


Procedures for AnalysisFBD Draw the FBD of the entire structure, a

portion or each of its members Choice is dependent on the most direct

solution to the problem When the FBD of a group of members of a

structure is drawn, the forces at the connected parts are internal forces and are not shown

Forces common to two members which are in contact act with equal magnitude but opposite sense on their respective FBD


Procedures for AnalysisFBD Two force members, regardless of their

shape, have equal but opposite collinear forces acting at the ends of the member

In many cases, the proper sense of the unknown force can be determined by inspection

Otherwise, assume the sense of the unknowns

A couple moment is a free vector and can act on any point of the FBD


Procedures for AnalysisFBD A force is a sliding vector and can act at

any point along its line of action

Equations of Equilibrium Count the number of unknowns and

compare to the number of equilibrium equations available

In 2D, there are 3 equilibrium equations written for each member


Procedures for AnalysisEquations of Equilibrium Sum moments about a point that lies at

the intersection of the lines of action of as many unknown forces as possible

If the solution of a force or couple moment magnitude is found to be negative, it means the sense of the force is the reserve of that shown on the FBD


Example 6.14Determine the horizontal and

vertical components of the force which the

pin C exerts on member CB of the frame.


SolutionMethod 1 Identify member AB as two force

member FBD of the members AB and BC

Solution

NC

CNN

F

NC

C

F

NF

mFmN

M

y

y

y

x

x

x

AB

AB

C

1000

0200060sin7.1154

;0

577

060cos7.1154

;0

7.1154

0)4(60sin)2(2000

;0



SolutionMethod 2 Fail to identify member AB as

two force member

SolutionMember AB

0

;0

0

;0

0)60cos3()60sin3(

;0

yy

y

xx

x

yx

A

BA

F

BA

F

mBmB

M


SolutionMember BC

NCNCNBNB

CNB

F

CB

F

mBmN

M

yxxy

yy

y

xx

x

y

C

1000;577;577;1000

02000

;0

0

;0

0)4()2(2000

;0



Example 6.15The compound beam is pin connected at B. Determine the reactions at its support. Neglect its weight and thickness.


Solution FBD of the entire frame Dismember the beam into two

segments since there are 4 unknowns but 3 equations of equilibrium

SolutionSegment BC

08

;0

0)2()1(8

;0

0

;0

yy

y

y

B

x

x

CkNB

F

mCmkN

M

B

F


SolutionMember AB

kNCkNBBmkNMkNAkNA

BkNA

F

mBmkNM

M

BkNA

F

yyxAyx

yy

y

yA

A

xx

x

4;4;0;.32;12;6

054

)10(

;0

0)4()2(54

)10(

;0

053

)10(

;0



Example 6.16Determine the horizontal and vertical components of the force which the pin

at C exerts on member ABCD of the frame.


Solution Member BC is a two force

member FBD of the entire frame FBD of each member

SolutionEntire Frame

NA

NAF

NA

NAF

ND

mDmNM

y

yy

x

xx

x

xA

981

0981;0

7.700

07.700;0

7.700

0)8.2()2(981;0


SolutionMember CEF

NC

NNCF

NC

NCF

NF

mFmNM

y

yy

x

xx

B

BC

245

0981)45sin2.1734(;0

1226

0)45cos2.1734(;0

2.1734

0)6.1)(45sin()2(981;0



Example 6.17The smooth disk is pinned at D and has a weight

of 20N. Neglect the weights of others member, determine the horizontal and vertical components of the reaction at pins B and D


Solution FBD of the entire frame FBD of the members

SolutionEntire Frame

NA

NAF

NA

NAF

NC

cmCcmNM

y

yy

x

xx

x

xA

20

020;0

1.17

01.17;0

1.17

0)5.3()3(20;0


SolutionMember AB

NB

BNNF

NN

cmNcmNM

NB

BNF

y

yy

D

DA

x

xx

20

04020;0

40

0)3()6(20;0

1.17

01.17;0


SolutionDisk

ND

DNN

F

D

F

y

y

y

x

x

20

02040

;0

0

;0



Example 6.18Determine the tension in the

cables and also the force P required to support the 600N force using the frictionless pulley system.


Solution FBD of each pulley Continuous cable and

frictionless pulley = constant tension P

Link connection between pulleys B and C is a two force member

SolutionPulley A

Pulley B

Pulley C

NR

TPRF

NT

PTF

NP

NPF

y

y

y

800

02;0

400

02;0

200

06003;0



Example 6.19A man having a weight of 750N supports himself by means of the cable and pulley system. If the seat has a weight of 75N, determine the force he must exert on the cable at A and the force he exerts on the seat. Neglect the weight of the cables and pulleys.


SolutionMethod 1 FBD of the man, seat and pulley

C

SolutionMan

Seat

Pulley C

NNNTT

TTF

NNTF

NNTF

EEA

AEy

SEy

SAy

200;275;550

02;0

075;0

0750;0



SolutionMethod 2 FBD of the man, seat and pulley

C as a single system

Solution

NNT

TTF

NNTF

NT

NNTF

EA

AEy

SEy

E

Ey

200;550

02;0

075;0

275

0750753;0



Example 6.20The hand exerts a force of 35N on the grip of

the spring compressor. Determine the force in the spring needed to maintain equilibrium of the mechanism.


Solution FBD for parts DC and ABG

SolutionLever ABG

Pin E

NF

NFF

FFF

FFF

NF

mmNmmFM

x

EFED

EFEAy

EA

EAB

140

014060cos2;0

060sin60sin;0

140

0)100(35)25(;0


SolutionArm DC

NF

mmmmF

M

s

s

C

62.60

0)75(30cos140)150(

;0



Example 6.21The 100kg block is held in equilibrium by means of the pulley and the continuous cable system. If the cable is attached to the pin at B, compute the forces which this pin exerts on each of its connecting members


Solution FBD of each member of

the frame Ad and CB are two force

members

View Free Body Diagram

SolutionPulley B

NB

NNBF

NB

NBF

y

yy

x

xx

3.837

05.49045sin5.490;0

8.346

045cos5.490;0


SolutionPin E

Two force member BC subjected to bending as caused by FBC

Better to make this member straight so that the force would only cause tension in the member

NF

NNFF

NF

NNFF

AB

ABx

CB

CBy

1343

08.346)1660(5

3;0

1660

05.4903.8375

4;0


Chapter SummaryChapter Summary

Truss AnalysisA simple truss consists of triangular

elements connected by pin jointsThe force within determined by

assuming all the members to be two force member, connected concurrently at each joint

Method of JointsFor the truss in equilibrium, each of its

joint is also in equilibrium


Method of JointsFor a coplanar truss, the concurrent force

at each joint must satisfy force equilibriumFor numerical solution of the forces in the

members, select a joint that has FBD with at most 2 unknown and 1 known forces

Once a member force is determined, use its value and apply it to an adjacent joint

Forces that pull on the joint are in tension


Method of JointsForces that push on the joint are in

compressionTo avoid simultaneous solution of

two equations, sum the force in a direction that is perpendicular to one of the unknown

To simplify problem-solving, first identify all the zero-force members


Method of SectionsFor the truss in equilibrium, each

section is also in equilibriumPass a section through the member

whose force is to be determinedDraw the FBD of the sectioned part

having the least forces on itForces that pull on the section are in

tension


Method of Sections Forces that push on the section are in

compression For a coplanar force system, use the three

equations of equilibrium for solving If possible, sum the force in a direction that

is perpendicular to two of the three unknown forces

Sum the moment about a point that passes through the line of action of two of the three unknown forces


Frames and MachinesThe forces acting at the joints of a

frame or machine can be determined by drawing the FBD of each of its members or parts

Principle of action-reaction should be observed when drawing the forces

For coplanar force system, there are three equilibrium equations for each member

Chapter ReviewChapter Review

6.5 space trusses a space truss consists of members joined together at their ends to form a stable...

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