truss-beer & johnston 9th ed

7/22/2019 Truss-beer & Johnston 9th Ed

1/15

Ninth EditionNinth Edition

VECTOR MECHANICS FOR ENGINEERS:

STATICSSTATICSCHAPTER

Ferdinand P. BeerFerdinand P. Beer

E. Russell Johnston Jr.E. Russell Johnston Jr.

Lecture Notes:Lecture Notes:

Analysis of Structures:

..

Texas Tech UniversityTexas Tech University Trusses

2010 The McGraw-Hill Companies, Inc. All rights reserved.


2/15

Vector Mechanics for Engineers: StaticsVector Mechanics for Engineers: StaticsNinth

Edition

Introduction For the equilibrium of structures made of several

connected parts, the internal forces as well the external

orces are considered.

In the interaction between connected parts, Newtons 3rd

Law states that theforces of action and reaction

between bodies in contact have the same magnitude,

same line of action, and opposite sense.

a) Frames: contain at least one one multi-force

member, i.e., member acted upon by 3 or more

.

b) Trusses: formed from two-force members, i.e.,

straight members with end point connections

2010 The McGraw-Hill Companies, Inc. All rights reserved. 6 - 2


3/15


Edition

Definition of a Truss A truss consists of straight members connected at

joints. No member is continuous through a joint.

Most structures are made of several trusses joinedtogether to form a space framework. Each truss

Bolted or welded connections are assumed to be

be treated as a two-dimensional structure.

pinned together. Forces acting at the member ends

reduce to a single force and no couple. Only two-

force members are considered.

When forces tend to pull the member apart, it is in

tension. When the forces tend to compress the


member, it is in compression.


4/15


Edition

Definition of a Truss



5/15


Edition

Analysis of Trusses by the Method of J oints Dismember the truss and create a freebodydiagram for each member and pin.

e wo orces exer e on eac mem er are

equal, have the same line of action, andopposite sense.

Forces exerted by a member on the pins or

joints at its ends are directed along the member

and equal and opposite.

Conditions of equilibrium on the pins provide

2n equations for 2n unknowns. For a simple

=, .

forces and 3 reaction forces at the supports.

Conditions for equilibrium for the entire truss


provide 3 additional equations which are not

independent of the pin equations.


6/15


Edition

J oints Under Special Loading Conditions Forces in opposite members intersecting intwo straight lines at a joint are equal.

equal when a load is aligned with a thirdmember. The third member force is equal

to the load includin zero load .

The forces in two members connected at a

joint are equal if the members are alignedand zero otherwise.

Recognition of joints under special loading

conditions simplifies a truss analysis.



7/15


Edition

Sample Problem 6.1Using the method of joints, determine the force

in each member of the truss.

SOLUTION:

Based on a free-body diagram of the entire

0M =

truss, solve the 3 equilibrium equations for the

reactions atEandC.

( )( ) ( )( ) ( )ft6ft12lb1000ft24lb2000 E+=

= lb000,10E

== xx CF 0 0=xC

++==F lb10 000lb1000-lb20000


= lb7000yC


8/15


Edition

Sample Problem 6.1

member forces. Determine these from the

joint equilibrium requirements.

534

lb2000 ADAB FF ==CF

TF

AD

AB

lb2500

lb1500

=

=

There are now only two unknown memberforces at joint D.

=

2010 The McGraw-Hill Companies, Inc. All rights reserved. 6 - 8( ) DADE FF 53

2=

CFDE

DB

lb3000

=

=


9/15


Edition

Sample Problem 6.1 There are now only two unknown member

forces at joint B. Assume both are in tension.

( )lb3750

250010000 54

5

4

=

==

BE

BEy

FFF

CFBE lb3750=

( ) ( )

lb5250

375025001500053

53

+=

==

BC

BCx

F

FF

TFBC lb5250=

There is one unknown member force at joint

E. Assume the member is in tension.

( )

lb8750

37503000053

53

=

++==

EC

ECx

F

FF

CFEC lb8750=



10/15


Edition

Sample Problem 6.1

All member forces and support reactions are

known at joint C. However, the joint equilibrium

re uirements ma be a lied to check the results.

( ) ( )checks087505250

4

53

==

=+= xF

5y



11/15


Edition

Analysis of Trusses by the Method of Sections

When the force in only one member or the

forces in a very few members are desired, the

.

To determine the force in memberBD, pass a

section throu h 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.



12/15


Edition

Sample Problem 6.3

SOLUTION:

Take the entire truss as a free body.

Apply the conditions for static equilib-

rium to solve for the reactions atA andL.

Pass a section throu h members FH

GH, andGIand take the right-hand

section as a free body.

pp y e con ons or s a c

equilibrium to determine the desired

member forces.

Determine the force in members FH,GH, andGI.



13/15


Edition

Sample Problem 6.3

SOLUTION:

Take the entire truss as a free bod .

Apply the conditions for static equilib-rium to solve for the reactions atA andL.

== kN6m15kN6m10kN6m50M

( )( ) ( )( ) ( )

=

+

kN5.7

m25kN1m25kN1m20

L

L

=++==

kN5.12

kN200

A

ALFy



14/15


Edition

Sample Problem 6.3

Pass a section through members FH, GH, andGI

and take the right-hand section as a free body.

Apply the conditions for static equilibrium to

determine the desired member forces.

( )( ) ( )( ) ( )

kN13.13

0m33.5m5kN1m10kN7.50

0

+=

=

=

GI

H

F

F

M

TFGI kN13.13=



15/15


Edition

Sample Problem 6.3

0

07.285333.0m15

m8tan

=

====

GMGL

FG

( )( ) ( )( ) ( )( )( )( ) 0m8cosm5kN1m10kN1m15kN7.5

=+

FHF

.=FHCFFH kN82.13=

( )15.439375.0

m8

m5tan

32

====HI

GI

( )( ) ( )( ) ( )( ) 0m10cosm5kN1m10kN1

0

=++

=GH

L

F

M


.=GHCFGH kN371.1=

truss-beer & johnston 9th ed

Documents