Chapter 3Chapter 3TrussesTrusses

Germany

Trusses in BuildingTrusses in Building

Trusses Types for BuildingTrusses Types for Building

٦ Chapter 2

Truss BridgeTruss Bridge

Truss Bridge TypesTruss Bridge Types

Classification of coplanar TrussClassification of coplanar TrussSimple TrussSimple Truss

Compound Trussp

Complex Trussp

DeterminacyDeterminacyFor plane trussp

If b+r = 2j statically determinatej yIf b+r > 2j statically indeterminate

Where b = number of barsfr = number of external support reactionj = number of jointsj f j

StabilityStabilityFor plane trusspIf b+r < 2j unstableThe truss can be statically determinate or indeterminate y(b+r >=2j) but unstable in the following cases:External Stability: truss is externally unstable if all of y y f fits reactions are concurrent or parallel

Internal stabilit Internal stability:

Internally stable Internally unstable

Internally unstable

Classify each of the truss as stable , unstable, y , ,statically determinate or indeterminate.

11319 jb

22)11(2222

11 3 19=+

===

jrb

jrb KKKK

18)9(2219

9 4 15=+

===

jrb

jrb KKKK

Stable & edeterminat Statically22)11(22 ==j

Stable & ateindetermin Statically18)9(22 ==j

639 jb 8312 === jrb

12)6(2212

6 3 9=+

===

jrb

jrb KKKK

16)8(2215

8 3 12=+

===

jrb

jrb KKKK

Stable & edeterminat Statically12)6(22 ==j

Unstable16)8(22 ==j

Stability of Compound TrussStability of Compound Truss

The Method of JointsSTEPS FOR ANALYSIS

1. If the support reactions are not given, draw a FBD of the entire truss and determine all the support reactions using entire truss and determine all the support reactions using the equations of equilibrium.

2 Draw the free body diagram of a joint with one or two 2. Draw the free-body diagram of a joint with one or two unknowns. Assume that all unknown member forces act in tension (pulling the pin) unless you can determine by (pu g p ) u y u a d yinspection that the forces are compression loads.

3. Apply the scalar equations of equilibrium, ∑ FX = 0 and 3. Apply the scalar equations of equilibrium, ∑ FX 0 and ∑FY = 0, to determine the unknown(s). If the answer is positive, then the assumed direction (tension) is correct, otherwise it is in the opposite direction (compression).

4. Repeat steps 2 and 3 at each joint in succession until all the p p jrequired forces are determined.

The Method of JointsThe Method of Joints

Example 1Example 1Solve the following truss

;0F =∑

0C 8030sin4

;0

FKNFF

F

AGAG

y

−=⇒=+

=

∑

∑

T 93.6030cos8;0

KNFFF

ABAB

x

=⇒=−

=∑

C620303

;0

KNFF

Fy

⇒

=∑

;0C 6.2030cos3

FKNFF

x

GBGB

=

−=⇒=−−

∑C 50.6030sin38 KNFF GFGF −=⇒=+−

∑T 6.2060sin6.260sin

;0

KNFF

F

BFBF

y

=⇒=−

=∑

093.660cos6.260cos6.2;0

FF

BC

x

=−++

=∑

T 33.4 KNFBC =⇒

Group WorkGroup WorkDetermine the force in each member

Zero Force MemberZero Force Member1- If a joint has only two non-colinear members and there is 1- If a joint has only two non-colinear members and there is no external load or support reaction at that joint, then those two members are zero-force members

Zero Force MemberZero Force Member2- If three members form a truss joint for which two of the members are collinear and there is no external load or reaction at that joint, then the third non-collinear member is a zero force

bmember.

Example 2Example 2Find Zero force member of the following truss

Method of Section

The Method of SectionThe Method of Section

The Method of SectionThe Method of SectionSTEPS FOR ANALYSIS

1. Decide how you need to “cut” the truss. This is based on: a) where you need to determine forces, and, b) where the total ) y , , )number of unknowns does not exceed three (in general).

2. Decide which side of the cut truss will be easier to work with (minimize the number of reactions you have to find).

3. If required, determine the necessary support reactions by d i th FBD f th ti t d l i th E fEdrawing the FBD of the entire truss and applying the EofE.

4. Draw the FBD of the selected part of the cut truss. We need to indicate the unknown forces at the cut members. Initially ywe assume all the members are in tension, as we did when using the method of joints. Upon solving, if the answer is

i i h b i i i i f positive, the member is in tension as per our assumption. If the answer is negative, the member must be in compression. (Please note that you can also assume forces to be either tension or (Please note that you can also assume forces to be either tension or compression by inspection as was done in the previous example above.))

5. Apply the equations of equilibrium (EofE) to the selected cut section of the truss to solve for the unknown member forces. Please note that in most cases it is possible to write one equation to solve for one unknown directly.

Example 2Example 2Solve the CF & GC members in the truss

;0M E =∑C 4.3460)93.6(300)12(30sin IbFF CFCF −=⇒=+

;0M A =∑T 4.3460)12(30sin4.346)93.6(300)12( IbFF GCGC

A

=⇒=−−∑

Example 3Example 3Solve the GF & GD members in the truss

C8370)3(7)3(6260 kNFFM∑C 8.10)6(2)3(7)6(3.56sin0

C 83.70)3(7)3(6.26cos0

kNFFM

kNFFM

GFGDo

GFGFD

−=⇒=+−−→=

−=⇒=+→=

∑∑

Example 4Example 4Solve the ED & EB members in the truss

Example 5Example 5Solve the BC & MC members in the K-truss

T 21750)20()15(29000 IbFFM BCBCL =⇒=+−→=∑

TIbFF MBy 12000 BJoint At =→=∑ y∑

tttit t lF thTIbFFF MLMLy 29001200120029000

part cutting totalFor the=⇒−+−→=∑

MJoint At

FFF

FFF

MKMCy

MKMCx

0120029000

00

132

132

133

133

=+−−→=

=+→=

∑∑

TIbFCIbFMK

MKMCy

1532 1532

1313

==

∑

TIbFMC 1532=

Group work 1Group work 1Solve All members

Group work 2Group work 2Solve All members

Group work 3Group work 3Solve members CH , CI

Example 4 Compound TrussExample 4__Compound TrussSolve the truss

C4630)60i4()2(4)4(50 kNFFM ⇒++→∑ C 46.30)60sin4()2(4)4(50 kNFFM HGHGC =⇒=++−→=∑

& FF AIAB⇒AJoint

& FF & FF

IBIJ

HJHI

AIAB

⇒⇒

IJoint HJoint

F & FF

JC

BJBC

IBIJ

⇒⇒

JJoint BJoint

JC⇒JJo

Example 5 Compound TrussExample 5__Compound Truss

Example 5 Compound TrussExample 5__Compound TrussSolve the truss

F& FF

EB

ABAE

⇒⇒

BJoint AJoint

AEFSolveAfter

AGAF

AE

& FFF

⇒AJoint SolveAfter

Complex TrussComplex Truss

iii xsSS += '

?0' =+=

xxsSS ECECEC

Force in members?=x

Complex Trusses P

DF

Er + b = 2j,

3 9 2(6)DF 3 9 (6)

SAD

A C•Determinate•Stable

=

B

F´ PE E

=F´ECP

DFsEC

DF

๑

๑

+ X xA C A C

S´EC + x sEC = ๐

x = S´EC

A

B

C

= FAD

A

B

C

sEC

Si = S´i + x si

?0'

==+=

xxsSS ECECEC

xsSS += '

?x

iii xsSS +=

Member

AB

'iS is ixsiS

ABACAFFEBEEDFCEC

Example 5 Complex TrussExample 5__Complex Truss

Space TrussP• Determinacy and Stability

Space TrussDeterminacy and Stability

b + r < ๓junstable truss b ju stab e t ussb + r = 3jstatically determinate-check stability

statically indeterminate-check stability b + r > 3jy y j

Space Truss•x, y, z, Force Components

Space Trussx, y, z, Force Components

222 zyxl ++ zyxl ++=

)(lxFFx = l

)(lyFFy =

)(lzFFz =

222zyx FFFF ++=

z z

Support Reactions

y yFy

x xz

short link

z

y

x F

y

ll x

zFzx roller

z

x

y

F

Fxyslotted roller constrained in a cylinder x

zFzx in a cylinder

z

x

yFx

Fz

Fyy

x ball-and -socket

Zero Force MemberZero Force Member1- If all but one of the members connected to a joint lie on the

l d id d l l d h j i same plane, and provided no external load act on the joint, then the member not lying in the plane of the other members must must subjected to zero force.

ΣFz = ๐ , FD = ๐z ๐ , D ๐

Zero Force MemberZero Force Member2- If it has been determined that all but two of several

b d j i f h h members connected at a joint support zero force, then the two remaining members must also support zero force, provided they don’t lie a long the same line and no external load act on they don t lie a long the same line and no external load act on the joint.

ΣF = ๐ F = ๐ΣFz = ๐ , FB = ๐

ΣFy = ๐ , FD = ๐

Example 6 Space TrussExample 6__Space Truss

Example 7 Space TrussExample 7__Space Truss