files 2- handouts lecture-24 to 26

8/2/2019 Files 2- Handouts Lecture-24 to 26

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COURSE:COURSE: CE 201 (STATI CS)CE 201 (STATI CS)

LECTURE NO.:LECTURE NO.: 24 to 2624 to 26

FACULTY:FACULTY: DR. SHAMSHAD AHMADDR. SHAMSHAD AHMAD

DEPARTMENT:DEPARTMENT: CI VI L ENGI NEERI NGCI VI L ENGI NEERI NG

UNI VERSI TY:UNI VERSI TY: KI NG FAHD UNI VERSI TY OF PETROLEUMKI NG FAHD UNI VERSI TY OF PETROLEUM

& MI NERALS, DHAHRAN, SAUDI ARABI A& MI NERALS, DHAHRAN, SAUDI ARABI A

TEXT BOOK:TEXT BOOK: ENGI NEERI NG MECHANI CSENGI NEERI NG MECHANI CS--STATI CSSTATI CS

by R.C. HI BBELER, PRENTI CE HALLby R.C. HI BBELER, PRENTI CE HALL


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LECTURE NO. 24 t o 26LECTURE NO. 24 to 26

THE METHOD OF JOI NTS ANDTHE METHOD OF JOI NTS AND

ZERO FORCE MEMBERSZERO FORCE MEMBERS

Objectives:Objectives:

To show how to determine the forces in theTo show how to determine the forces in themembers of a t russ using t he method of j ointsmembers of a t russ using t he method of j oints

To show how to identify the zeroTo show how to identify the zero -- forceforce

members of a t russmembers of a t russ


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THE METHOD OF JOI NTSTHE METHOD OF JOI NTS

The method of joints for the analysis of a truss consists ofThe method of joints for the analysis of a truss consists of

t he follow ing steps:t he follow ing steps:

Determine the support reactions considering the equilibrium ofDetermine the support reactions considering the equilibrium of

t russ as a wholet russ as a whole

Consider only one j oint at a t imeConsider only one j oint at a t ime

Draw the free body diagram for the joint into consideration,Draw the free body diagram for the joint into consideration,

indicating the magnitudes, directions, and the senses of theindicating the magnitudes, directions, and the senses of the

know n ext ernal forces, react ions, and m ember forcesknow n ext ernal forces, react ions, and m ember forces

Apply the conditions of equil ibrium for t he joint int o consideraApply the conditions of equil ibrium for t he joint int o consideration,t ion,

as:as:

FFxx == 00 Eq.1Eq.1

FFyy == 00 Eq.2Eq.2

Solve the equations of equilibrium (Eq.1 & 2) to determine theSolve the equations of equilibrium (Eq.1 & 2) to determine the

forces in the members meet ing at t he j oint int o considerationforces in t he members meet ing at t he joint int o considerat ion

Go on considering the other joints of the truss one by one ti l lGo on considering the other joints of the truss one by one ti l l t hetheforces in each of t he members of t he t russ are determ inedforces in each of t he members of t he t russ are determ ined


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THE METHOD OF JOI NTSTHE METHOD OF JOI NTS

I mport ant PointsI mportant Points

Always assume the sense of the unknown member forcesAlways assume the sense of the unknown member forces

act ing on the jointact ing on the joint s f rees f ree--body diagram asbody diagram as tensiontension

I f aft er analysis t he magnitude of a member force is foundI f aft er analysis t he magnitude of a member force is foundto be negative, reverse its sense, i.e. the sense is to beto be negative, reverse its sense, i.e. the sense is to be

compression.compression.

I n all cases, t he analysis should st art at a j oint having atI n all cases, t he analysis should st art at a j oint having atleast one know n force or react ion and at most tw o unknow nleast one know n force or react ion and at most tw o unknow n

forces, for example j ointforces, for example joint BB in t he above figure.in t he above figure.


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The zeroThe zero-- force members of a truss supportforce members of a truss support nonoloadingloading and are used to increase the stability ofand are used to increase the stability of

the truss during construction and to providethe truss during construction and to provide

suppor t if t he applied load is changed.suppor t if t he applied load is changed.

I dent if icat ion of t he zeroIdentif ication of the zero-- force members of aforce members of a

truss greatly simplifies the analysis of the truss,truss greatly simplifies the analysis of the truss,

using the method of j oints.using the method of j oints.

Following two general rules may be helpful inFollowing two general rules may be helpful in

ident ifying t he zeroident ifying t he zero-- force members:force members:


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RuleRule-- II I f only two members form aIf only two members form atruss joint and no external loadtruss joint and no external loador support reaction is applied toor support reaction is applied tothe joint, the members must bethe joint, the members must bezerozero-- force membersforce members ..

For example:For example:I nit ial ly, 9 members but reduced toI nit ial ly, 9 members but reduced to

only 5 members after removing theonly 5 members after removing the

zero force members.zero force members.

0; 0

0; 0

x AB

y AF

F F

F F

+

= =

+ = =

+Fy = 0 FDC sin = 0 FDC = 0

+Fx = 0 FDE + 0 = 0 FDE = 0


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RuleRule-- I II I If three members form a truss jointI f t hree members form a truss j ointfor which two of the members arefor which two of the members are

collinear, t he third member is a zerocollinear, t he third member is a zero--

force member provided no externalforce member provided no external

force or support react ion is applied t oforce or support react ion is applied t othe j ointt he j oint ..

For example:For example:I nit ial ly, 7 members but reduced toI nit ial ly, 7 members but reduced to

only 3 members after removing theonly 3 members after removing the

zero force members.zero force members.

+ Fx=0 FDA = 0+ Fy = 0 FDC= FDE

+ Fx=0 FCA sin = 0 FCA = 0+ Fy = 0 FCB= FCD


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PROBLEM SOLVI NG:PROBLEM SOLVI NG: Example # 1Example # 1

Determine the force in each member of the truss and state ithe members are in tension or compression. Set P1 = 500-lb

andP2 = 100-lb.


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The free-body diagram of the

entire truss is shown below:

AxA C

CyAy

P2= 100lb

B

P1= 500lb

AxA C

CyAy

P2= 100lb

B

P1= 500lb

The support reactions can be calculatedby applying equilibrium conditions to the

above free-body diagram, as follows:

Fx

= 0 Ax 100 = 0 (1) Ax= 100 lbMabout A = 0 14 Cy + 500 6 + 100 8 = 0 (2) C = 271.428 lbFy = 0 Ay + Cy 500 = 0 Ay + Cy = 500 (3) A = 500 271.428 = 228.572 lb.


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The magnitude and sense of the forces

in each member of the truss may be

determined by considering free-body

diagram of the joints, as follows:

Joint C:

The free-body diagram of joint Cis as

follows:

45o

FCA

FCB

F.B.D. of joint C

45o

FCA

FCB

45o

FCA

FCB

F.B.D. of joint C

271.43 lb

45o

FCA

FCB

F.B.D. of joint C

45o

FCA

FCB

45o

FCA

FCB

F.B.D. of joint C

45o

FCA

FCB

F.B.D. of joint C

45o

FCA

FCB

45o

FCA

FCB

F.B.D. of joint C

271.43 lb

Fy = 0 271.43 FCB sin 45 = 0FCB = 271.43sin 45 = 383.86 lb (T) Ans.Fx= 0 FCA FCB cos 45 = 0 FCA = FCB cos 45 = 383.86 cos 45o= 271.43 lb = 271.43 lb (C) Ans.

Joint A:

The FBD of jointA is as follows:

271.43

FAB228.53 lb

100 lb

271.43

FAB228.53 lb

100 lb 1 8tan 59.036

= =

sin = 0.8 and cos = 0.6

Fx= 0 100 271.43 + FAB cos = 0FAB = 285.71 (T) Ans.


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Determine the force in each member othe truss and state if the members are

in tension or compression. Set P = 4

KN.


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Analysis of the given truss may bestarted from joint A without

calculating the support reactions

because at joint A there are only two

unknowns.Joint A:

The FBD of jointA is as follows:

Applying equilibrium conditions atointA,

Fy = 0 4 FAEsin = 0 FAE= 40.447 = 8.948 kN

= 8.948 kN (C) Ans.

Fx= 0 FAB + FAEcos = 0 FAB = 0.894 FAE = 0.894 ( 8.948)

= 8 kN (T) Ans.

Joint B:

The FBD of jointB is as follows:

Fx= 0 8 + FBC= 0 FBC= 8 kN (T) Ans.

Fy = 0

8 FBE= 0 FBE= 8 kN FBE= 8 kN (C) Ans.


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Joint E:The FBD of jointEis as follows:

Fx= 08.948 cos+FECcos+ FED cos= 0FEC+ FED = 8.948 (1)

Fy = 0 88.948sin+FECsinFED sin = 0FECFED= 8 8.948sinsin + = 26.836 kN (2)Eq. (1) + Eq. (2) 2FEC= 17.88 FEC= 8.9458 kN (T) Ans.From Eq. (1)

FED = 8.948 8.948 = 17.896 kN FED = 17.896 kN ( C ). Ans.Joint D:

The FBD of jointD is as follows:Fy= 0 FDC 17.896 sin = 0 FDC= 8 kN (T) Ans.Fx= 0 17.896 cos -Dx = 0 Dx= 17.896 0.894 = 16 kN


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Joint C:

The FBD of joint Cis as follows:

Fx= 0 8 8.948 cosCx= 0 Cx= 16kN = 16 kN( )Fy = 0 4 8 8.948 sin + Cy = 0 C = 16 kN( )

The magnitude and sense of the force

in each member of the truss is shownbelow:

4 kN 8 kN 4 kN

8 kN8 kNA B C

D

E

8 kN

8 kN

8.948

kN

17.896kN

8.948kN

Cx = 16 kN

Dx = 16 kN

Cy = 16 kN

4 kN 8 kN 4 kN

8 kN8 kNA B C

D

E

8 kN

8 kN

8.948

kN

17.896kN

8.948kN

Cx = 16 kN

Dx = 16 kN

Cy = 16 kN


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Determine the force in each member of the trussand state if the members are in tension or

compression considering only the self-weight o

the members @ 4 kg/m. SetP = 0.


16/28


Masses of members are calculated

below using the unit weight and the

length of the members:

mAB = 4 4 = 16kg = 157NmAE = 4 ( )

2 24 2+ = 17.88 kg = 175 N

mBE = 4 2 = 8 kg = 78 NmBC = 4 4 = 16 kg = 157 NmCE = 4

( )

2 24 2+ = 17.88 kg = 175 N

Loads at joints of the truss are

calculated as follows:

PA = (mAB + mAE)= (157 + 175) = 166 N

PB = (mAB + mBE + mBC)

= (157 + 78 + 157) = 196 N

PC = (mBC + mCE + mCD)= (157 + 175 +157) = 245 N

PE = (mAE + mBC + mEC + mED)

= (175 + 78 + 175 + 175) = 302 N

PD = (mED + mCD)

= (175 + 157) = 166 N

FBD of the truss showing loads at

oints is given below:

166N 196N 245N

AB C

D

E

Cx

Dx

Cy

166N

302N

166N 196N 245N

AB C

D

E

Cx

Dx

Cy

166N

302N

166N 196N 245N

AB C

D

E

Cx

Dx

Cy

166N

302N

166N 196N 245N

AB C

D

E

Cx

Dx

Cy

166N

302N


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PROBLEM SOLVI NG:PROBLEM SOLVI NG: Example # 3Example # 3 166N 196N 245N

AB C

D

E

Cx

Dx

Cy

166N

302N

166N 196N 245N

AB C

D

E

Cx

Dx

Cy

166N

302N

166N 196N 245N

AB C

D

E

Cx

Dx

Cy

166N

302N

166N 196N 245N

AB C

D

E

Cx

Dx

Cy

166N

302N

Analysis of the given truss may be

started from joint A without

calculating the support reactionsbecause at joint A there are only two

unknowns.Joint A:

The free-body diagram of joint A is asfollows:

Fy = 0

166 FAEsin

= 0 FAE= 1660.449 = 371.36 N FAE= 371.36 N (C) Ans.Fx= 0 FAB + FAEcos = 0FAB = FAEcos= (371.36)0.894 = 332 N (T) Ans.

Joint B:The FBD of jointB is as follows:

Fx= 0 FBC 332 = 0 FBC= 332 N (T) Ans.Fy = 0 196 FBE= 0 FBE= 196 FBE= 196 N (C) Ans.


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PROBLEM SOLVI NG:PROBLEM SOLVI NG: Example # 3Example # 3 166N 196N 245N

AB C

D

E

Cx

Dx

Cy

166N

302N

166N 196N 245N

AB C

D

E

Cx

Dx

Cy

166N

302N

166N 196N 245N

AB C

D

E

Cx

Dx

Cy

166N

302N

166N 196N 245N

AB C

D

E

Cx

Dx

Cy

166N

302N

Joint C:


Fx= 0 Cx 332 557 cos = 0Cx= 830 NFy = 0 245 581557 sin +Cy = 0 C = 1075 N



166N 196N 245N

332N332NA

B C

D

E

196N

581N

557N

928N

371.36N

Cx = 830N

Dx =

Cy = 1075N

166N

302N

166N 196N 245N

332N332NA

B C

D

E

196N

581N

557N

928N

371.36N

Cx = 830N

Dx =

Cy = 1075N

166N

302N

830 N

166N 196N 245N

332N332NA

B C

D

E

196N

581N

557N

928N

371.36N

Cx = 830N

Dx =

Cy = 1075N

166N

302N

166N 196N 245N

332N332NA

B C

D

E

196N

581N

557N

928N

371.36N

Cx = 830N

Dx =

Cy = 1075N

166N

302N

830 N


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Determine the force in each member of the truss andstate if the members are in tension or compression.

SetP1 = P2 = 4 kN.


21/28


Analysis of the given truss may bestarted from joint C without

calculating the support reactions

because at joint C there are only two

unknowns.Joint C:


Applying equilibrium conditions atoint C,

Fy = 0 FCB sin4 = 0 FCB = 8 kN (T) Ans.Fx= 0 FCD FCB cos 30 = 0FCD= 6.93 kN = 6.93 kN (C) Ans.Joint D:

The free-body diagram of joint D is asfollows:

Fx= 0 FDE 6.93 = 0 FDE= 6.93 kN= 6.93 kN ( C) Ans.

D

Fy = 0 FDB 4 = 0 FDB = 4 kN ( T) Ans.


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Joint B:The FBD of jointB is as follows:

Fx= 0 FBA cos30 FBEcos30+ 8 cos30 = 0 FBA + FBE= 8 (1)

Fy = 0 +FBA sin 30 FBEsin 30 4 8 sin 30= 0 FBA FBE= 4 40.5+ = 16 (2)Solving, Eqs. (1) and (2),FBA = 12 kN (T) and FBC= 4 kN (C) Ans.

Support A

The FBD of supportA, as follows:

By Sine Rule,12

120 150 90

yxAA

Sin Sin Sin= =

From the above equation, Ax and Ay

can be determined as:x= 10.392 kN andA = 6 kN


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Support E

FBD of supportE, as follows:

Fx= 0 Ex 6.93 4 cos30 = 0 Ex= 10.392 kNFy = 0 Ey 4 sin30 = 0 E = 2 kN



4kN4kN

6.93kN 6.93kN

A

B

CDE

4kN4kN

12kN

8kN

Ey=2kN

Ay=6kN

Ax=10.392kN

Ex=10.392kN

4kN4kN

6.93kN 6.93kN

A

B

CDE

4kN4kN

12kN

8kN

Ey=2kN

Ay=6kN

Ax=10.392kN

Ex=10.392kN


24/28


For the given loading, determine the zero-force

members in the Pratt roof truss. Explain your

answers using appropriate joint free-body diagrams.


25/28


Since, there is no reaction or external

force at joint L and FLA and FLK are

collinear, FLB = 0 (according to Rule II)

FBC

FBK

F.B.D. of Joint B

B

FBA

FBC

FBK

F.B.D. of Joint B

B

FBA

FLB

FLK

F.B.D. of Joint L

L

FLA

FLB

FLK

F.B.D. of Joint L

L

FLA

Since, there is no external force or

reaction at jointB and FBA and FBCare

collinear, FBK = 0 (according to Rule

II)FKC

FKS

F.B.D. of Joint K

K

FKL

FKC

FKS

F.B.D. of Joint K

K

FKL

Since, there is no external force or

reaction at joint K, and FKL

and FKSare collinear, FKC = 0 (according to

Rule II)


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Similarly, considering joints H, F, I,

andE,

FHF= 0FFI= 0

FIE= 0

FEJ= 0

Therefore, the zero-force members o

the given of truss are as follows:

B,BK,KC,HF, FI,IEandEJ Ans.

After removing the zero-force

members, the truss diagram is shown

below:

Multiple Choice Problems


27/28


1. The zero-force members of the truss as shown in

the following figure are

(a)EA andED (b)DA and CA

(c) CA and CB (d)BA andBC

Ans: (b)

Feedback:

The members DA and CA are the zero-force

members by applying Rule II at jointsD and C



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Multiple Choice Problems2.The magnitude and sense of the force in member

CEof the truss as shown in the following figure are

(a) 0 (b) 5 kN (T)

(c) 5 kN (C) (d) none of these

Ans: (c)Feedback:

0 (at joint C)x

F = 5 0

5 kN = 5 kN (C)

CE

CE

F

F

+ = =

files 2- handouts lecture-24 to 26

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