Simple Truss Analysis Method of Jointsmigueltbahia/Mecanica dos...Engineering Mechanics: Statics * Dr. Yiheng Wang Question 3: For the following truss structure (with members AB, BC,

Simple Truss AnalysisMethod of Joints

Engineering Mechanics: Statics * Dr. Yiheng Wang

Objective of this video:

To provide a step-by-step problem-solving strategy for simple truss analysis using the method of joints.


F2

F1

F3

F4

F2

F1

F3

F4

particle equilibrium


Compression: push towards the joint.

Tension: pull away from the joint.

Reminder: 2-D particle equilibrium:

enables solving for 2 unknowns at a time.

0

0

y

x

F

F


200 lb

A BC

D

E

F

Example 1: Determine the force in each member of the truss and indicate if the member is in tension or compression.

45º45º

1 ft

1 ft


200 lb

A BC

D

E

F

Step 1: Remove zero-force member by inspection.

45º45º

1 ft

1 ft

0

0

BE

DE

F

F


200 lb

A BC

D

E

F

Step 2: Decide if support reactions need to be determined.

45º45º

1 ft

1 ft

hint: from each joint you can solve for maximum 2 reactions.

PAUSEPLEASE


200 lb

A BC

D

E

F

Step 3: Solve each joint from easy to hard. Apply particle equilibrium.

45º45º

1 ft

1 ft45º


200 lb F

Joint F:

67.5ºFEF

FDF

x

y

05.67cos

0lb2005.67sin

DFEFy

EFx

FFF

FF

Tlb216

Clb8.82

EF

DF

F

F

200 lb F

82.8 lb

216 lb

Tip: always draw unknown reactions as tensions.


200 lb

A BC

D

E

F

45º45º

1 ft

1 ft


216 lbE

Joint E:

FCEFAE

67.5º

05.67coslb216

05.67sinlb216

AEy

CEx

FF

FF

Clb8.82

Tlb200

AE

CE

F

F

216 lb

E

200 lb82.8 lb

x

y


200 lb

A BC

D

E

F

45º45º

1 ft

1 ft

Cy

CxFBC

PAUSEPLEASE

Ay


x

y

82.8 lb

A

Joint A:

FAB

Ay

045sinlb8.82lb8.82

045coslb8.82

yy

ABx

AF

FF

lb141

Tlb3.58

y

AB

A

F

82.8 lb45º

82.8 lb

A82.8 lb

58.3 lb

141 lb


200 lb

A BC

D

E

F

45º45º

1 ft

1 ft


200 lb

A BC

D

E

F

45º45º

1 ft

1 ft

Tlb3.58

Clb8.82

Tlb200

Clb8.82

Tlb216

0

0

BCAB

AE

CE

ADDF

EF

BE

DE

FF

F

F

FF

F

F

F


Question 1: For the following truss structure (with members AB, BC, AD, DE, BE, CF

and FG all of 60 cm length each), determine the force in member CE.

(a) (b)

(c) (d)(T) N 257 (C) N 633

(T) N 79.9N 0

200 N

AB

C

DE

F

G

120 N160 N

45º


(a) (b)

(c) (d)(T) N 257 (C) N 633

(T) N 79.9N 0

200 N

AB

C

DE

F

G

120 N160 N

45º



and FG all of 60 cm length each), determine the force in member DE.

(a) (b)

(c) (d)(T) N 257 (C) N 633

(T) N 79.9N 0

200 N

AB

C

DE

F

G

120 N160 N

45º



and FG all of 60 cm length each), determine the force in member AE.

CN283EGF

TN200FGF

TN9.87CFF

CN113EFF

CN120ADF

N0 EDBE FF

TN9.79CEF

CN363AEF

TN257 BCAB FF

N377yA

N313xC

N144yC

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Simple Truss Analysis Method of Jointsmigueltbahia/Mecanica dos...Engineering Mechanics: Statics * Dr. Yiheng Wang Question 3: For the following truss structure (with members AB, BC,