structural analysis ii structural analysis trigonometry concepts vectors equilibrium reactions...

Structural Analysis II

• Structural Analysis• Trigonometry Concepts• Vectors• Equilibrium • Reactions • Static Determinancy and Stability • Free Body Diagrams• Calculating Bridge Member Forces

Learning Objectives

• Generate a free body diagram

• Calculate internal member forces using the Method of Joints

Free Body Diagram

• Key to structural analysis

1) Draw a simple sketch of the isolated structure, dimensions, angles and x-y coordinate system

2) Draw and label all loads on the structure

3) Draw and label reactions at each support

Structural Analysis Problem

• Calculate the internal member forces on this nutcracker truss if the finger is pushing down with a force of eight newtons.

Structural Analysis SolutionDraw the Free Body Diagram

Nutcracker truss formed by tied ends

Step 1: Draw simple sketch with dimensions, angles, and x-y coordinate system

Corresponding sketch

70o

40o

a b70o

12 cm

c

x

y

Structural Analysis SolutionDraw the Free Body Diagram

Nutcracker truss

with 8N load

Step 2: Draw and label all loads on the structure

Added to free body diagram

70o 70o

40o

8N

a b

c

12 cmx

y

Structural Analysis Solution Draw the Free Body Diagram

• The truss is in equilibrium so there must reactions at the two supports. They are named Ra and Rb.

Step 3: Draw and label all reactions at each support

70o 70o

40o

8N

Ra Rb

a b

c

12 cmx

y

Structural Analysis Solution Method of Joints

• Use the Method of Joints to calculate the internal member forces of the truss

1. Isolate one joint from the truss2. Draw a free body diagram of this joint3. Separate every force and reaction into x

and y components4. Solve the equilibrium equations5. Repeat for all joints


Step 1: Isolate one jointStep 2: Draw the free body diagram

70o 70o

40o

8N

Ra = 4N Rb = 4N

a b

cx

y

12 cm

70o

Ra = 4Na x

y

b

c

Fac

Fab


First analyse Ra

• x-component = 0N

• y-component = 4N

Step 3: Separate every force and reaction into x and y components

a x

y

Ra = 4N


Next analyse Fab

• x-component = Fab

• y-component = 0N


a x

y

bFab


Lastly, analyse Fac• x-component = Fac*cos70˚ N

• y-component = Fac*sin70˚ N


70o

a x

yc

Fac

Summary of Force Components, Node ‘a’

Force Name Ra Fab Fac

Free Body Diagram

x- component 0N Fab Fac * cos70˚ N

y-component 4N 0NFac * sin70˚ N

a x

y

Ra = 4N

x

y

Fab

70o

a x

y cFac


• The bridge is not moving, so ΣFy = 0

• From the table,ΣFy = 4N + Fac * cos70˚ = 0

• Fac = ( -4N / cos70˚ ) = -4.26N

• Internal Fac has magnitude 4.26N in compression

Step 4: Solve y-axis equilibrium equations


• The bridge is not moving, so ΣFx = 0

• From the table,ΣFx = Fab + Fac * sin70˚ = 0

Fab = - ( -4.26N / sin70˚ ) = 1.45N

• Internal Fab has magnitude 1.45N in tension

Step 4: Solve x-axis equilibrium equations


Tabulated Force Solutions

Member Force Magnitude

AB 4.26N, compression

BC 1.45N, tension

AC (not yet calculated)


Step 5: Repeat for other jointsStep 1: Isolate one jointStep 2: Draw the free body diagram

70o 70o

40o

8N

Ra Rb

a b

c

12 cm

x

y

b

40o

8N

c

a

Fac = -4.26N Fbc

x

y


First analyse Rc

• y-component is -8N• x-component is 0N


8N

c x

y


Next analyse Fac

• x-component is –(Fac * sin20˚)= - (-4.26N * 0.34)= 1.46N

• y-component is –(Fac * cos20˚)= - (-4.26N * 0.94)= 4.00N


20o

c

a

Fac

x

y


Lastly analyse Fbc

• y-component = –(Fbc * cos20˚)

• x-component = (Fbc * sin20˚)


20o

c

Fbc

b

x

y

Summary of Force Components, Node ‘c’

Force Name Rc Fac Fbc

Free Body Diagram

x- component 0.00 N 1.46 N Fbc * sin20˚ N

y-component -8.00 N 4.00 N-Fbc * cos20˚ N

8Nc

20o

c

a

Fac

20o

cFbc

b


• The bridge is not moving, so ΣFy = 0

• From the table,ΣFy = -8.00N + 4.00N - Fbc * cos20˚ =

0Fbc = -4.26N

• Internal Fbc has magnitude 4.26N in compression

Step 4: Solve y-axis equilibrium equations


• The bridge is not moving, so ΣFx = 0

• From the table,ΣFx = 1.46N + Fbc * sin20˚ = 0Fbc = -4.26N

• This verifies the ΣFy = 0 equilibrium equation and also the symmetry property

Step 4: Solve x-axis equilibrium equations


Tabulated Force Solutions

Member Force Magnitude

AB 4.26N, compression

BC 1.45N, tension

AC 4.26N, compression

Acknowledgements

• This presentation is based on Learning Activity #3, Analyze and Evaluate a Truss from the book by Colonel Stephen J. Ressler, P.E., Ph.D., Designing and Building File-Folder Bridges

structural analysis ii structural analysis trigonometry concepts vectors equilibrium reactions...

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