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

Structural Analysis Solution Method of 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

Structural Analysis Solution Method of Joints

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

Structural Analysis Solution Method of Joints

Next analyse Fab

• x-component = Fab

• y-component = 0N

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

a x

y

bFab

Structural Analysis Solution Method of Joints

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

• y-component = Fac*sin70˚ N

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

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

Structural Analysis Solution Method of Joints

• 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

Structural Analysis Solution Method of Joints

• 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

Structural Analysis Solution Method of Joints

Tabulated Force Solutions

Member Force Magnitude

AB 4.26N, compression

BC 1.45N, tension

AC (not yet calculated)

Structural Analysis Solution Method of Joints

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

Structural Analysis Solution Method of Joints

First analyse Rc

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

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

8N

c x

y

Structural Analysis Solution Method of Joints

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

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

20o

c

a

Fac

x

y

Structural Analysis Solution Method of Joints

Lastly analyse Fbc

• y-component = –(Fbc * cos20˚)

• x-component = (Fbc * sin20˚)

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

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

Structural Analysis Solution Method of Joints

• 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

Structural Analysis Solution Method of Joints

• 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

Structural Analysis Solution Method of Joints

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