Virtual Work Truss ExampleTemperature and Fabrication Errors

StevenVukazichSanJoseStateUniversity

A

BC

3 m 3 m

4 m

D E

Example Using the Principle of Virtual Work

Consider the idealized truss structure from the previous example. Truss members AD and DE increase in temperature 40℃. Member CE decreases in temperature 30℃. In addition, member BE is fabricated 0.5 cm too short.

Find the vertical displacement of point B using the Principle of Virtual Work

For all truss members use:A = 25 cm2

E = 210 GPa𝛼& = 12×10+,/℃

𝛿/0

Virtual System to Measure 𝛿/0

1. Remove all loads (if any) from the structure;

2. Apply a unit, dimensionless virtual load in-line with the real displacement, 𝛿/0, that we want to find;

3. Perform a truss analysis to find all truss member virtual axial forces, FQi

1

A B C

3 m 3 m

4 m

D E

A

BC

3 m 3 m

4 m

D E

Ax

Ay Dy

1𝑀3

�

�

= 0+

Find Support Reactions

Dy = -0.5

1

A

BC

3 m 3 m

4 m

D E

Ax

Ay Dy

Find Support Reactions

1

1𝐹7

�

�

= 0+

Ay = -0.51𝐹8

�

�

= 0+

A

BC

3 m 3 m

4 m

D E

0.5

Virtual System Support Reactions

0.5 1

A

0.5

FBD of Joint A

FQAB

3

45

𝜃

FQAD

A

3

45

𝜃

35𝐹;3<

1𝐹7

�

�

= 0+

A

0.5

FQAB

45𝐹;3<

FQAD = 0.625

FBD of Joint A

A

3

45

𝜃

35𝐹;3<

A

0.5

FQAB

45𝐹;3<

FQAB = – 0.375

FBD of Joint A

1𝐹8

�

�

= 0+

A

BC

3 m 3 m

4 m

D E

Virtual System Results on a FBD of the Entire Truss

Tension is Positive(0

.5)

(–0.

5)

(0)(– 0.375)

(0.375)

B

0.50.5 1

Virtual truss member forces, FQi

1 ⋅ 𝛿/0 =1𝐹;&𝛼&∆𝑇&𝐿& + 1 𝐹;&∆𝐿&BCDE

FGHIJ

&KL

FM

&KL

Step 2 – Use the Principle of VirtualWork to Find 𝛿/0

From Step 1–virtual analysis

A

BC

3 m 3 m

4 m

D E

For all truss members use:A = 25 cm2

E = 210 GPa𝛼& = 12×10+,/℃

𝛿/0

Use a Table to Organize VirtualWork Calculations

Member 𝜶×𝟏𝟎𝟔 /℃ ∆𝑻 ℃ ∆fabr (cm) L (m) FQ UQ (cm)

AD 12 40 0 5 0.625 0.15AB 12 0 0 3 – 0.375 0BD 12 0 0 4 – 0.5 0DE 12 40 0 3 0.375 0.054BE 12 0 –0.5 5 – 0.625 0.3125BC 12 0 0 3 0 0CE 12 –30 0 4 0.5 – 0.072Total 0.4445

𝐹;3< ∝3< ∆𝑇3<𝐿3< = 0.625 12×10+,/℃ 40℃ 5m100cmm

= 0.15cm

Sample Calculations

𝐹;/X∆𝐿BCDE3< = −0.625 −0.5 = 0.3125cm

R𝐞𝐬𝐮𝐥𝐭𝐬𝐟𝐨𝐫𝜹𝑩𝒗

Positive sign indicates that deflection is in the same direction of the virtual force

𝛿/0 = 0.4445cmupward

AB

C

3 m 3 m

4 m

D E

0.4445cm