ECS448 - Topic 5 (Deformation - Virtual Work for Frame & Truss)

Virtual Work

ECS 448 - DETER DETERMINA MINATE TE STRUCTU STRUCTURES RES

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Topic 5 – Def Deformat ormation ion

Example 7 Determine the horizontal displacement displacement of point C of the beam as shown. Take E = 200 GPa, I = 150 x 106 mm4 15 kN 4m C D 5m A B 8m

Step 1. Draw free free body diagram for Real load and Virtual Load. Real Load, M

15 kN

Vistual Load, m 1 kN

HA

MA

VA ECS 448 - DETER DETERMINA MINATE TE STRUCTU STRUCTURES RES

HA

MA

VA Slide Sli de 64 of 95 95


Example 7 Step 2. real load oad at all salient poi Calculate Moment equa quation for the frame with real points 0
0
15 kN

15 kN

ΣMX ↵+ = 0

ΣMX ↵+ = 0

4m

MX = -15 (X) X

M X

∴M = - 15X

MX = 15( 15(X X–4) # #

M X

0
X

15 kN 4m

ΣMX ↵+ = 0

X M X

MX = - 15 (4) (4) ∴M X = - 60 kNm kNm


#



Example 7 Step 3. virtua uall load load at all salient poi Calculate Moment equa quation for the frame with virt points 0
ΣMX ↵+ = 0

1 kN

MX = - 1 (X) (X) M X

X

∴M X = - X

#

0
1 kN 5m

ΣMX ↵+ = 0

MX = - 1(5) (5) ∴M X = - 5 kNm kNm#

M X X




Example 7 Step 4.

Calc Ca lcul ulat ate e Hori Horizo zont ntal al Disp Displa lace ceme ment nt at poin pointt C usi using virt irtual work ork eq equ uatio ation n

1.∆ =

∫

mM dx EI

L

0

4

(-15x)(0) dx + EI

(-60)(-X) dx + EI

∫

1kN.∆ =

1 4 (-15x)(0)d x +  0  EI

0

1kN.∆ C =

∫

5

1kN.∆ =

0

∫

∫

5

0

∫

8

0

(-60)(-X)d x +

(15X - 60)(-5) dx EI

∫ (15X - 60)(-5)dx  8

0

750 (200x10 6 )(150x10 −6 )

∆ C = 0.025m = 25.0mm( ←)




Example 8 Determine the vertical displacement at points C of the two-member frame shown in the figure below. Take E = 200 GPa., I = 160 x 106 mm4 40 kN B

C

m / N k 0 2

A

60 0

5m


2m


3m


Example 8 Step 1. Draw free body diagram for Real load and Virtual Load. Vistual Load, m

Real Load, M 40 kN

1 kN B

C

B C

VB

VB

m N k 0 2

A VA

60 0

A

HA

5m

2m


VA

3m


60 0

HA

5m

2m

3m


Example 8 Step 2. real load oad at all salient poi Calculate Moment equa quation for the frame with real points Memb Member er AB (0
x

MX

ΣMX ↵ + = 0

- MX + 173.21 Sin 60 0(x) – 63 Sin 30 0(x) - 15 (x)(x/2)= 0 173.21

∴ MX = 118.50X - 7.5X2

#

63 Memb Member er BC (3
Memb Member er BC (0
ΣMX ↵ + = 0

x

MX

ΣMX ↵ + = 0

x

MX - 103 (x)= (x)= 0

MX + 40(x-3) - 103 (x)= 0

∴ MX = 103. 103.0x 0x #

∴ MX = 63x + 120

103


MX


#

103


Example 8 Step 3. virtua uall load load at all salient poi Calculate Moment equa quation for the frame with virt points Memb Member er BC (0
Memb Member er AB (0
x

MX x 0

MX

0.5

0.5

ΣMX ↵ + = 0

ΣMX ↵ + = 0

MX - 0.5 (x)= (x)= 0

- MX + (0.5) X Cos 60 0(x)= 0 ∴ MX = 0.25X

∴ MX = 0.5x

#



#


Example 8 Step 4.

Calc Ca lcul ulat ate e Ve Vert rtic ical al Disp Displa laccem emen entt at poin pointt C usin using g virt virtua uall work work eq equa uati tion on

1.∆ =

L

mM

0

EI

∫

10

1kN.∆ =

∫

1kN.∆ =

1 EI

(0.25X)(11 8.50X - 7.5X 2 ) EI

0

1kN.∆ C =

dx

10

∫

0

dx +

3

(0.5X)(103 X)

0

EI

∫

(0.25X)(11 8.50X - 7.5X 2 )dx +

1 EI

∫

3

0

dx +

5

(0.5X)(63X + 120)

3

EI

∫

(0.5X)(103 X)dx +

1 EI

dx

5

∫ (0.5X)(63X + 120)dx 3

7160 (200x10 6 )(160x10 −6 )

∆ C = 0.22375m = 22.375mm( ↓ )




Example 9 Determine the slope at points C of the two-member frame shown in the figure below. The support at A is fixed. Take E = 200 GPa., I = 235 x 106 mm4 30 kN/m

B

C 3.6 m

60 0

A




Example 9 Step 1. Draw free body diagram for Real load and Virtual Load. Vistual Load, m

Real Load, M 30 kN/m

B

1 kN.m

C

B

3.6 m

60 0

HA VA

MA


C 3.6 m

60 0

HA VA


MA


Example 9 Step 2. real load oad at all salient poi Calculate Moment equa quation for the frame with real points Step 3.

Calculate Moment equation for the frame with virt virtua uall load load at all salient points Step 4.

Calc Ca lcul ulat ate e An Angu gula larr Disp Displa lace ceme ment nt at poin pointt C usin using g virt virtua uall work work eq equa uati tion on

1.θ =

L

mM

0

EI

∫

dx

3

(54X - 356.4)(1)

0

EI

1kNm.θ =

∫

1kNm.θ =

- 1059.48 (200x10 6 )(235x10 −6 )

θ = 0.0225 rad (

dx +

∫

3.6

0

(-15X 2 )(1) EI

dx

)




Review r o e ms




Problem 1 Calculate vertical displacement and slope at point D of the frame shown in the Figure P1 using the Virtual Work Method. Take E = 200 GPa., I = 60 x 10 6 mm4 . 15 kN/m C

D

3m

1m

4m

A

B

3m

3m

6m

Figure P1




Problem 2 Figure P2 shows a rigid-jointed frame that is subjected to wind loads. The wind load is transferred to the members at the girts and purlins from the roof segments (AB) and simply supported wall (BC). The frame is roller supported at A and pinned at C. Using the Virtual work work me metthod hod, deter etermi min ne horiz orizon onta tall disp isplacem aceme ent .

2m

2m

A Purlins 30 0 B

4m

1 5 k N / m

Girts

Take E = 200 GPa., I = 60 x 106 mm4 . 4m

C

Figure P2




Problem 3 Calculate vertical displacement, horizontal displacement and and slope at point C of the frame shown in the Figure P3 using the Virtual Work Method. Take E = 200 GPa., I = 60 x 106 mm4 . 40 kN A

C B

2 0 k N / m

8m

D

3m

3m

6m

Figure P3 ECS 448 - DETER DETERMINA MINATE TE STRUCTU STRUCTURES RES



Virtual Work




Example 1 The cross-sectional area of each member of the truss is A = 400 mm2 and E = 200 GPa. 1. Determ Determine ine the the vert vertica icall displ displace acemen mentt of of joint joint C if a 8-kN force is applied to the truss at C . 2. If no loads act on on the the truss, truss, what what would be the the vertic vertical al displac displacement ement of joint C if member AB were 5 mm too short?

8 kN




Example 1 Step 1.

Calculate all membe mber forces of the truss ΣFY ↑+ = 0

F AC C

-3 + 0.6F AC = 0

8 kN

∴F AC = 5 kN

8 kN A F AB

3m

ΣFX→+ = 0

-3 kN

H

AB

V A

4m

4m

ΣM A ↵+ = 0

ΣFY ↑+ = 0

8VB = 8 (3)

V A + VB = 0

∴ V B = 3 kN

#

∴ V A = - 3 kN

ΣFX→+ = 0

B

F AB

8 - H A = 0 ∴ H A = 8 kN


+ .

AC –

∴ F AB = 4 kN

FBC

VB

#

#

3 kN

= #

ΣFY ↑+ = 0

3 + 0.6FBC = 0 ∴FBC = - 5 kN

#

#



Example 1 Step 2.

Apply 1 kN unit vertical load at C Ca Calc lcu ulat late all all me memb mber er forc orces

ΣFY ↑+ = 0

0.5 + 0.6F AC = 0 1 kN

∴F AC = -0.833 kN

F AC

#

ΣFX→+ = 0

C

0 kN A F AB 0.5 kN

H A A V A

∴

AB

= .

#

B

4m

4m

ΣM A ↵+ = 0

ΣFY ↑+ = 0

8VB = 1 (4)

V A + VB = 1

∴ V B = 0.5 kN

F AB + 0.8F AC = 0

∴ V A = 0.5 kN

FBC

VB ΣFX→+ = 0

B

F AB

H A = 0 #

∴ H A = 0 kN

#

#



0.5 kN

ΣFY ↑+ = 0

0.5 + 0.6FBC = 0 0.833 3 kN kN ∴FBC = - 0.83 #


Example 1 Step 3.

Apply Virtual Work Method in Tabulation form

1.∆

=∑

AE

n (kN)

L (m)

nNL(kN2.m)

0. 0.667

8

21.344

5

-0.833

5

-20.825

-

- .

Member

N (kN)

AB

4

AC

. Total =

1kN.∆ CV =

nNL

21.344

21.344kN 2 .m (400x10 −6 m 2 )(200x10 6 kN/m 2 )

∆ CV = 2.668x10 − 4 m = 0.267mm( ↓ )




Example 1 Step 4.

If no loads act on the truss, what would be the vertical displacement of joint too short?

1.∆

C

if member AB were 5 mm

= ∑ n∆L

1kN.∆ VC = ∑ n AB ∆L AB 1kN.∆ VC = (0.667kN)( −0.005m) VC

−3

=− .

= .

Step 5.

If applied loads act on the truss is considered, and member AB were 5 mm too short, then vertical displacement at C is,

1.∆ VC = ∑ ∆ VC

nNL

+ ∑ n∆L

AE = 0.267mm − 3.335mm = 3.068mm( ↑)




Example 2 A pin-jointed plane truss ABCDE, pinned supported at A and E as shown. The truss is subjected to a vertical concentrated load of 10 kN at B and 15 kN at C. 1. Used Method of Virtual work and determine the vertical deflection at C. Member EB has been fabricated 5mm too short. Take E = 200 GPa. 2. Remove the loads on the truss and determine the vertical displacement of point B if members AB and BC experienced a temperature increase of ∆T = 1100C. Take E = 200GPa and α = 1.8 x 10-6 / 0C 3. Remove the loads on the truss and determine the vertical displacement of point B if member EB is fabricated 19mm too long.



E

1200 mm2

D

1 8 0

.

m m 2

A 1800 mm2

1800 mm2 B 10 kN

1.2 m

15 kN

C

1.2 m


Example 2 Step 1.

Used Method of Virtual work and determine the vertical deflection at C. Member EB has been fabricated 5mm too short. Take E = 200 GPa.

Real Load FBD VA

HE

Virtual Load FBD

D

VA

HE

E

D

E

1.2 m

HA

A VA

B 1.2 m

C

1.2 m

HA

1.2 m 10 kN


A VA

15 kN

B 1.2 m

C 1.2 m 1 kN



Example 2 Step 2.

Tabul bulate the member forces (Real loads and Virtual Loads) in tabul bulation form. M emb e r

L (m)

E(kN/m2)

A(m2)

N (kN)

n (kN)

nNl/EA(kN.m)

AB

1.2

200X106

1.8X10-3

-40

-2

0.000267

BC

1.2

200X106

1.8X10-3

-15

-1

0.00005

BD

1.2

200X106

1.8X10-3

-15

-1

0.00005

BE

1.697

200X106

1.2X10-3

35.355

1.414

0.000353

CD

1.697

200X106

1.2X10-3

21.213

1.414

0.000212

DE

1.2

200X106

1.2X10-3

15

1

0.000075

Total =

0.001007

1.∆ VC = ∑

nNL + ∑ n ∆L AE

1kN.∆ VC = 1.007x10 −3 kN.m − (1.414x5x1 0 −3 ) ∆ VC = −6.063x10 -3 m = 6.063mm( ↑) ECS 448 - DETER DETERMINA MINATE TE STRUCTU STRUCTURES RES



Example 2 Step 3.

Remove the loads on the truss and determine the vertical displacement of point B if members AB and BC expe experi rien ence ced d a temp temper erat atur ure e incr increa ease se of ∆T = 1100C. Take E = 200GPa and α = 1.8 x 10-6 / 0C

1.∆

Virtual Load FBD VA

HE

= ∑ nα∆TL

D

E

1kN .∆ BV

= ∑n

∆TL B + ∑ n BC α ∆TL BC

α

B

1.2 m

HA

A VA

B 1.2 m

C 1.2 m

1kN.∆BV = ( −1)(1.8x10 −6 )(110)(1.2 ) + 0 ∆ BV = −2.376x10 − 4 m = 0.2376mm( ↑)

1 kN




Example 2 Step 4.

Remove the loads on the truss and determine the vertical displacement of point B if member EB is fab abrricat icated ed 19mm 9mm too lon long.

1.∆

Virtual Load FBD VA

HE

= ∑ n∆ L

D

E

1kN .∆

= ∑ n EB ∆L EB

1.2 m

HA

A VA

B 1.2 m

C 1.2 m

1kN.∆ BV = (1.414kN)( 19mm) ∆ BV = 26.87mm( ↓ )

1 kN




Review r o e ms




Problem 1 A pin-jointed plane truss ABCDE, pinned supported at A and roller supported at E as shown. The truss is subjected to a vertical concentrated load of 40 kN and horizontal concentrated load of 10 kN at C. 40 kN

1. Used Used Meth Method od of Virtu irtual al work work and and dete determ rmin ine e the vertical deflection at D when member AD has been fabricated 5mm too short and member BD experienced a temperature increase of ∆T = 110 110 C. Take Take E = 200 GPa and α = 1.8 x 10-6/ 0C

10 kN 2

3m B

1200

m m 0 2 1 mm2 1200

3. Remov Remove e the loads on the truss truss and determi determine ne the horiz rizontal displacem cement of point B if member BD is fabricated 19mm too t oo long. ECS 448 - DETER DETERMINA MINATE TE STRUCTU STRUCTURES RES


4m

mm2

E

D

2

2. Remov Remove e the loads on the truss truss and determi determine ne the horiz rizontal displacem cement of point B if members AB and BC experienced a temperature increase of ∆T = 1100C. Take E = 200GPa and α = 1.8 x 10-6/ 0C

C

m m 0 0 2 1

1 1 A 4m

4m


Problem 2 A pin-jointed plane truss ABCD, pinned at A and supported on rollers at D as shown. The truss is subjected to a uniformly distributed load of 5 kN/m acting vertically downward on member BC and a horizontal concentrated load of 20 kN at B. AE is cons onstant for all members. 1. Determine the horizontal displacement of the truss at C using virtual work. .

na on o e oa s s own, mem er BD is cooled 300C, re-ca e-callculate the horizontal displacement at C. Given the thermal expansion coefficient, α = 1x105/0C and the axial axial rigidity, rigidity, AE = 12,000 kN.

C 20 kN B

30 0

m 4

60 0

A D 3m




Problem 3 A pin-jointed plane truss ABCDE, pinned at A and B as shown. The truss is subject subjected ed to 450 inclined loads at D and E with 10 kN and 20 kN respectively. Given A = 1800 mm2 and E = 200 GPa. 10 kN

E

1. Used Used Meth Method od of Virtu irtual al work work and and dete determ rmin ine e the Determine the horizontal displacement of the truss at E using virtual work. 2. Commen Commentt for the hori horizo zonta ntall dis lacem lacemen entt at E when the member CE is fabricated 10 mm too long, while the external loads are still in place.

45

0

2.1 m D

C

45

0

2.1 m

B

A

1.8 m



1.8 m


References Hibbeler R.C, (2012), “Structural Analysis, 8th Edition in S.I. Units”, Pearson, Singapore Hibbeler R.C, (2009), “Structural Analysis, 7th Edition in S.I. Units”, Pearson, Singapore Hibbeler R.C, (2006), “Structural Analysis, 6th Edition in S.I. Units”, Pearson, Singapore UiTM Structural Division (2003), “Basic Structural Analysis”, Cerdik Publications Sdn. Bhd. UiTM Structural Divisi vision on (2003), “Ba “Basic Structural Mec Mechanics”, Cerdik Publication ions Sdn. Bhd. Mc Cormac N., (1999), “Structural Analysis”, 2nd Edition, John Wiley and Sons.




ECS448 - Topic 5 (Deformation - Virtual Work for Frame & Truss)

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