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Deflection: Virtual
Work Method; Trusses
Theory of Structure – I
2
Contents
External Work and Strain Energy
Principle of Work and Energy
Method of Virtual Work:
Trusses
3
Ue
Eigen work
External Work and Strain Energy
Most energy methods are based on the conservation of
energy principle, which states that the work done by all the
external forces acting on a structure, Ue, is transformed into
internal work or strain energy, Ui.
Ue = Ui
L
F
D x
F
P
xP
FD
FdxdUe
As the magnitude of F is gradually increased
from zero to some limiting value F = P, the final
elongation of the bar becomes D.
• External Work-Force.
D
x
e FdxU0
D
D
0
)( dxxP
U e
DD
D
PxP
Ue2
1)
2(
0
2
Eigen work
4
P
L
F´
Displacement work
x
F
D
P
(Ue)Total = (Eigen Work)P + (Eigen Work)F´
+ (Displacement work) P
)'()')('(2
1))((
2
1)( DDD PFPU Totale
D
D´
L
D´
F ´ + P
5
1 cm
L
20 kN
L
x (m)
F
0.01 m
20 kN
mNUe 100)1020)(01.0(2
1 3
6
Displacement work
5 kN
x (m)
F
L
0.25 cm
15 kN
0.0075
)1015)(0025.0()105)(0025.0(2
1)1015)(0075.0(
2
1 333 W
mN 10050.3725.625.56
L
15 kN
0.75 cm
L
15 kN
0.75 cm 0.01
20 kN
7
• External Work-Moment.
dq M
qMddUe
Displacement work
q
M
q
M
'''2
1
2
1)( qqq MMMU Totale
q´
M ´ + M
q
q0
MdU e-----(8-12)
qMUe2
1 -----(8-13)
)')('(2
1)( qq MMU Totale
-----(8-14)
Eigen work
8
2
1oU
• Strain Energy-Axial Force.
L
N
D
dVUU oi
dV)2
1(
dVE
)(2
1 2
dVA
N
E
2)(2
1
A
N Adx
A
N
E
2)(2
1
dxEA
NU
L
i 0
2
2
E
9
• Strain Energy-Bending
M M
dx
dq
I
My
L
oi dVUU
L
dV)2
1(
L
dVE
)(2
1 2
dVI
My
EL
2)(2
1
dAdxI
yM
EL
)(2
12
22
dxAdyI
M
EAL
))((2
1 2
2
2
L
dxEI
IM)
2(
2
2
L
i dxEI
MU
0
2
)2
(
2
1oU
x dx
w
P
L
10
Principle of Work and Energy
P
L
-PL
M diagram
+ SMx = 0: 0 PxM
PxM
ie UU
D
L
EI
dxMP
0
2
22
1
x
D
L
EI
dxPxP
0
2
2
)(
2
1
L
EI
xPP
062
1 32
D
EI
PL
3
3
D
P
x V
M
11
B
Method of Virtual Work : Truss
• External Loading.
N1
N3 N5 N7 N8 N9
D
1kN
n1
n3 n5 n7 n8 n9
DAE
nNL1
Where:
1 = external virtual unit load acting on the truss joint in the stated direction of D
n = internal virtual normal force in a truss member caused by the external virtual
unit load
D = external joint displacement caused by the real load on the truss
N = internal normal force in a truss member caused by the real loads
L = length of a member
A = cross-sectional area of a member
E = modulus of elasticity of a member
P1
P2
B
12
• Temperature
DD LTn )(1
Where:
D = external joint displacement caused by the temperature change
= coefficient of thermal expansion of member
DT = change in temperature of member
• Fabrication Errors and Camber
DD Ln1
Where:
D = external joint displacement caused by the fabrication errors
DL = difference in length of the member from its intended size as
caused by a fabrication error
13
Example 8-15
The cross-sectional area of each member of the truss shown in the figure is
A = 400 mm2 and E = 200 GPa.
(a) Determine the vertical displacement of joint C if a 4-kN force is
applied to the truss at C.
(b) If no loads act on the truss, what would be the vertical displacement
of joint C if member AB were 5 mm too short?
(c) If 4 kN force and fabrication error are both accounted, what would
be the vertical displacement of joint C.
A B
C
4 m 4 m
4 kN
3 m
14
A B
C 4 kN
N(kN)
A B
C
n (kN)
SOLUTION
•Virtual Force n. Since the vertical displacement of joint C is to be
determined, only a vertical 1 kN load is placed at joint C. The n force in
each member is calculated using the method of joint.
1 kN
0.667 2
1.5 kN 1.5 kN
4 kN
0.5 kN 0.5 kN
0
•Real Force N. The N force in each member is calculated using the
method of joint.
Part (a)
15
DAE
nNLkN Cv ))(1(
DCv = 0.133 mm,
0.667 2
8
10.67
A B
C
n (kN)
1 kN
A B
C 4 kN
N (kN)
A B
C
L (m)
= A B
C
nNL (kN2•m)
)10200)(610400(
67.10)67.1041.1041.10(
1
2
62
m
kNm
mkN
AEC
D
16
Part (b): The member AB were 5 mm too short
5 mm
DD )())(1( LnCv
)005.0)(667.0( DCv
DCv = -3.33 mm,
Part (c): The 4 kN force and fabrication error are both accounted.
DCv = 0.133 - 3.33 = -3.20 mm
DCv = -3.20 mm,
A B
C
n (kN)
1 kN
0.667
17
Example 8-16
Determine the vertical displacement of joint C of the steel truss shown. The
cross-section area of each member is A = 400 mm2 and E = 200 GPa.
4 m 4 m 4 m
A B C
D
E F
4 m
4 kN 4 kN
18
4 m 4 m 4 m
A B C
D
E F
4 m
n (kN)
4 m 4 m 4 m
A B C
D
E F
4 m
4 kN 4 kN
N(kN)
SOLUTION
•Virtual Force n. Since the vertical displacement of joint C is to be
determined, only a vertical 1 kN load is placed at joint C. The n force in
each member is calculated using the method of joint.
•Real Force N. The N force in each member is calculated using the
method of joint.
1 kN
0.667 0.667 0.333 0.333
1
-0.333
4 4 4 4 4
-4
0.667 kN 0.333 kN
0
4 kN 4 kN
0
19
DAE
nNLkN Cv ))(1(
)10200)(10400(
4.72)]18.3016)67.10(2)33.5(307.15[
1
2
626
m
kNm
mkN
AECv
D
DCv = 1.23 mm,
0.667 0.667 0.333 0.333
1
-0.333
A B C
D
E F
n (kN) 1 kN
4 4 4 4 4
-4
A B C
D
E F
4 kN 4 kN N(kN)
4 4 4 4 4
4
A B C
D
E F
L(m)
A B C
D
E F
nNL(kN2•m) =
10.67 10.67 5.33
5.33 16
5.33
20
Example 8-17
Determine the vertical displacement of joint C of the steel truss shown. Due to
radiant heating from the wall, members are subjected to a temperature change:
member AD is increase +60oC, member DC is increase +40oC and member AC is
decrease -20oC.Also member DC is fabricated 2 mm too short and member AC
3 mm too long. Take = 12(10-6) , the cross-section area of each member is A =
400 mm2 and E = 200 GPa.
2 m
A B
C D
3 m
20 kN
10 kN wall
21
2 m A B
C D
3 m
n (kN)
SOLUTION
1 kN
0.667
0
0 1
13.33 kN
23.33 kN
20 kN
23.33
0
20 20
0.667 kN
0.667 kN
1 kN
• Due to loading forces.
)12.10413.3160()200)(400(
1DCv
DCv= 2.44 mm,
DAE
nNLkN Cv ))(1(
2 m A B
C D
3 m
20 kN
10 kN
N (kN)
2
2
3 3
A B
C D
L (m)
31.13
0
0 60
A B
C D
nNL(kN2•m)
22
• Due to temperature change.
DD LTnkN Cv )())(1(
)]61.3)(20)(2.1()2)(40)(667.0()3)(60)(1)[(1012( 6 D
Cv = 3.84 mm,
• Due to fabrication error.
DD )())(1( LnkN Cv
)003.0)(2.1()002.0)(667.0( DCv = -4.93 mm,
• Total displacement . 93.484.344.2)( D TotalCv = 1.35 mm,
1 kN 0.667
0
0 1
A B
C D
n (kN)
+40
+60
A B
D
DT (oC)
C 2
2
3 3
A B
C D
L (m)
Fabrication error (mm)
-2
A B
D C
23
