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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 Principle of Virtual Work 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
x
F
P
xP
F
FdxdU e
As the magnitude of F is gradually increasedfrom zero to some limiting value F = P, the finalelongation of the bar becomes .
• External Work-Force.
x
e FdxU0
0
)( dxxP
U e
PxP
U e 2
1)
2(
0
2
Eigen work
4
P
L
F´
Displacement work
x
F
P Eigen w
ork
(Ue)Total = (Eigen Work)P + (Eigen Work)F´
+ (Displacement work) P
)'()')('(2
1))((
2
1)( PFPU Totale
´
L
´
F ´ + P
5
1 cm
L
20 kN
L
x (m)
F
0.01 m
20 kN
mNU e 100)1020)(01.0(2
1 3
6
Displacement work
5 kN
x (m)
F
L
0.25 cm
15 kN
0.0075
Eigen w
ork
)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.
dM
MddU e
Displacement work
M
M Eigen w
ork
'''2
1
2
1)( MMMU Totale
´
M ´ + M
0
MdU e -----(8-12)
MU e 2
1 -----(8-13)
)')('(2
1)( MMU Totale -----(8-14)
Eigen work
8
2
1oU
• Strain Energy-Axial Force.
L
N
dVUU oi
dV)2
1(
dVE
)(2
1 2
dVA
N
E2)(
2
1
A
N Adx
A
N
E2)(
2
1
dxEA
NU
L
i 0
2
2
E
9
• Strain Energy-Bending
M M
dx
d
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
E AL
))((2
1 22
2
L
dxEI
IM)
2(
2
2
L
i dxEI
MU
0
2
)2
(
2
1oU
x dx
wP
L
10
dVG
)(2
1 2
dVJ
T
G2)(
2
1
dxdAJ
T
G)()(
2
1 22
2
dxGJ
TU i 2
2
dV)2
1(
dVUU oi
J
T
G
dx
c
d
J
TT
• Strain Energy-Torsion
2
1oU
For reference:
Strength of Material by Singer, Fourth Edition, Page 67-68
11
VV
dx
dy
dVG
)(2
1 2
dVIt
VQ
G2)(
2
1
dxdAIt
Q
G
V)(
2
22A
K
dV)2
1( It
VQ
• Strain Energy-Shear
dVUU oi
dxGA
VKU i 2
2
G
2
1oU
For reference:
Strength of Material by Singer, Fourth Edition, Page 161-163
12
Principle of Work and Energy
P
L
-PL
M diagram
+ Mx= 0: 0 PxM
PxM
ie UU
L
EI
dxMP
0
2
22
1
x
L
EI
dxPxP
0
2
2
)(
2
1
L
EI
xPP
062
1 32
EI
PL
3
3
P
xV
M
13
Then apply real load P1.
Au
u
L
Principle of Virtual Work
Apply virtual load P´ first
P1
A
P´ = 1
1 • u • dL
Real displacements
Virtual loadings
1 • u • dL
Real displacements
Virtual loadings
In a similar manner,
dVUdLuP o1)2
1( 11u
u
L
dL
ie UU
14
B
Method of Virtual Work : Truss
• External Loading.
N 2
N1
N3N 4
N5
N6
N7 N8 N9
1kN
n 2
n1
n3n 4
n5
n6
n7 n8 n9
AE
nNL1
Where:1 = external virtual unit load acting on the truss joint in the stated direction of
n = internal virtual normal force in a truss member caused by the external virtual unit load = external joint displacement caused by the real load on the trussN = internal normal force in a truss member caused by the real loadsL = length of a memberA = cross-sectional area of a memberE = modulus of elasticity of a member
P1
P2
B
15
• Temperature
LTn )(1
Where: = external joint displacement caused by the temperature change
= coefficient of thermal expansion of member T = change in temperature of member
• Fabrication Errors and Camber
Ln1
Where: = external joint displacement caused by the fabrication errors
L = difference in length of the member from its intended size as caused by a fabrication error
16
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
17
A B
C4 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
0.667-0.833 -0.833
2+2.5
-2.5
1.5 kN1.5 kN
4 kN
0.5 kN0.5 kN
0
•Real Force N. The N force in each member is calculated using the method of joint.
Part (a)
18
AE
nNLkN Cv ))(1(
Cv = 0.133 mm,
0.667-0.833 -0.833
2+2.5
-2.5
8
5 5
10.67
-10.41 10.41
A B
C
n (kN)
1 kN
A B
C4 kN
N (kN)
A B
C
L (m)
=A B
C
nNL (kN2•m)
)10200)(610400(
67.10)67.1041.1041.10(
1
262
mkN
m
mkN
AEC
19
Part (b): The member AB were 5 mm too short
5 mm
)())(1( LnCv
)005.0)(667.0( Cv
Cv = -3.33 mm,
Part (c): The 4 kN force and fabrication error are both accounted.
Cv = 0.133 - 3.33 = -3.20 mm
Cv = -3.20 mm,
A B
C
n (kN)
1 kN
0.667-0.833 -0.833
20
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
AB C
D
EF
4 m
4 kN4 kN
21
4 m 4 m 4 m
AB C
D
EF
4 m
n (kN)
4 m 4 m 4 m
AB C
D
EF
4 m
4 kN4 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
.471
-0.47
1
-0.943
0.6670.333
0.33
3
1
-0.333
4-5
.66 0
-5.66
444 4
-4
0.667 kN0.333 kN
0
4 kN4 kN
0
22
AE
nNLkN Cv ))(1(
)10200)(10400(
4.72)]18.3016)67.10(2)33.5(307.15[
1
2626
mkN
m
mkN
AECv
Cv = 1.23 mm,
0.667-0
.471
-0.47
1
-0.943
0.6670.3330.
333
1
-0.333
AB C
D
EF
n (kN) 1 kN
4-5
.66 0
-5.66
444 4
-4
AB C
D
EF
4 kN4 kN N(kN)
45.6
65.6
6
5.66
444 4
4
AB C
D
EF
L(m)
AB C
D
EF
nNL(kN2•m)=
10.6715
.07 030.18
10.675.33
5.33 16
5.33
23
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
AB
CD
3 m
20 kN
10 kNwall
24
2 mAB
CD
3 m
n (kN)
SOLUTION
1 kN
0.667
0
-1.2
01
13.33 kN
23.33 kN
20 kN
23.33
0
-24.0
4
2020
0.667 kN
0.667 kN
1 kN
• Due to loading forces.
)12.10413.3160()200)(400(
1Cv
Cv= 2.44 mm,
AE
nNLkN Cv ))(1(
2 mAB
CD
3 m
20 kN
10 kN
N (kN)
2
2
3.61
33
AB
CD
L (m)
31.13
0
104.1
20
60
AB
CD
nNL(kN2•m)
25
• Due to temperature change.
LTnkN Cv )())(1(
)]61.3)(20)(2.1()2)(40)(667.0()3)(60)(1)[(1012( 6 Cv = 3.84 mm,
• Due to fabrication error.
)())(1( LnkN Cv
)003.0)(2.1()002.0)(667.0( Cv = -4.93 mm,
• Total displacement . 93.484.344.2)( TotalCv = 1.35 mm,
1 kN0.667
0
-1.2
01
AB
CD
n (kN)
+40
-20
+60
AB
D
T (oC)
C 2
2
3.61
33
AB
CD
L (m)
Fabrication error (mm)
-2
+ 3
AB
D C
26