Structure Analysis II. Deflection Energy Method When a force F undergoes a displacement dx in the...
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Transcript of Structure Analysis II. Deflection Energy Method When a force F undergoes a displacement dx in the...
Structure Analysis II
Structure Analysis II
DeflectionEnergy Method
Energy Method
PU
dxFU
e
x
e
21
0
When a force F undergoes a displacement dx in the same direction as the force, the work done is
If the total displacement is x the work become
dxFdU e
External Work
The force applied gradually
The work of a moment is defined by the product of the magnitude of the moment M and the angle then
If the total angle of rotation is the work become:
dMdU e
d
MU
dMU
e
e
21
0
The moment applied gradually
Energy MethodStrain Energy – Axial Force
L
A
N
E
AE
NL
AE
LNU
PU
NP
i 2
2
21
N = internal normal force in a truss member caused by the real loadL = length of memberA = cross-sectional area of a memberE = modulus of elasticity of a member
Energy MethodStrain Energy – Bending
MU
dxEI
Md
21
L
i
i
EI
dxMU
EI
dxMdU
0
2
2
2
2
Principle of Virtual Work
uPWork of External Loads
Work of InternalLoads
dLu..1
Virtual Load
Real displacement
Method of Virtual Work: Trusses
AE
NLn . .1
dLu..1
1 = external virtual unit load acting on the truss joint in the stated direction ofn = 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 loadL = length of memberA = cross-sectional area of a memberE = modulus of elasticity of a member
Example 1The cross sectional area of each member of the truss show, is A = 400mm2 & E = 200GPa. a) Determine the vertical displacement of joint C if a 4-kN force is applied to the truss at C
A virtual force of 1 kN is applied at C in the vertical direction
AE
NLn . .1Soluti
on
Membern (KN)N (KN)L (m)nNLAB0.6672810.67AC-0.8332.55-10.41CB-0.833-2.5510.41
Sum 10.67
mmm
AEAE
nNL
mkNm
kN
133.0 000133.0
1020010400
)67.1067.10 .1
)/(6
)(6
(
22
Group work 1Text book Example 8-14Determine vertical displacement at CA = 0.5 in2
E = 29 (10)3 ksi
Method of Virtual Work: Beam
dxEI
Md dLu..1
dxEI
Mm
L
0
.1
1 = external virtual unit load acting on the truss joint in the stated direction ofm = internal virtual moment in a truss member caused by the external virtual unit load = external joint displacement caused by the real load on the trussM = internal moment in a beam caused by the real loadL = length of memberI = moment of inertia of cross-sectionalE = modulus of elasticity of a member
Method of Virtual Work: Beam
dxEI
MmL
mKN 0
).( .1
Similarly the rotation angle at any point on the beam can be determine, a unit couple moment is applied at the point and the corresponding internal moment have to be determine
m
Example 2Determine the displacement at point B of a steel beamE = 200 Gpa , I = 500(106) mm4
Solution
mEI
EI
x
EI
dxx
EI
dxxxdx
EI
Mm
L
15.0)10)(10(500)10(200
)10(15)10(15
4
66)6()1( .1
1266
33
10
0
410
0
310
0
2
0
mB
B
15.0
5.7)10)(10(500)10(200
20001266
Another Solution
Real Load
Virtual Load
Example 3Determine the Slope and displacement at point B of a steel beamE = 200 Gpa , I = 60(106) mm4
Solution
Virtual Load
rad 0094.0)10(60)10(2002
)5(3)10(3
2
3 3)3()1()3()0( .1
66
22
10
5
210
5
10
5
5
00
B
L
EI
x
EI
dxx
EI
dxx
EI
dxxdx
EI
Mm
Real Load
radEIEIB
0094.0
1121
112
69 1060)10(200112
Another Solution
Real Load
Virtual Load
Determine both the horizontal deflection at A
Example 4
Solution
Real Load
Virtual Load
mEIEIEIA
031.0
12505.2
5000
100
69 10200)10(2001250
Group Work 2
Determine both the Vertical deflection at C