Chapter-9 unit load
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Transcript of Chapter-9 unit load
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Structure Analysis IStructure Analysis IChapter 9Chapter 9
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DeflectionDeflection
Energy MethodEnergy Method
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Energy MethodEnergy MethodExternal Work
When a force F undergoes a displacement dxi h di i h f h k
External Work
in the same direction as the force, the work done is dxFdUe =
If the total displacement is x the work become
= ∫ dxFUx
e0
∆= PUe 21 The force applied gradually
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Energy MethodEnergy MethodExternal Work
If P is already applied to the bar and that h f F` i li d
External Work
another force F` is now appliedThe work done by P when the bar undergoes the further deflection ∆` is thenthe further deflection ∆ is then
' 'U P= ∆'eU P∆
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The work of a moment is defined by the product of the magnitude of the moment Mproduct of the magnitude of the moment Mand the angle thenθd
If h l l f i i h k
θdMdUe =
θIf the total angle of rotation is the work become:
θ
θθ
dMUe0
= ∫
θMUe 21= The moment applied gradually
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Energy MethodEnergy MethodStrain Energy – Axial ForceStrain Energy Axial Force
E= εσ
AN
∆
=σAENL
=∆
L∆
=ε
NPPU
NP
21 ∆=
=
N = internal normal force in a truss member caused
AELNUi 2
2
=N = internal normal force in a truss member caused by the real loadL = length of memberA i l f bA = cross-sectional area of a memberE = modulus of elasticity of a member
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Energy MethodEnergy Method
Strain Energy BendingStrain Energy – Bending
M
θ
θ
MU
dxEIMd
21=
=
=i EIdxMdU
2
2
∫=L
i EIdxMU
0
2
20
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Principle of Virtual WorkPrinciple of Virtual Work
∑∑ ∆ δP ∑∑ =∆ δuPWork of E t rn l
Work of Int rn lExternal
LoadsInternalLoads
Virtual Load
∑=∆ dLu1 ∑=∆ dLu..1
Real displacementReal displacement
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Method of Virtual Work: TrussesExternal Loading
∑∑=∆ dLu..1
∑=∆AENLn . .1AE
1 = t l i t l it l d ti th t j i t i th t t d di ti f ∆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
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Method of Virtual Work: Trusses Temperature
1. T Ln α∆ = ⋅ ⋅∆ ⋅∑∑
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 temperature change.α = coefficient of thermal expansion of member∆T = change in temperature of member∆T change in temperature of memberL = length of member
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Method of Virtual Work: Trusses Fabrication Errors and Camber
1. Ln∆ = ⋅∆∑∑
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 fabrication errors∆ L = difference in length of the member from its intended size as caused by a
fabrication error.fabrication error.
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Example 1Example 1
The cross sectional area of each member of the truss show is A = 400mm2The 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
h Cto the truss at C
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∑=∆NLn ..1Solution
A virtual force of 1 kN is applied at C in the vertical direction
∑∆AE
n ..1
pp
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Member n (KN) N (KN) L (m) nNLMember n (KN) N (KN) L (m) nNLAB 0.667 2 8 10.67AC ‐0.833 2.5 5 ‐10.41CB ‐0.833 ‐2.5 5 10.41
Sum 10.67
AEAEnNL kN
1020010400)67.1067.10.1 66
(===∆ −∑mmm
AEAE mkNm
133.0 000133.01020010400 )/(
6)(
6 22
==∆×××∑
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Example 2Text book Example 8-14
Example 2Text book Example 8 14Determine vertical displacement at CA = 0.5 in2
E = 29 (10)3 ksiE = 29 (10) ksi
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Example 3Example 3
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Method of Virtual Work: BeamMethod of Virtual Work: BeamM∑∆ dLu1 dxEIMd =θ∑=∆ dLu..1
dxMmL
∫=∆.1 dxEI
m∫∆0
.1
1 = t l i t l it l d ti th b i th t t d di ti f ∆1 = external virtual unit load acting on the beam in the stated direction of ∆m = internal virtual moment in a beam caused by the external virtual unit load∆ = external joint displacement caused by the real load on the truss
l b d b h l l dM = internal moment in a beam caused by the real loadL = length of beamI = moment of inertia of cross-sectionalE = modulus of elasticity of the beam
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Method of Virtual Work: BeamMethod of Virtual Work: Beam
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
MmL
∫
corresponding internal moment have to be determineθm
dxEI
MmmKN ∫=
0).( .1 θθ
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Example 4Example 4Determine the displacement at point B of a steel beamp pE = 200 Gpa , I = 500(106) mm4
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Solution
xdxxdxxxdxMmL 66)6()1(.1
10410 310 2
⎥⎤
⎢⎡
==−×−
==∆ ∫∫∫
m
EIEIEIdx
EIm
150)10(15)10(15
4 .1
330000
∆
⎥⎦
⎢⎣
∆ ∫∫∫
mEI
15.0)10)(10(500)10(200
)()(1266 =
×==∆ −
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Another Solution
Real Load
Virtual LoadVirtual Load
572000×
−=∆
mB
B
15.0
5.7)10)(10(500)10(200 1266
=∆
−××
=∆ −
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Example 5Example 5
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Example 6Example 6Determine the Slope θ and displacement at point B of a steel p p pbeamE = 200 Gpa , I = 60(106) mm4
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Solution
Virtual Load
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Real LoadReal Load
3 3)3()1()3()0(110210105
⎥⎤
⎢⎡
==−×−
+−×
== ∫∫∫∫L xdxxdxxdxxdxMmθ θ
d00940)5(3)10(3
2.1
2255500
−
⎥⎦
⎢⎣
==+== ∫∫∫∫ EIEIEIEIdx
EIm
θ
θ θ
rad0094.0)10(60)10(2002
)()(66 =
××= −Bθ
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Another Solution
R l L dReal Load
Virtual LoadVirtual Load
1121112−θ
radEIEIB
0094.0
1
69 1060)10(200112 ==
=−×=
−××
θ
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Example 7Example 7
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Example 8Example 8
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Example 9Example 9
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Example 9bExample 9b
Determine the horizontal deflection at A
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Solution
Real Load
Virtual Load
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1250500100 −−−
mEIEIEIA
0310
12505.25000100
1250 −==
=×+×=∆
− m031.069 10200)10(200== −××
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Virtual Strain Energy gyCaused by: Axial Load
nNLU =nUAE
=
n = internal virtual axial load caused by the external virtual unit loadN = internal normal force in the member caused by the real loadL = length of memberL length of memberA = cross-sectional area of the memberE = modulus of elasticity of the material
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Virtual Strain Energy gyCaused by: Shear
d d/
dy dxGγ
γ τ=
=
( )/
/G
dy G dxγ τ
τ=
( )/K V Aτ =
⎛ ⎞L Vν⎛ ⎞∫
Vdy K dxGA⎛ ⎞= ⎜ ⎟⎝ ⎠
0S
VU K dxGAν⎛ ⎞= ⎜ ⎟⎝ ⎠∫( )/S
GAdU dy K V A dxν ν
⎝ ⎠= =
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Virtual Strain Energy gyCaused by: Shear
L VU K dxν⎛ ⎞= ⎜ ⎟∫0
SU K dxGA
= ⎜ ⎟⎝ ⎠∫
ν= internal virtual shear in the member caused by the external virtual unit loadV = internal shear in the member caused by the real loadG= shear modulus of elasticity for the materialA = cross-sectional area of the memberK = form factor for the cross-sectional area
K=1.2 for rectangular cross sectionsK=10/9 for circular cross sectionsK=1.0 for wide-flange and I-beams where A is the area of the web. g
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Virtual Strain Energy gyCaused by: Torsion
/d dθ //
cd dxG
γ θγ τ==
Tcγ
τ =J
Td d d d
τ
γ τθ
tTL
d dx dx dxc Gc GJ
T
γ τθ = = =
ttTLUGJ
=t
tTdU td dxGJ
θ= =
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Virtual Strain Energy gyCaused by: Torsion
ttTLUGJ
=
i l i l d b h l i l i l d
GJ
t= internal virtual torque caused by the external virtual unit loadT= internal shear in the member caused by the real loadG= shear modulus of elasticity for the materialJ = polar moment of inertia of the cross-sectionalL = member length
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Virtual Strain Energy gyCaused by: Temperature
mTd dαθ ∆ m
L
d dxcm T
θ
α
=
∆∫0
mtemp
m TU dxc
α∆= ∫
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Virtual Strain Energy gyCaused by: Temperature
Lm
tempm TU dx
cα⋅ ⋅∆
= ∫0
p c∫
m= internal virtual moment in the beam caused by the external virtual unit load or unit moment
α = coefficient of thermal expansion∆Tm = change in temperature between the mean temperature
and the temperature at the top or the bottom of the beamand the temperature at the top or the bottom of the beamc = mid-depth of the beam
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Example 10Example 10
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Example 11Example 11
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