Structure Analysis IStructure Analysis IChapter 9Chapter 9

DeflectionDeflection

Energy MethodEnergy Method

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

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∆

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

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

Energy MethodEnergy Method

Strain Energy BendingStrain Energy – Bending

M

θ

θ

MU

dxEIMd

21=

=

=i EIdxMdU

2

2

∫=L

i EIdxMU

0

2

20

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

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

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

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.

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

∑=∆NLn ..1Solution

A virtual force of 1 kN is applied at C in the vertical direction

∑∆AE

n ..1

pp

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

==∆×××∑

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

Example 3Example 3

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

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

Example 4Example 4Determine the displacement at point B of a steel beamp pE = 200 Gpa , I = 500(106) mm4

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 =

×==∆ −

Another Solution

Real Load

Virtual LoadVirtual Load

572000×

−=∆

mB

B

15.0

5.7)10)(10(500)10(200 1266

=∆

−××

=∆ −

Example 5Example 5

Example 6Example 6Determine the Slope θ and displacement at point B of a steel p p pbeamE = 200 Gpa , I = 60(106) mm4

Solution

Virtual Load

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

Another Solution

R l L dReal Load

Virtual LoadVirtual Load

1121112−θ

radEIEIB

0094.0

1

69 1060)10(200112 ==

=−×=

−××

θ

Example 7Example 7

Example 8Example 8

Example 9Example 9

Example 9bExample 9b

Determine the horizontal deflection at A

Solution

Real Load

Virtual Load

1250500100 −−−

mEIEIEIA

0310

12505.25000100

1250 −==

=×+×=∆

− m031.069 10200)10(200== −××

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

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

⎝ ⎠= =

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

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= =

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

Virtual Strain Energy gyCaused by: Temperature

mTd dαθ ∆ m

L

d dxcm T

θ

α

=

∆∫0

mtemp

m TU dxc

α∆= ∫

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

Example 10Example 10

Example 11Example 11