Structure Analysis II. Deflection Energy Method When a force F undergoes a displacement dx in the same direction as the force, the work done is If 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