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. . . . . .

Principle of Virtual Work Unit Load Method Applications

.

.

Lecture : Energy Methods (II) — Principle of Virtual

Work and Unit Load Method

Yubao Zhen

Dec ,

Energy Methods (II) — Virtual Work and Unit Load Method



. . . . . .


.

Review: Elastic energy associated with deformations

elastic energy densities for basic and general D stress state

U e = W i = ∫ V [σ xє x +

σ y є y +

σ z є z +

τ xy γxy +

τ yz γ yz +

τ xz γxz ] dV

elastic energy stored due to basic loads

Axial load N (x ):

U e =

L

N

EA

dx prismatic: U e =N L

EABending moment M (x ):

U e = L

M

EI dx prismatic: U e =

M L

EI Transverse shear V (x ):

U e=

L

f sV

GAdx prismatic: U

e=

f sV L

GA Torsional moment T (x ):

U e = L

T

GI pdx prismatic: U e =

T L

GI p

principle of conservation of energy: W = U e




. . . . . .


.



U e = W i = ∫ V [σ xє x +

σ y є y +

σ z є z +

τ xy γxy +

τ yz γ yz +

τ xz γxz ] dV


Axial load N (x ):

U e =

L

N

EA



U e = L

M


M L


U e=

L

f sV

GAdx prismatic: U

e=

f sV L


U e = L

T


T L

GI p





. . . . . .


.



U e = W i = ∫ V [σ xє x +

σ y є y +

σ z є z +

τ xy γxy +

τ yz γ yz +

τ xz γxz ] dV


Axial load N (x ):

U e =

L

N

EA



U e = L

M


M L


U e=

L

f sV

GAdx prismatic: U

e=

f sV L


U e = L

T


T L

GI p



l f l k d h d l



. . . . . .


.



U e = W i = ∫ V [σ xє x +

σ y є y +

σ z є z +

τ xy γxy +

τ yz γ yz +

τ xz γxz ] dV


Axial load N (x ):

U e =

L

N

EA



U e = L

M


M L


U e =

L

f sV

GAdx prismatic: U e =

f sV L


U e = L

T


T L

GI p



P i i l f Vi t l W k U it L d M th d A li ti



. . . . . .


.



U e = W i = ∫ V [σ xє x +

σ y є y +

σ z є z +

τ xy γxy +

τ yz γ yz +

τ xz γxz ] dV


Axial load N (x ):

U e =

L

N

EA



U e = L

M


M L


U e =

L

f sV


f sV L


U e = L

T


T L

GI p






. . . . . .


.



U e = W i = ∫ V [σ xє x +

σ y є y +

σ z є z +

τ xy γxy +

τ yz γ yz +

τ xz γxz ] dV


Axial load N (x ):

U e = L

N

EA



U e = L

M


M L


U e =

L

f sV


f sV L


U e = L

T


T L

GI p






. . . . . .


.

Outline

. principle of virtual work ——虚功原理

concepts of virtual quantities, the principle

. unit load method ——单位力法theory, Mohr integration (摩尔 /莫尔积分), procedure

. applications

bending, truss





. . . . . .


.

Outline



.

unit load method ——单位力法theory, Mohr integration (摩尔 /莫尔积分), procedure

. applications

bending, truss





. Principle of virtual work

(虚功原理)

. . . . . .

Principle of Virtual Work Unit Load Method Applications Virtual work



. . . . . .

p pp

.

Limitations on direct application of energy conservation

ext. work: W e =

F ∆, strain energy: U e =∑k(U ke)requires a force at the point of interest where ∆ is needed.

non-applicable if . direction of ∆ required is not along the force. there is no such an applied force. if there are more than one such applied forces

(∵ only eqn. for energy conservation)

solution: principle of virtual work (虚功原理)

to find displacement or rotation at any point in a structure





. . . . . .

.


ext. work: W e =










. . . . . .

.

Principle of virtual work in rigid body mechanics: review

Principle of virtual work (principle of virtual displacement)

At equilibrium, any infinitesimal virtual displacement in configuration

space, consistent with the constraints, requires NO work .

. A virtual displacement means an instantaneous (即时的，瞬间

的), imaginary and small change in coordinates

.

Mathematically: i F iδ r i =

r i: generalized coordinates; F i: generalized force

k

x

friction free

F

δ initial

final forces at x : spring kx ; external F A small perturbation δ :

Total virtual work (based on position atx ):

W = F δ − kx δ = (F − kx )δ . if x ≠ F /k, non-equilibrium

. if x = F

/k, equilibrium





. . . . . .

.







.



k

x

friction free

F

δ initial




. if x = F

/k, equilibrium





. . . . . .

.

Virtual displacement

.

Virtual displacement.

.

Any imaginary displacement consistent with the constraints of thestructure, i.e., displacement boundary conditions at the supports are

satisfied

before deformation

(equilibrium without load)

after deformation

(equilibrium with load)

a virtual deformation

(with load)

P 1

P 2

P 3

virtual displacement



f



. . . . . .

.

Virtual force

.

Virtual force.

.Any imaginary system of forces in equilibrium.

before deformation(equilibrium without load)

after deformation


a virtual forceP 1

P 2

P 3

induced


before deformation


P ′∆

induced reactions

real loading system virtual loading system



Vi l k



. . . . . .

.

Virtual work

.

Virtual work .

.

Work done by a real force acting through a virtual displacement or a

virtual force acting through a real displacement.

Remarks

. virtual forces and/or displacements can be arbitrary

. generalized sense for virtual quantities

force —— translation, moment —— rotation

. in practical applications

force→ real, then displacement→ virtual

displacement→ real, then force→ virtual



Vi l k



. . . . . .

.

Virtual work

.

Virtual work .

.



Remarks









Vi t l k



. . . . . .

.

Virtual work

.

Virtual work .

.



Remarks










. P.V.W. for

deformable bodies

(变形体虚功原理)

. . . . . .


Principle of virtual work for deformable bodies



. . . . . .

.

Principle of virtual work for deformable bodies

.

Principle of Virtual work for deformable bodies.

.

External virtual work is equal to internal virtual work when

equilibrated forces and stresses undergo unrelated but consistent

displacements and strains. i.e.,W virtuale = W virtuali

Note: case for deformable bodies includes the case for rigid bodies in

which the internal virtual work becomes zero.



Remarks on P V W



. . . . . .

.

Remarks on P.V.W.

equilibrated forces (平衡力系): surface/body forces, internal stresses σ

consistent displacements (协调位移): u and consistent internal strains є

virtual components —— displacements and strains

imaginary displacements and compatible strains

equivalent to:

equilibrium equations + stress BCs

for given loads, any displacement field is a candidate of equilibrium

state, and can be regarded as virtual

one and only one satisfies both equilibrium and compatibility —the equilibrium (real) state where makes system potential

Π = U − W minimum



Remarks on P V W



. . . . . .

.

Remarks on P.V.W.





equivalent to:








Remarks on P V W



. . . . . .

.

Remarks on P.V.W.





equivalent to:









. Internal virtual work

——内力虚功

. . . . . .


Internal virtual work for various loadings



. . . . . .

.


General strategy:

for one set of internal forces, use another set’s corresponding

deformations as the virtual displacements

dθ′

M T

dx

V N N

dx

V

M

dφ′

∆dx′

n

vm

t

T

γ ′

(y)

notations:

‘Real’ loads: N ,V , M ,T ;

‘Virtual’ quantities

. loads: n, v , m, t ; . displacements: ∆dx ′

, γ′

dx , d θ ′

, d ϕ′






. . . . . .

.


General strategy:

for one set of internal forces, use another set’s corresponding

deformations as the virtual displacements

dθ′

M T

dx

V N N

dx

V

M

dφ′

∆dx′

n

vm

t

T

γ ′

(y)

notations:

‘Real’ loads: N ,V , M ,T ;

‘Virtual’ quantities

. loads: n, v , m, t ; . displacements: ∆dx ′

, γ′

dx , d θ ′

, d ϕ′



Elementary virtual work



. . . . . .

.

y

γ

′

(y)

dx

V N N

dx

V

∆dx′

n

v

(2)(1)

. axial load: ∆dx ′

=ndx

AE, dW V iN = N ∆dx

′

=Nn

AEdx ,

W V iN = L

nN

AE

dx

. transverse shear force: γ′

dx by v , dW V iV = ∫ A τγ′

dxdA = f svV

GAdx ,

W V iV = L

f svV

GAdx



Elementary virtual work



. . . . . .

.

y

γ

′

(y)

dx

V N N

dx

V

∆dx′

n

v

(2)(1)

. axial load: ∆dx ′

=ndx

AE, dW V iN = N ∆dx

′

=Nn

AEdx ,

W V iN = L

nN

AE

dx

. transverse shear force: γ′

dx by v , dW V iV = ∫ A τγ′

dxdA = f svV

GAdx ,

W V iV = L

f svV

GAdx



Elementary Virtual work (cont.)



. . . . . .

.

y

dθ′

M T M

dφ′

m

t

T

(4)(3)

. bending moment: d θ ′

=m

EI dx , dW V iM = Md θ

′

=mM

EI dx ,

W V iM = L

mM

EI dx

. torque: d ϕ′

=t

GI pdx , dW V iT = Td ϕ

′

=tT

GI pdx ,

W V iT = L

tT

GI pdx



Elementary Virtual work (cont.)



. . . . . .

.

y

dθ′

M T M

dφ′

m

t

T

(4)(3)

. bending moment: d θ ′

=m

EI dx , dW V iM = Md θ

′

=mM

EI dx ,

W V iM = L

mM

EI dx

. torque: d ϕ′

=t

GI pdx , dW V iT = Td ϕ

′

=tT

GI pdx ,

W V iT = L

tT

GI pdx



On the form of internal virtual work



. . . . . .

.

inW

V

i,

L

nN

EAdx

L

f svV

GAdx

L

mM

EI dx

L

tT

GI pdx

internal loads (n,

N ), (v ,

V ), (m, M ) and (t

,T ) are symmetric.

two explanations:

equilibrated forces: from real load system

consistent displacements: from virtual load system

equilibrated forces: from virtual load system

consistent displacements: from real load system



On the form of internal virtual work



. . . . . .

.

inW

V

i,

L

nN

EAdx

L

f svV

GAdx

L

mM

EI dx

L

tT

GI pdx

internal loads (n,

N ), (v ,

V ), (m, M ) and (t

,T ) are symmetric.

two explanations:

equilibrated forces: from real load system

consistent displacements: from virtual load system

equilibrated forces: from virtual load system

consistent displacements: from real load system



Strain energy .vs. internal virtual work



. . . . . .

.

Deformation caused by Strain energy (U e) Internal virtual work (W V i )

axial load N L

N

EAdx

L

nN

EAdx

transverse shear V L

f sV

GAdx

L

f svV

GAdx

bending moment M L

M

EI dx

L

mM

EI dx

torsional moment T

L

T

GI p dx

L

tT

GI p dx

real/physical internal loads: N , V , M , T ;

virtual/imaginary internal loads: n, v , m, t



Internal virtual work for combined loadings



. . . . . .

.

By principle of superposition (for N , V , M ,T only):

W

V

i=

∫

L

nN

EA dx +

f svV

GA dx +

mM

EI dx +

tT

GI p dx For a complete list, if all internal loads are non-zero (x coincides with N )

W V i = ∫

L

nN

EAdx +

f sv y V y

GAdx +

f sv z V z

GAdx +

mz M z

EI z dx +

m y M y

EI y dx +

tT

GI pdx

Remarks:

. In general, these terms are NOT in the same order in magnitude.

For most cases, N , V terms≪ M , T terms

. For system with only axial loads (e.g., trusses), all M , V , T terms vanish

.

For small eccentric loading, N term is in the same order of M term; for largeeccentricity, N term can be ignored.

. for springs in the system, virtual internal work form: F f

k



.




. . . . . .


W

V

i=

∫

L

nN

EA dx +

f svV

GA dx +

mM

EI dx +

tT


W V i = ∫

L

nN

EAdx +

f sv y V y

GAdx +

f sv z V z

GAdx +

mz M z

EI z dx +

m y M y

EI y dx +

tT

GI pdx

Remarks:




.



k



.




. . . . . .


W V

i=

∫

L

nN

EAdx +

f svV

GAdx +

mM

EI dx +

tT

GI pdx

For a complete list, if all internal loads are non-zero (x coincides with N )

W V i = ∫

L

nN

EAdx +

f sv y V y

GAdx +

f sv z V z

GAdx +

mz M z

EI z dx +

m y M y

EI y dx +

tT

GI pdx

Remarks:




.



k




. Unit load method

(or) Mohr’s method

——单位力法/摩尔方法

. . . . . .


.

Displacement/rotation anywhere along any direction



. . . . . .

before deformation


after deformation



(with load) for the unit load

P 1

P 2

P 3


before deformation


1

∆

. to get ∆, apply a unit load (a force or a moment) at exactly the same

location and along the same direction as requested.

. Apply principle of virtual work W virtuale = W virtuali .



.

Displacement/rotation anywhere along any direction



. . . . . .

before deformation


after deformation




P 1

P 2

P 3


before deformation


1

∆

. to get ∆, apply a unit load (a force or a moment) at exactly the same

location and along the same direction as requested.

. Apply principle of virtual work W virtuale = W virtuali .



.

Unit Load Method (单位力法，Mohr’s method)



. . . . . .

before deformation


after deformation




P 1

P 2

P 3


before deformation


1

∆

. take the physical state with the real loadings t o b e a ‘virtual state’

. superpose it to the equilibrium state of the unit loaded system (having its own

deformation, of no interest though).

⋅ ∆ = ∫ nN

AEdx +∫ mM

EI dx + ∫

f svV

GAdx + ∫ tT

GI pdx



.

Unit Load Method (单位力法，Mohr’s method)



. . . . . .

before deformation


after deformation




P 1

P 2

P 3


before deformation


1

∆

. take the physical state with the real loadings t o b e a ‘virtual state’

. superpose it to the equilibrium state of the unit loaded system (having its own

deformation, of no interest though).

⋅ ∆ = ∫ nN

AEdx +∫ mM

EI dx + ∫

f svV

GAdx + ∫ tT

GI pdx



.

Mohr Integration (摩尔积分)



. . . . . .

∆ =

nN

AE

dx +

mM

EI

dx +

f svV

GA

dx +

tT

GI pdx

Mohr Integration

. extend to include terms if needed.

. N , M ,V , T :

internal loads on cross sections of the real load system with ∆ to be

solved

. n, m, v , t :

internal loads on cross sections of the unit load system

. if spring is present, addFf

kto the R.H.S.



.




. . . . . .

∆ =

nN

AE

dx +

mM

EI

dx +

f svV

GA

dx +

tT

GI pdx

Mohr Integration


. N , M ,V , T :


solved

. n, m, v , t :



kto the R.H.S.



.




. . . . . .

∆ =

nN

AE

dx +

mM

EI

dx +

f svV

GA

dx +

tT

GI pdx

Mohr Integration


. N , M ,V , T :


solved

. n, m, v , t :



kto the R.H.S.



.

Procedure to apply Mohr’s method



. . . . . .

.

set up the unit load system (单位载荷系统)an identical structure with all physical external loadings removed

. apply a generalized unit load at the location where generalized

displacement is requested

for translation, apply a unit force

for rotation, apply a unit couple moment

. use method of sections to solve the internal loads

N (), V (), M (), T () in the real load system

n(), v (), m(), t () in the unit load system

for real/virtual internal loads in a segment, use the same coordinateand the same sign conventions



.




. . . . . .

.












.




. . . . . .

.












.

Procedure (cont.)



. . . . . .

. evaluate term by term in the Mohr integration.

if with bending/torsion, ignore (in general) N , V terms;

for trusses, only N terms are available.

.

use (arbitrary) consistent energy units system in evaluation.n, v , m, t absorb (吸收) the unit (单位) for the unit load (单位载荷)

. judge the direction of generalized displacement.

if ∆ > : along the prescribed direction of the unit load;

if ∆ < : along the opposite prescribed direction of the unit load.



.

Procedure (cont.)



. . . . . .




.







.

Procedure (cont.)



. . . . . .




.







.

Application of unit load method to beams



. . . . . .

in beams, analysis on the magnitude of contributions

load is perpendicular to beam→N is negligibletransverse shear effect≪ bending effect→V is negligible

no torsion T

conclusion: (in general) in bending problems, only bending moment term

is considered in the unit load method.

for deflection ∆: a virtual unit load is applied

⋅ ∆ = L

mM

EI dx

for slope θ : a virtual unit couple moment is applied

⋅ θ = L

mθ M

EI dx



.




. . . . . .


load is perpendicular to beam→N is negligibletransverse shear effect≪ bending effect→V is negligible

no torsion T




⋅ ∆ = L

mM

EI dx


⋅ θ = L

mθ M

EI dx



.




. . . . . .


load is perpendicular to beam→

N is negligibletransverse shear effect≪ bending effect→V is negligible

no torsion T




⋅ ∆ = L

mM

EI dx


⋅ θ = L

mθ M

EI dx



.




. . . . . .




no torsion T




⋅ ∆ = L

mM

EI dx


⋅ θ = L

mθ M

EI dx



.




. . . . . .




no torsion T




⋅ ∆ = L

mM

EI dx


⋅ θ = L

mθ M

EI dx



.

On the sign convention



. . . . . .

an absolute one —— the traditional rule. N : tensile/compressive. M : concave/convex (straight beam);

tensile/compressive side (curved beam). V : clockwise/counter-clockwise.

T : right-hand rule

a relative one —— practically very useful. treat the internal loads in the real load system as positive. use it to judge the sign of corresponding internal loads in the unit

load system

(positive: same direction, negative: opposite direction)



.




. . . . . .



T : right-hand rule


load system




.




. . . . . .



T : right-hand rule


load system





. Applications

. . . . . .


.

Example — bending

Given: cantilevered beam, distributed load w

D t i ∆ d θ



. . . . . .

Determine: ∆B and θ B

virtual unit load

Solution:. x runs to the right.

unit force downward, unit couple moment

counter-clockwise

.

unit force: m= −

x ; unit moment: mθ = −

. bending moment by real load M (x ) = −w

x

. Mohr integration:

∆B = ∫ mM EI

dx = ∫ L (−

x )(−

w

x

)EI dx = wL

EI

θ B = ∫ mθ M

EI dx = ∫

L

(−)(−w

x )

EI dx =

wL

EI



.

Example — bending


Determine ∆ and θ



. . . . . .


virtual unit load



counter-clockwise

.

unit force: m= −



x

. Mohr integration:

∆B = ∫ mM EI

dx = ∫ L (−

x )(−

w

x

)EI dx = wL

EI

θ B = ∫ mθ M

EI dx = ∫

L

(−)(−w

x )

EI dx =

wL

EI



.

Example — bending


Determine: ∆ and θ



. . . . . .


virtual unit load



counter-clockwise

.

unit force: m= −



x

. Mohr integration:

∆B = ∫ mM EI

dx = ∫ L (−

x )(−

w

x

)EI dx = wL

EI

θ B = ∫ mθ M

EI dx = ∫

L

(−)(−w

x )

EI dx =

wL

EI



.

Example — bending


Determine: ∆B and θB



. . . . . .


virtual unit load



counter-clockwise

.

unit force: m= −



x

. Mohr integration:

∆B = ∫ mM EI

dx = ∫ L (−

x )(−

w

x

)EI dx = wL

EI

θ B = ∫ mθ M

EI dx = ∫

L

(−)(−w

x )

EI dx =

wL

EI



.

Example — bending





. . . . . .


virtual unit load



counter-clockwise

.

unit force: m= −



x

. Mohr integration:

∆B = ∫ mM EI

dx = ∫ L (−x

)(−

w

x

)EI dx = wL

EI

θ B = ∫ mθ M

EI dx = ∫

L

(−)(−w

x )

EI dx =

wL

EI



.

Example — bending





. . . . . .


virtual unit load



counter-clockwise

.

unit force: m= −



x

. Mohr integration:

∆B = ∫ mM EI

dx = ∫ L (−x

)(−

w

x

)EI dx = wL

EI

θ B = ∫ mθ M

EI dx = ∫

L

(−)(−w

x )

EI dx =

wL

EI



.

Example — bending

Given: cantilevered beam, concentrated load P

Determine: θB and ∆B



. . . . . .

Determine: θ B and ∆B

virtual unit moment load

Solution:. unit loads

a virtual unit couple moment at B for θ B(counter-clockwise)

a virtual unit load at B for ∆B (downward)

. calculation of mθ in CA piecewise

mθ = (for AB, ≤ x ≤L

)

mθ = (for BC , ≤ x ≤L

)

. calculation of m in CA piecewise

m = (for AB, ≤ x ≤L

)

m = −x (for BC , ≤ x ≤L

)



.

Example — bending





. . . . . .

Dete e: θB a d B







)


)



)

m = −x (for BC , ≤ x ≤L

)



.

Example — bending





. . . . . .

B B







)


)



)

m = −x (for BC , ≤ x ≤L

)



.

Example — bending





. . . . . .

B B


Solution:

. unit loads





)


)



)

m = −x (for BC , ≤ x ≤L

)



.

Example — bending





. . . . . .


Solution:

. unit loads





)


)



)

m = −x (for BC , ≤ x ≤L

)

Energy Methods (II) Virtual Work and Unit Load Method


.

Example — bending





. . . . . .


Solution:

. unit loads





)


)



)

m = −x (for BC , ≤ x ≤L

)



.

Example (cont.)

Solution:

l l i f M i CA i i



. . . . . .

real load

. calculation of M in C A piecewise

M = −Px (for AB, ≤ x ≤ L

)

M = −P (L

+ x ) (for BC , ≤ x ≤

L

)

. virtual work eqn.

θ B = ∫ mmomentθ M

EI dx = ∫

L/

()(−P (L

+ x ))

EI dx = −

PL

EI

∆B =∫ m

force

θ M EI

dx = ∫ L/

(−x

)(−P

(L

+ x

))EI dx

=

PL

EI +

PL

EI =

PL

EI

double check with the elastic curve in Appendix G: v = −Px (L − x )/EI



.

Example (cont.)

Solution:

l l ti f M i CA i i



. . . . . .

real load



)

M = −P (L

+ x ) (for BC , ≤ x ≤

L

)

. virtual work eqn.


EI dx = ∫

L/

()(−P (L

+ x ))

EI dx = −

PL

EI

∆B =∫ m

force

θ M EI

dx = ∫ L/

(−x

)(−P

(L

+ x

))EI dx

=

PL

EI +

PL

EI =

PL

EI




.

Example (cont.)

Solution:

l l ti f M i CA i i



. . . . . .

real load



)

M = −P (L

+ x ) (for BC , ≤ x ≤

L

)

. virtual work eqn.


EI dx = ∫

L/

()(−P (L

+ x ))

EI dx = −

PL

EI

∆B =∫ m

force

θ M EI

dx = ∫ L/

(−x

)(−P

(L

+ x

))EI dx

=

PL

EI +

PL

EI =

PL

EI




.

Example (cont.)

Solution:

calculation of M in CA piecewise



. . . . . .

real load



)

M = −P (L

+ x ) (for BC , ≤ x ≤

L

)

. virtual work eqn.


EI dx = ∫

L/

()(−P (L

+ x ))

EI dx = −

PL

EI

∆B =∫ m

force

θ M EI dx = ∫ L/

(−x

)(−P

(L

+ x

))EI dx

=

PL

EI +

PL

EI =

PL

EI




.

Example (cont.)

Solution:

calculation of M in CA piecewise



. . . . . .

real load



)

M = −P (L

+ x ) (for BC , ≤ x ≤

L

)

. virtual work eqn.


EI dx = ∫

L/

()(−P (L

+ x ))

EI dx = −

PL

EI

∆B =∫ m

force

θ M EI dx = ∫ L/

(−x

)(−P

(L

+ x

))EI dx

=

PL

EI +

PL

EI =

PL

EI




.

Example —— coincident locations of F and ∆

Given: () circular arc, fixed at A; () a force F at B; () EI , R

Determine: Horizontal and vertical displacements at B



. . . . . .

p

θ

F

AO

B

θ

1

AO

B

θ

AO

B 1

F

B

θ

V N

M

B

θv

n

m

1

B

θ

1

O

O

O

v

n

m

. set up unit load systems for vertical/

horizontal displacements

. N , V , M co-exist, ignore N , V

. sign convention: the relative one

. Internal bending moments:

Real load system:

M = FR sin θ Vertical unit load system:

m=

×

R sin θ =

R sin θ Horizontal unit load system:

m = × (R − R cos θ ) = R( − cos θ )



.






. . . . . .

p

θ

F

AO

B

θ

1

AO

B

θ

AO

B 1

F

B

θ

V N

M

B

θv

n

m

1

B

θ

1

O

O

O

v

n

m






Real load system:


m= ×

Rsin

θ =

Rsin

θ Horizontal unit load system:

m = × (R − R cos θ ) = R( − cos θ )



.






. . . . . .

θ

F

AO

B

θ

1

AO

B

θ

AO

B 1

F

B

θ

V N

M

B

θv

n

m

1

B

θ

1

O

O

O

v

n

m






Real load system:


m= ×

Rsin

θ =

Rsin


m = × (R − R cos θ ) = R( − cos θ )



.






. . . . . .

θ

F

AO

B

θ

1

AO

B

θ

AO

B 1

F

B

θ

V N

M

B

θv

n

m

1

B

θ

1

O

O

O

v

n

m






Real load system:


m= ×

Rsin

θ =

Rsin


m = × (R − R cos θ ) = R( − cos θ )



.

Example (cont.)

F

B

F

BM

. vertical and horizontal displacements



. . . . . .

θ

AO

θ

1

AO

B

θ

AO

B 1

θV N

B

θv

n

m

1

B

θ

1

O

O

O

v

n

m

∆B y =∫ s

mM

EI ds = ∫

π /

FRsin

θ

EI d θ

=

FR

EI

θ −

sinθ ∣π /

=

π FR

EI

∆Bx =∫ s

mM

EI ds = ∫

π /

FRsin θ −

FR

sinθ

EI Rd θ

=

FR

EI

cosθ − cos θ

∣

π /

=

π FR

EI



.

Example (cont.)

F

B

F

BM




. . . . . .

θA

O

θ

1

AO

B

θ

AO

B 1

θV N

B

θv

n

m

1

B

θ

1

O

O

O

v

n

m

∆B y =∫ s

mM

EI ds = ∫

π /

FR sin θ

EI d θ

=

FR

EI

θ −

sinθ ∣π /

=

π FR

EI

∆Bx =∫ s

mM

EI ds = ∫

π /

FRsin θ −

FR

sinθ

EI Rd θ

=

FR

EI

cosθ − cos θ

∣

π /

=

π FR

EI



.

Example (cont.)

F

B

F

BM




. . . . . .

θA

O

θ

1

AO

B

θ

AO

B 1

θV N

B

θv

n

m

1

B

θ

1

O

O

O

v

n

m

∆B y =∫ s

mM

EI ds = ∫

π /

FR sin θ

EI d θ

=

FR

EI

θ −

sinθ ∣π /

=

π FR

EI

∆Bx =∫ s

mM

EI ds = ∫

π /

FRsin θ −

FR

sinθ

EI Rd θ

=

FR

EI

cosθ − cos θ

∣

π /

=

π FR

EI



.

Example —— different locations of F and ∆

Given: () structure as shown () load at C , D a force F ; () EI , F , R;

Determine: relative displacement ∆ A−B, and horizontal ∆ Ax



. . . . . .

AC

E

G

F D B

θ

R

R RF

F



.

Example (cont.)

AC

E

θ

R R

F .

set up unit load systems

. consider only bending moment M ;



. . . . . .

G

F D B

R

AC

E

G

F D B

1

1

θ

R

R R

F

a vertical pair for ∆ A−B

and a half structure due to symmetry. internal bending moments:

real load system:

M = ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

, ( AC , ≤ x ≤ R)

F (x − R), (CE, R ≤ x ≤ R)

FR( + sin θ ), (EG, ≤ θ ≤ π /)

vertical unit load system:

m =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x , ( AC , ≤ x ≤ R)

x , (CE, R ≤ x ≤ R)

R + R sin θ , (EG, ≤ θ ≤ π /)



.

Example (cont.)

AC

E

θ

R R

F .





. . . . . .

G

F D B

R

AC

E

G

F D B

1

1

θ

R

R R

F



real load system:

M = ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

, ( AC , ≤ x ≤ R)

F (x − R), (CE, R ≤ x ≤ R)

FR( + sin θ ), (EG, ≤ θ ≤ π /)


m =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x , ( AC , ≤ x ≤ R)

x , (CE, R ≤ x ≤ R)

R + R sin θ , (EG, ≤ θ ≤ π /)



.

Example (cont.)

AC

E

θ

R R

F .





. . . . . .

G

F D B

R

AC

E

G

F D B

1

1

θ

R

R R

F



real load system:

M = ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

, ( AC , ≤ x ≤ R)

F (x − R), (CE, R ≤ x ≤ R)

FR( + sin θ ), (EG, ≤ θ ≤ π /)


m =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x , ( AC , ≤ x ≤ R)

x , (CE, R ≤ x ≤ R)

R + R sin θ , (EG, ≤ θ ≤ π /)



.Example (cont.)

AC

E

θ

R R

F .





. . . . . .

G

F D B

R

AC

E

G

F D B

1

1

θ

R

R R

F



real load system:

M = ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

, ( AC , ≤ x ≤ R)

F (x − R), (CE, R ≤ x ≤ R)

FR( + sin θ ), (EG, ≤ θ ≤ π /)


m =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

x , ( AC , ≤ x ≤ R)

x , (CE, R ≤ x ≤ R)

R + R sin θ , (EG, ≤ θ ≤ π /)



.Example (cont.)

AC

E

1

θ

R R



. . . . . .

G

F D B

1

R

. relative displacement between A and B

∆ A−B = [ + R

R

F (x − R)x

EI dx +

π /

FR( + sin θ )R( + sin θ )EI

Rd θ ]=

F

EI [− R

RRxdx +

R

Rx dx + R

π /

( + sin θ + sin θ )d θ ]=

FR

EI (

+

π )



.Example (cont.)

AC

E1

θ

R R



. . . . . .

G

F D B

R

1

.

Horizontal unit load system:special handling: removing of the horizontal translation by

symmetry

m =

⎧⎪⎪⎨⎪⎪⎩

, ( AE)

R( − cos θ ), (EG, ≤ θ ≤ π /)

∆ Ax = + π /

FR( + sin θ )R( − cos θ )EI

Rd θ =π −

FR

EI Energy Methods (II) — Virtual Work and Unit Load Method


.Applications to trusses (桁架)

Application to trusses: due to axial load only

aim: to get the nodal displacement along any direction



. . . . . .

structural load

∆ = ∑ nNL

AE

temperature change

∆ =∑n(α∆TL)Fabrication errors

∆ = ∑n(∆L)idea:

take displacement of the real load system as ’virtual’ displacement








. . . . . .

structural load

∆ = ∑ nNL

AE

temperature change











. . . . . .

structural load

∆ = ∑ nNL

AE

temperature change






.General procedure for truss problems

k id i l i h ll l l d d



. . . . . .

make an identical truss with all real loads removed;

apply the unit load along the same direction in which ∆ is requested

apply equilibrium condition to calculate the ni and N i

apply the principle of virtual work

important tricks:

. numbering (编号) of the truss element bars and

. organization (组织) —— tabulating intermediate results



. General procedure for truss problems

k id i l i h ll l l d d



. . . . . .





important tricks:






k id ti l t ith ll l l d d



. . . . . .





important tricks:









. . . . . .





important tricks:





. Example : Truss example —— final exam (A)

Given: truss with stiffness EA for all members, horizontal P at B

Solve: vertical displacement at D



. . . . . .


F

30◦

30◦

A

B

a

a

a

a

D

C

analysis:

. no other method applicable, except

energy method

.

statically determinate structure, applyunit load at D

. bars, a long expression. Use table to

organize








. . . . . .


F

30◦

30◦

A

B

a

a

a

a

D

C

analysis:


energy method

.



organize








. . . . . .


F

30◦

30◦

A

B

a

a

a

a

D

C

analysis:


energy method

.



organize



. Example : Truss example —— solution

Internal loads in bars due to F and unit force (upward) at D

F

30◦

B

a

bar i N

F i ni l i N F i nil i

F a Fa



. . . . . .

30◦

30

A

a

a

a

a

D

C

12

34

567

F √ / a √ Fa/ −√ / a

√

/ a

−√

F

√

a

a

−√

F √

a

F a

by Mohr integration:

∆D =√

Fa

EA(upward)





F

30◦

B

a

bar i N


F a Fa



. . . . . .

30◦

30

A

a

a

a

a

D

C

12

34

567 √ / √ /

−√ / a

√

/ a

−√

F

√

a

a

−√

F √

a

F a


∆D =√

Fa

EA(upward)





F

30◦

B

a

bar i N


F / a Fa//



. . . . . .

30◦

30

A

a

a

a

a

D

C

12

34

567 √ / √ /

−√ / a

√

/ a

−√

F

√

a

a

−√

F √

a

F a


∆D =√

Fa

EA(upward)



. Example : Truss example

Given: three-bar steel truss, AB has ∆T = ○C, αst = × −/○C,

Est = GPa, A = mm

Determine: horizontal displacement of joint B

Solution:



. . . . . .

. apply horizontal unit load kN to the left

(note: the unit)

. calculation of n.

n AB = −/(cos○) = −. kN.

Note n AC = nBC = .

. calculation of N :combined effects of axial load and

temperature.

from node C , N BC cos ○ + kN = →

N BC = − kN

from node B, N AB = N BC = − kN





Est = GPa, A = mm


Solution:



. . . . . .


(note: the unit)

. calculation of n.

n AB = −/(cos○) = −. kN.

Note n AC = nBC = .


temperature.


N BC = − kN






Est = GPa, A = mm


Solution:



. . . . . .


(note: the unit)

. calculation of n.

n AB = −/(cos○) = −. kN.Note n AC = nBC = .


temperature.


N BC = − kN






Est = GPa, A = mm


Solution:



. . . . . .


(note: the unit)

. calculation of n.



temperature.


N BC = − kN






Est = GPa, A = mm


Solution:



. . . . . .


(note: the unit)

. calculation of n.



temperature.


N BC = − kN




. Example : Truss example (cont.)

Solution (cont.):

.

Application of PVW.nNL



. . . . . .

⋅ ∆Bh = ∑nNL

AE+∑n(α∆TL) =

n ABN ABL AB

AE+ n AB(α∆TL AB) =

(−.

)×

(−

)×

( × −) × ( × ) +(−. × × −/○C × )→∆Bh = −. m

final displacement at B: . mm to

the right




Solution (cont.):

.

Application of PVW.nNL ( )



. . . . . .

⋅ ∆Bh = ∑nNL

AE+∑n(α∆TL) =

n ABN ABL AB


(−.

)×

(−

)×

( × −) × ( × ) +(−. × × −/○C × )→∆Bh = −. m


the right




Solution (cont.):

.

Application of PVW.

∆ ∑ nNL ∑ ( ∆T )



. . . . . .

⋅ ∆Bh = ∑n

AE+∑n(α∆TL) =

n ABN ABL AB


(−.

)×

(−

)×

( × −) × ( × ) +(−. × × −/○C × )→∆Bh = −. m


the right




Solution (cont.):

.

Application of PVW.

∆ ∑ nNL ∑ ( ∆TL)



. . . . . .

⋅ ∆Bh = ∑ AE+∑n(α∆TL) =

n ABN ABL AB


(−.

)×

(−

)×

( × −) × ( × ) +(−. × × −/○C × )→∆Bh = −. m


the right



. Summary

principle of virtual work /virtual displacement

calculation of internal virtual work for various loadings

axial load: L nN

dx shear: L f svV

dx



. . . . . .

axial load: EA

dx shear: GA

dx

bending moment: L

mM

EI dx torque:

L

tT

GI pdx

Unit load method /Mohr integration∆ = nN

AEdx + mM

EI dx + f svV

GAdx + tT

GI pdx

. physics; . procedure; . units and signs;

Most important cases:

beams (bending moments only, in general)trusses (axial loads only)



. Summary



axial load: L nN

dx shear: L f svV

dx



. . . . . .

axial load: EA

dx shear: GA

dx

bending moment: L

mM

EI dx torque:

L

tT

GI pdx

Unit load method /Mohr integration

∆ = nN

AEdx + mM

EI dx + f svV

GAdx + tT

GI pdx






. Summary



axial load: L nN

dx shear: L f svV

dx



. . . . . .

axial load: EA

dx shear: GA

dx

bending moment: L

mM

EI dx torque:

L

tT

GI pdx


∆ = nN

AEdx + mM

EI dx + f svV

GAdx + tT

GI pdx






. Summary



axial load: L nN

Adx shear:

L f svV

GAdx



. . . . . .

EA

GA

bending moment: L

mM

EI dx torque:

L

tT

GI pdx


∆ = nN

AEdx + mM

EI dx + f svV

GAdx + tT

GI pdx






. Summary



axial load: L nN

EAdx shear:

L f svV

GAdx



. . . . . .

EA

GA

bending moment: L

mM

EI dx torque:

L

tT

GI pdx


∆ = nN

AEdx + mM

EI dx + f svV

GAdx + tT

GI pdx






. Homework (problems repeated in HW )

Chap. - , ,



. . . . . .

ALL by unit load method

Due: .. (Fri.)


27 Uni Presentation

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