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Virtual Work for Frames
Applying the virtual work equations to a frame
structure is as simple as separating the frame
into a series of “beams” and summing the virtual
work for each section.
In addition, when evaluating the deformation of
a frame structure, you may have to consider
both bending and axial internal force
components.
Virtual Work for Frames
Compute the the deflectiondeflection at point C on the
frame shown below. Include only the effects
of bending in your virtual work equation (do not
consider axial work).
12 k 8 ft 8 ft
10 ft
A
B C
D
E
E = 29,000ksi
I = 3,500 in^4 A = 35 in^2
Virtual Work for Frames
The first step is to find the equation for moment in
each section of the frame due to the real loadsreal loads.
To do develop the moment expression we need the reaction a points A and E. 1212 kk 8 ft 8 ft
10 ft
A
B C
D
E
A (^) y (^) Ey
A (^) x
∑Fy=^0 = −^12 k^ +^ EY^ +Ay
E (^) y= 6 k
∑MA=^0 = −^12 k^ (8ft^ )^ +EY^ (16ft^ )
Ay = 6 k
∑Fx=^0 =Ax
Virtual Work for Frames
The next step is to find the equation for moment in each section of the frame. Consider section AB
1212 k k ∑Mt=^0 =MAB MAB^ =^0 8 ft 8 ft
10 ft
A
B C
D
E
6 k (^6) k
∑Mcut^0 MAB AB
MAB y
Virtual Work for Frames
Consider section BC
1212 k k∑ M t=^0 =^ M BC^ −^6 k x(^ ) MBC^ =^6 x 8 ft 8 ft
10 ft
A
B C
D
E
6 k (^6) k
MBC
x
∑Mcut^0 MBC^6 k x(^ ) BC
Virtual Work for Frames
Consider section DC
1212 kk 8 ft 8 ft
10 ft
A
B C
D
E
6 k (^6) k
MDC = 6 x
∑Mcut=^0 = −M^ DC^ +^6 k x(^ )
x
MDC
Virtual Work for Frames
Consider section ED
1212 kk 8 ft 8 ft
10 ft
A
B C
D
E
6 k (^6) k
MDE = 0
∑Mcut=^0 = −MDE
MDE y
Virtual Work for Frames
The next step is to find the equation for moment in
each section of the frame due to the virtual loadvirtual load.
To do develop the moment expression we need the reaction a points A and E. 1 8 ft 8 ft
10 ft
A
B C
D
E
A (^) y (^) Ey
A (^) x
∑Fy=^0 = −^1 +^ EY^ +Ay
Ey = 0.
∑MA=^0 = −1(8^ ft^ )^ +EY^ (16ft^ )
Ay = 0.
∑Fx=^0 =Ax
1
Virtual Work for Frames
In this problem, the virtual moments are the real moments divided by 12 (from superposition).
12 k 8 ft 8 ft (^) D
Section M m 8 ft 8 ft
10 ft
A
B C
D
E
AB 0 0
BC 6x x/ DC 6x x/
DE 0 0
Virtual Work for Frames
The virtual work equations are:
Δ = (^) ∫ + (^) ∫ + (^) ∫ +∫
B C C E AB AB BC BC DC DC DE DE C A B D D
M m M m M m M m dy dx dx dy EI EI EI EI
Substituting the moment expressions into the virtual work equation and integrating yields the following: 8 8
0 0
C
x x x x dx dx EI EI
Δ = (^) ∫ +∫
(^8 )
0
6 x dx EI
= (^) ∫
38
0
2 x EI
1, 024kft^3 EI
3 3 3 4
kft in ft
ksi in
= = 0.017in
Virtual Work for Frames
Compute the the slopeslope at point C on the frame shown
below. Include only the effects of bending in your virtual work equation (do not consider axial work).
12 k 8 ft 8 ft
10 ft
A
B C
D
E
E = 29,000psi
I = 3,500 in^4 A = 35 in^2
Virtual Work for Frames
Find the moments in the frame due to a virtual couple First, find the reaction in the frame to the virtual couple
8 ft 8 ft
10 ft
A
B C
D
E
A (^) y (^) Ey
A (^) x
∑Fy=^0 =^ EY^ +Ay
1 Ey = − 16
∑MA=^0 =^1 ft^ +EY^ (16ft^ )
1 Ay = 16
∑Fx=^0 =Ax
1 ft
Virtual Work for Frames
Repeat the previous example and include the effects of axial work. In order to compute the axial work, we need the axial force in the real and virtual loading systems
12 k 8 ft 8 ft
10 ft
A
B C
D
E
E = 29,000psi
I = 3,500 in^4 A = 35 in^2
Virtual Work for Frames
Find the axial force in each section of the frame. Consider section AB
1212 k k ∑^ F=^0 =^ F AB^ +^6 k FAB^ =^ −^6 k 8 ft 8 ft
10 ft
A
B C
D
E
6 k (^6) k
AB
F (^) AB
∑Fy^0 FAB^ +^6 k
y
Virtual Work for Frames
Consider section BC
1212 k k ∑^ F x=^0 =F BC FBC^ =^0 8 ft 8 ft
10 ft
A
B C
D
E
6 k (^6) k
x
∑Fx^0 FBC BC
F (^) BC
Virtual Work for Frames
Consider section DC
1212 kk 8 ft 8 ft
10 ft
A
B C
D
E
6 k (^6) k
FDC = 0
∑Fx=^0 = −FDC
x
F (^) DC
Virtual Work for Frames
Consider section ED
1212 k k ∑^ F=^0 =^ F DE^ +^6 k FDE=^ −^6 k 8 ft 8 ft
10 ft
A
B C
D
E
6 k (^6) k
F (^) DE
∑Fy^0 FDE^ +^6 k DE
y
Virtual Work for Frames
In this problem, the virtual axial forces are the real axial forces divided by 12 (from superposition).
12 k 8 ft 8 ft (^) D
Section N n
8 ft 8 ft
10 ft
A
B C
D
E
AB -6k -0.
BC 0 0
DC 0 0
DE -6k -0.
Virtual Work for Frames
The virtual work equations for axial forces are:
Δ (^) C= (^) ∑
nNL
AE
Substituting the values for the axial forces into theSubstituting the values for the axial forces into the virtual work equations yields the following:
720 k in
AE
k in
ksi in
= = 0.0007in
C
k in k in
AE AE
Virtual Work for Frames
The displacement at point C due to bending moment work and axial force work is:
ΔC = 0.017in + 0.0007in = 0.0177in
Δ f b di 12 k 8 ft 8 ft
10 ft
A
B C
D
E
E = 29,000psi
I = 3,500 in^4 A = 35 in^2
Δ from bending moment work
Δ from axial force work
Virtual Work for Frames
Compute the axial forces in the frame due to the virtual couple. Recall we already have the frame reactions due to the virtual couple
E = 29,000psi
I = 3,500 in^4 A = 35 in^2
8 ft 8 ft
10 ft
A
B C
D
E
1 ft
1 16 1 16
Virtual Work for Frames
The next step is to find the axial force in each section of the frame. Consider section AB
8 ft 8 ft
10 ft
A
B C
D
E
1 ft
1 16 1 16
nAB = − 161 n (^) AB
∑ Fy^ =^0 =^ nAB+ 161
y
Virtual Work for Frames
Consider section BC
∑ F x=^0 =n BC nBC^ =^0 8 ft 8 ft
10 ft
A
B C
D
E
1 ft
1 16
1 16
x
∑ Fx^0 nBC BC
n (^) BC
Virtual Work for Frames
Consider section DC
8 ft 8 ft
10 ft
A
B C
D
E
1 ft
1 16
1 16
x nDC=^0
∑ Fx^ =^0 = −nDC
n (^) DC
Virtual Work for Frames
Virtual Strain Energy From Shear
The shearing deformationdy caused by the real loads is
dy = λ dx, where λ is the shear strain.
V
V
dy
dx
λ
Since we are assuming the material is linear and elastic, then Hooke’s law applies
The shear strain is related to the
shear stress by λ = τ/G, where τ is
the shear stress andG is the
shearing modulus of elasticity.
Virtual Work for Frames
Virtual Strain Energy From Shear
The shear stress may be calculated as τ = K(V/A)dx,
whereK is a form factor that depends of the shape of
the beam’s cross–sectional areathe beam s cross sect onal areaA.A.
V
V
dy
dx
λ
Combining these two expressions
givesdy =KV/(GA) dy.
The internal virtual work done by
the virtual shear force v acting on
the beam before it is deformed by
the real loads is dU i =v dy
Virtual Work for Frames
Virtual Strain Energy From Shear
Integrating the expressiondU i =v dy over the entire
beam gives:
V
V
dy
dx
λ
0
L
shear
vV
U K dx
GA
Remember thatv is the shear due
to the virtual load andV is the
shear due to the real loads.
10 k
B C
0.5k/ft
Virtual Work for Frames
Compute the vertical deflectiondeflection and rotation at point C on the frame shown.
Include the effects of bending and both axial and shear force in your virtual work equations
A
20 ft
10 ft
E = 29,000ksi
G = 12,000ksi
I = 1,000 in^4 A = 25 in^2 K = 1.
virtual work equations.
End of Virtual Work - Frames
Any questions?
