Structure Analysis I
Chapter 9
Deflection
Energy Method
Energy Method
External Work
When a force F undergoes a displacement dx
i the
in
h same di
direction
i as the
h fforce, the
h workk
done is dU e = F dx
If the total displacement is x the work become
x
U e = ∫ F dx
0
U e = 12 P∆
The force applied gradually
Energy Method
External Work
If P is already applied to the bar and that
another
h fforce F` iis now applied
li d
The work done by P when the bar undergoes
the further deflection ∆
∆` is then
Ue' = P∆'
The work of a moment is defined by the
product of the magnitude of the moment M
and the angle dθ then
dU e = M dθ
If the
h totall angle
l off rotation
i is
i θ the
h workk
become:
θ
U e = ∫ M dθ
0
U e = 12 Mθ
The moment applied gradually
Energy Method
Strain Energy – Axial Force
σ = Eε
N
A
∆
ε=
L
σ=
NL
∆=
AE
P=N
U = 12 P∆
2
N L
Ui =
2 AE
N = internal normal force in a truss member caused
by the real load
L = length of member
A = cross-sectional
i l area off a member
b
E = modulus of elasticity of a member
Energy Method
Strain Energy – Bending
M
dx
EI
U = 12 Mθ
dθ =
M 2 dx
dU i =
2 EI
L
M 2 dx
Ui = ∫
2 EI
0
Principle of Virtual Work
∑ P∆ = ∑ u δ
Work of
E t rn l
External
Loads
Work of
Int rn l
Internal
Loads
Virtual Load
1.∆ = ∑ u.
u dL
Real displacement
Method of Virtual Work: Trusses
External Loading
1.∆ = ∑ u.dL
NL
1. ∆ = ∑ n .
AE
1 = external
t
l virtual
i t l unit
it load
l d acting
ti on the
th truss
t
joint
j i t in
i the
th stated
t t d direction
di ti off ∆
n = 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 truss
N = internal normal force in a truss member caused by the real load
L = length of member
A = cross-sectional area of a member
E = modulus of elasticity of a member
Method of Virtual Work: Trusses
Temperature
1. ∆ = ∑ n ⋅ α ⋅ ∆T ⋅ L
1 = external virtual unit load acting on the truss joint in the stated direction of
n = 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
L = length of member
Method of Virtual Work: Trusses
Fabrication Errors and Camber
1. ∆ = ∑ n ⋅ ∆L
1 = external virtual unit load acting on the truss joint in the stated direction of
n = 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.
Example 1
The cross sectional area of each member of the truss show,
show is A = 400mm2
& E = 200GPa.
a) Determine the vertical displacement of joint C if a 4-kN force is applied
to the
h truss at C
Solution
NL
1. ∆ = ∑ n .
AE
A virtual force of 1 kN is applied
pp
at C in the vertical direction
Member
AB
AC
CB
n (KN)
n
(KN)
0.667
‐0.833
‐0.833
N (KN)
N
(KN)
2
2.5
‐2.5
LL (m)
(m)
8
5
5
nNL
10.67
‐10.41
10.41
Sum 10.67
10.67 ( kN )
nNL 10.67
=
=
1. ∆ = ∑
AE
AE
400 × 10 −6 ( m 2 ) × 200 × 106 ( kN / m 2 )
∆ = 0.000133 m = 0.133 mm
Example 2
Text book Example 88-14
14
Determine vertical displacement at C
A = 0.5 in2
E = 29 (10)3 ksi
Example 3
Method of Virtual Work: Beam
1.∆ = ∑ u.
u dL
M
dθ =
dx
EI
L
M
1.∆ = ∫ m dx
EI
0
1 = external
t
l virtual
i t l unit
it load
l d acting
ti on the
th beam
b
in
i the
th stated
t t d direction
di ti off ∆
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
M = internall moment in a beam
b
caused
db
by the
h reall lload
d
L = length of beam
I = moment of inertia of cross-sectional
E = modulus of elasticity of the beam
Method 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 mθ have to be determine
L
mθ M
1( KN .m ).θ = ∫
dx
EI
0
Example 4
Determine the displacement
p
at p
point B of a steel beam
E = 200 Gpa , I = 500(106) mm4
Solution
10
M
(−1x) × (−6 x )dx
6 x dx ⎡ 6 x ⎤
1.∆ = ∫ m dx = ∫
=∫
=⎢
⎥
EI
EI
EI
EI
4
⎣
⎦0
0
0
0
L
10
2
10
3
15(103 )
15(103 )
∆=
=
= 0.15m
6
6
−12
EI
200(10 ) × 500(10 )(10 )
4
Another Solution
Real Load
Virtual Load
− 2000
× −7.5
6
6
−12
200(10 ) × 500(10 )(10 )
∆ B = 0.15m
∆B =
Example 5
Example 6
Determine the Slope
p θ and displacement
p
at p
point B of a steel
beam
E = 200 Gpa , I = 60(106) mm4
Solution
Virtual Load
Real Load
10
M
(0) × (−3 x)dx
(−1) × (−3 x)dx
3 x dx ⎡ 3 x ⎤
1.θ = ∫ mθ
dx = ∫
+∫
=∫
=⎢
⎥
EI
EI
EI
EI
EI
2
⎣
⎦5
0
0
5
5
L
5
10
3(10 2 ) − 3(52 )
θB =
= 0.0094 rad
d
6
−6
2 × 200(10 ) × 60(10 )
10
2
Another Solution
R lL
Real
Load
d
Virtual Load
− 112
112
× −1 =
EI
EI
= 200 (109112
= 0.0094 rad
)×60×10 −6
θB =
Example 7
Example 8
Example 9
Example 9b
Determine the horizontal deflection at A
Solution
Real Load
Virtual Load
− 100
− 500
− 1250
∆A =
×0+
× 2 .5 =
EI
EI
EI
= 200 (10−9 1250
= −0.031 m
)×200×10 −6
Virtual Strain Energy
gy
Caused by: Axial Load
nNL
Un =
AE
n = internal virtual axial load caused by the external virtual unit load
N = internal normal force in the member caused by the real load
L = length of member
A = cross-sectional area of the member
E = modulus of elasticity of the material
Virtual Strain Energy
gy
Caused by: Shear
d = γ ddx
dy
γ = τ /G
dy = (τ / G ) dx
τ = K (V / A )
⎛V ⎞
dy = K ⎜
⎟ dx
⎝ GA ⎠
dU S = ν dy = ν K (V / A ) dx
⎛ νV
US = ∫K ⎜
⎝ GA
0
L
⎞
⎟ dx
⎠
Virtual Strain Energy
gy
Caused by: Shear
⎛ νV
US = ∫K ⎜
⎝ GA
0
L
⎞
⎟ dx
⎠
ν= internal virtual shear in the member caused by the external virtual unit load
V = internal shear in the member caused by the real load
G= shear modulus of elasticity for the material
A = cross-sectional area of the member
K = form factor for the cross-sectional area
K=1.2 for rectangular cross sections
K=10/9 for circular cross sections
g and I-beams where A is the area of the web.
K=1.0 for wide-flange
Virtual Strain Energy
gy
Caused by: Torsion
γ = cdd θ / dx
d
γ = τ /G
Tc
τ=
J
γ
τ
T
dθ = d
dx =
ddx =
ddx
c
Gc
GJ
tT
T
dU t = td θ =
dx
GJ
tTL
Ut =
GJ
Virtual Strain Energy
gy
Caused by: Torsion
tTL
Ut =
GJ
t= internal
i
l virtual
i
l torque causedd by
b the
h externall virtual
i
l unit
i lload
d
T= internal shear in the member caused by the real load
G= shear modulus of elasticity for the material
J = polar moment of inertia of the cross-sectional
L = member length
Virtual Strain Energy
gy
Caused by: Temperature
dθ =
α ∆T m
c
m α ∆T m
=∫
dx
c
0
L
U temp
d
dx
Virtual Strain Energy
gy
Caused by: Temperature
m ⋅ α ⋅ ∆T m
=∫
dx
c
0
L
U tempp
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 beam
c = mid-depth of the beam
Example 10
Example 11