16.21 Techniques of Structural Analysis and
Design
Spring 2003
Unit #7 - Concepts of work and energy
Strain energy and potential energy of a beam
Figure 1: Kinematic assumptions for a beam
ˉ
Kinematic assumptions for a beam: From the ?gure: AA
?
= u
3
(x
1
).
Assume small de?ections: B
?
～ B
??
,B
ˉ
B
??
= u
3
+ du
3
. C
ˉ
C
?
= u
3
(x) +
u1(x
1
,x
3
). Assume planar sections normal to the neutral axis remain planar
1
?
? ?
?? ?
after deformation. Then:
u
3
= u
3
(x
1
) (1)
du
3
u
1
(x
1
,x
3
) = ?x
3
(2)
dx
1
u
3
(x
1
) is the only primary unknown of the problem (3)
From these kinematic assumptions we can derive a theory for beams.
Strains:
du
1
d
2
u
3
?
11
=
dx
1
= ?x
3
dx
2
(4)
1
?
22
= ?
33
= ?ν?
11
, plane stress (5)
1
?
du
1
du
3
?
1
?
du
3
du
3
?
?
13
= + = ? + = 0 (6)
2 dx
3
dx
1
2 dx
1
dx
1
Constitutive:
d
2
u
3
σ
11
= E?
11
= ?Ex
3
dx
2
(7)
1
Equilibrium: Apply equilibrium (in the undeformed con?guration) to in-
tegral quantities (moment M and shear force V ). De?nitions of integral
quantities as forces “equivalent” to the internal stresses:
V (x
1
) + σ
13
d
A
= 0 (8)
?
A(x
1
)
M (x
1
) + σ
11
x
3
dA = 0 (9)
A(x
1
)
replacing σ
11
:
?
d
2
u
3
2
?
d
2
u
3
2
M (x
1
) = ?E
dx
2
x
3
dA = E
dx
2
x
3
dA (10)
A(x
1
) 1 1
?
A(x
1
)
d
2
u
3
M (x
1
) = EI(x
1
)
dx
2
(11)
1
2
Also note:
Mx
3
σ
11
= ? (12)
I
With these de?nitions we can apply equilibrium as shown in the ?gure:
?
dV
F
x
3
= 0 : V ? qdx
1
? V ? dV = 0 → = ?q (13)
dx
1
?
dx
2
1
M
B
= 0 : ?M + M + dM ? V dx
1
+ q = 0 →
dM
dx
1
= V (14)
2
d
?
dM
?
d
2
M
dx
1
dx
1
= ?q →
dx
2
= ?q (15)
1
Replacing equation (11) in the last expression:
d
2 ?
d
2
u
3
?
dx
2
EI
dx
2
+ q(x
1
) = 0 (16)
1 1
Fourth order di?erential equation governing the de?ections of beams. Needs
4 boundary conditions. Examples:
a) b)
L L
x1 x1
? case a u
3
(0) = 0, u
?
3
(L) = 0, u
???
3
(0) = 0, u
??
3
(L) = 0.
? case a u
3
(0) = 0, u
??
3
(L) = 0.
3
(0) = 0, u
3
(L) = 0, u
??
3
?
? ?
?
?
Strain energy of a beam Start from the general de?nition of strain
energy density:
?
?
ij
U = σ
ij
d?
ij
(17)
0
for a linear elastic material we concluded:
?
=
1
σ
ij
?
ij
(18)U
2
Classical beam theory: σ
11
?= 0, all other stress components are zero.
U
?
=
1
σ
11
?
11
=
1
E?
2
(19)
2 2
11
d
2
u
3
?
1
2
?
d
2
u
3
?
2
?
11
= ?x
3
dx
2
→ U = Ex
3
dx
2
(20)
1
2
1
(21)
?
1
2
?
d
2
u
3
?
2
U = UdV =
V
2
Ex
3
dx
2
dV (22)
V 1
?
L ?
d
2
u
3
?
2
?
1
2
=
2
0
E
dx
2
x
3
dAdx
1
(23)
1 A(x)
U =
1
2
?
L
0
EI(x)
?
d
2
u
3
dx
2
1
?
2
dx
1
(24)
also note: (25)
(26)U =
1
2
L
0
M (x
1
)
d
2
u
3
dx
2
1
dx
1
Complementary strain energy of a beam
Complementary strain energy density:
?
?
?
11
1 σ
2
11
1
U
c
= ?
11
dσ
11
= =
0
2 E 2E
The complementary strain energy is then
?
1
?
L
M
2
?
U
c
= U
c
dV =
V
2
0
EI
2
A(x
1
)
?
?Mx
3
?
2
(27)
I
2
x
3
dAdx
1
(28)
U
c
=
1
2
?
L
0
M
2
EI
dx
1
(29)
Potential of the external forces:
4
? ?
?
?
?
S
t
i
u
i
dS ?
V
V = ? f
i
u
i
dV (30)
?
L
0
V = ? q(x
1
)u
3
(x
1
)dx
1
? Pu
3
(L) ? Mu
3
(L) (31)
The total potential energy of the beam is:
Π(u
3
) =
?
L
0
?
1
2
?
d
2
u
3
EI
dx
2
?
2
dx
1
+ qu
3
+ Pu
3
(L) + Mu
3
(L) (32)
This is the expression we gave the ?rst day of class!.
5
