# Document 13475902 ```16.21 Techniques of Structural Analysis and
Design
Spring 2004
Unit #7 (continued)&shy; Concepts of work and
energy
Strain energy and potential energy of a beam
March 2, 2005
Figure 1: Kinematic assumptions for a beam
&macr; � = u3 (x1 ).
Kinematic assumptions for a beam: From the ﬁgure: AA
�
��
&macr; �� = u3 + du3 . CC
&macr; � = u3 (x) +
Assume small deﬂections: B ∼ B , BB
1
u1(x1 , x3 ). Assume planar sections normal to the neutral axis remain planar
after deformation. Then:
u3 = u3 (x1 )
(1)
du3
dx1
u3 (x1 ) is the only primary unknown of the problem
u1 (x1 , x3 ) = −x3
(2)
(3)
From these kinematic assumptions we can derive a theory for beams.
Strains:
�11 =
�13
du1
d2 u3
= −x3 2
dx1
dx1
(4)
�22 = �33 = −ν�11 , plane stress
1 � du1 du3 � 1 � du3 du3 �
=
+
+
= −
=0
2 dx1 dx1
2 dx3 dx1
(5)
(6)
Constitutive:
σ11 = E�11 = −Ex3
d2 u3
dx21
(7)
Equilibrium: Apply equilibrium (in the undeformed conﬁguration) to in&shy;
tegral quantities (moment M and shear force V ). Deﬁnitions of integral
quantities as forces “equivalent” to the internal stresses:
�
V (x1 ) +
σ13 dA = 0
(8)
A(x1 )
�
M (x1 ) +
σ11 x3 dA = 0
(9)
A(x1 )
replacing σ11 :
�
M (x1 ) =
A(x1 )
�
d2 u3
d2 u3 �
−E 2 x23 dA = E 2
dx1
dx1
M (x1 ) = EI(x1 )
2
d2 u3
dx21
�
x23 dA
A(x )
� 1�� �
(10)
(11)
Also note:
σ11 = −
M x3
I
(12)
With these deﬁnitions we can apply equilibrium as shown in the ﬁgure:
�
�
dV
= −q
dx1
Fx3 = 0 : V − qdx1 − V − dV = 0 →
M B = 0 : −M + M + dM − V dx1 + q
dx21
dM
=0 →
=V
2
dx1
d2 M
d � dM �
= −q →
= −q
dx1 dx1
dx21
(13)
(14)
(15)
Replacing equation (11) in the last expression:
d2 � d2 u3 �
EI 2 + q(x1 ) = 0
dx21
dx1
(16)
Fourth order diﬀerential equation governing the deﬂections of beams. Needs
4 boundary conditions. Examples:
L
x1
L
x1
a)
b)
• case a u3 (0) = 0, u�3 (0) = 0, u��3 (L) = 0, u���
3 (L) = 0.
• case a u3 (0) = 0, u��3 (0) = 0, u3 (L) = 0, u��3 (L) = 0.
3
Strain energy of a beam Start from the general deﬁnition of strain
energy density:
� �ij
U0 =
σij d�ij
(17)
0
for a linear elastic material we concluded:
1
U0 = σij �ij
2
(18)
Classical beam theory: σ11 �= 0, all other stress components are zero.
�11
1
1
U0 = σ11 �11 = E�211
2
2
� d2 u �2
1
d2 u3
3
= −x3 2 → U0 = Ex23
dx1
2
dx21
(19)
(20)
(21)
1 2 � d2 u3 �2
dV
Ex3
dx21
V
V 2
�
�
1 L � d2 u3 �2
=
E
2 0
dx21
A(x)
� L
� d2 u �2
1
3
U=
EI(x)
dx1
2
2 0
dx1
�
U=
�
U0 dV =
also note:
�
d2 u3
1 L
U=
M (x1 ) 2 dx1
2 0
dx1
Complementary strain energy of a beam
Complementary strain energy density:
� �11
2
1 � −M x3 �2
1 σ11
∗
U0 =
�11 dσ11 =
=
2 E
2E
I
0
The complementary strain energy is then
�
�
�
1 L M2
∗
Uc =
U0 dV =
2
2
EI
0
V
A(x1 )
� L 2
1
M
Uc =
dx1
2 0 EI
Potential of the external forces:
4
(22)
(23)
(24)
(25)
(26)
(27)
(28)
(29)
�
V =−
�
ti ui dS −
S
�
V =−
fi ui dV
(30)
V
L
q(x1 )u3 (x1 )dx1 − P u3 (L) − M u�3 (L)
(31)
0
The total potential energy of the beam is:
� L�
�
1 � d2 u3 �2
�
EI
dx
+
qu
Π(u3 ) =
1
3 + P u3 (L) + M u3 (L)
2
2
dx
0
This is the expression we gave the ﬁrst day of class!.
5
(32)
```