Transcript Slide 1

Mechanics of Solids I
Energy Method
Strain Energy
o A uniform rod is subjected to a slowly increasing load
o The elementary work done by the load P as the rod
elongates by a small dx is
dU  Pdx  elementary work
which is equal to the area of width dx under the
load-deformation diagram.
o The total work done by the load for deformation x1,
x1
U   P dx  total work  strain energy
0
which results in an increase of strain energy in the rod.
o In the case of a linear elastic deformation,
x1
U   kx dx  21 kx12  21 Px
1 1
0
Strain Energy Density
o To eliminate the effects of size, evaluate the strain
energy per unit volume,
x
U  1 P dx
V 0 A L
1
u    x d x  strain energy density
0
o The total strain energy density is equal to the area under the curve to 1.
o As the material is unloaded, the stress returns to zero but there is a
permanent deformation. Only the strain energy represented by the
triangular area is recovered.
o Remainder of the energy spent in deforming the material is dissipated as
heat.
Strain-Energy Density
o The strain energy density resulting from
setting 1 = R is the modulus of toughness.
o If the stress remains within the proportional
limit,
1
E12  12
u   E x d x 

2
2E
0
o The strain energy density resulting from
setting 1 = Y is the modulus of resilience.
uY 
 Y2
2E
 modulus of resilience
Elastic Strain Energy for Normal Stresses
o In an element with a nonuniform stress distribution,
u  lim U  dU
V 0 V
dV
U   u dV  total strain energy
o For values of u < uY , i.e., below the proportional
limit,
 x2
U
2E
dV  elastic strain energy
o Under axial loading,  x  P A
dV  A dx
L
2
P
U
dx
2
AE
0
o For a rod of uniform cross section,
2
P
U L
2 AE
Elastic Strain Energy for Normal Stresses
o For a beam subjected to a bending load,
 x2
M2y2
U   dV  
dV
2E
2EI2
o Setting dV = dA dx,
L
2
 2

M2 y 2
M
U 
dAdx

y
dA
dx



2
2 
0 A 2 EI
0 2EI  A

L
x 
My
I
L
2
M
U
dx
2
EI
0
o For an end-loaded cantilever beam,
M  Px
L
2 2
2 3
P
x
P
U
dx  L
6EI
0 2EI
Strain Energy for Shearing Stresses
o For a material subjected to plane shearing stresses,
 xy
u    xy d xy
0
o For values of xy within the proportional limit,
u  21 G
2
xy
 21  xy  xy 
2
 xy
2G
o The total strain energy is found from
U   udV

2
 xy
2G
dV
Strain Energy for Shearing Stresses
• For a shaft subjected to a torsional load,
2
 xy
T2 2
U
dV  
dV
2G
2GJ2
• Setting dV = dA dx,
L
2
 2

T2 2
T
U  
dAdx


dA
dx



2
2 
0 A 2GJ
0 2GJ  A

L
T
 xy 
J
L
2
T
U
dx
2
GJ
0
• In the case of a uniform shaft,
2
T
U L
2GJ
Example 11.1
b
M12
M2 2
U
dx  
dx
2
EI
2
EI
0
0
a
 

= 1   Pbx
2EI  0 L
a
2 2 2
P
U a b
6EIL
2
b
dx  
0
 
Pav
L
2

dx 

Strain Energy for a General State of Stress
o Previously found strain energy due to uniaxial stress and plane
shearing stress. For a general state of stress,
u
1
2
 
x x
  y y   z z   xy xy   yz yz   zx zx 
o With respect to the principal axes for an elastic, isotropic body,
u  1  a2   b2   c2  2  a b   b c   c a  
2E
 uv  ud
uv  1 2v  a   b   c   due to volume change
6E
2
2
2
ud  1  a   b    b   c    c   a    due to distortion

12G 
2
o Basis for the maximum distortion energy failure criteria,
 Y2
ud  ud Y 
6G
for a tensile test specimen
Work and Energy Under a Single Load
o Strain energy may also be found from
the work of the single load P1,
x1
U   P dx
0
o For an elastic deformation,
o Previously, we found the strain
energy by integrating the energy
density over the volume.
For a uniform rod,
2

U   udV  
dV
2E
L

0
 P1 A
2E
2
P12 L
Adx 
2 AE
x1
x1
0
0
U   P dx   kx dx  21 kx12  21 Px
1 1
o Knowing the relationship between
force and displacement,
PL
1
AE
P12 L
 PL
1 
1
U  2 P1 
  2 AE
AE


x1 
Work and Energy Under a Single Load
o Strain energy may be found from the work of other types of
single concentrated loads.
• Transverse load
y1
U   P dy  21 Py
1 1
0
3
 PL
 P12 L3
1
1
 2 P1 
  6EI
3
EI


• Bending couple
1
U   Md  M11
0
1
2
M12 L
 ML
1 
1
 2 M1 
  2EI
EI


• Torsional couple
1
U   T d  21 T11
0
2
TL
T
L


 21 T1  1   1
 JG  2 JG
Deflection Under a Single Load
o If the strain energy of a structure due to a
single concentrated load is known, then
the equality between the work of the load
and energy may be used to find the
deflection.
o Strain energy of the structure,
o From the given geometry,
LBC  0.6 l
LBD  0.8 l
o From statics,
FBC  0.6P FBD  0.8P
2
2
FBC
LBC FBD
LBD
U

2 AE
2 AE
3
3
P2 l  0.6    0.8  
2
P



 0.364 l
2 AE
AE
o Equating work and strain energy,
2
P
U  0.364 L  21 Py B
AE
y B  0.728 Pl
AE
Work and Energy Under Several Loads
o Deflections of an elastic beam subjected to
two concentrated loads,
x1  x11  x12  11P1  12P2
x2  x21  x22   21P1   22P2
o Compute the strain energy in the beam
by evaluating the work done by slowly
applying P1 followed by P2,
U
1
2

2
P  212PP
1 2   22P2
2
11 1

o Reversing the application sequence yields
U
1
2

P  2 21P2P1  11P12
2
22 2

o Strain energy expressions must be equivalent.
It follows that 12 = 21 (Maxwell’s reciprocal
theorem).
Castigliano’s Theorem
• Strain energy for any elastic structure
subjected to two concentrated loads,
U
1
2

2
P  212PP
1 2   22P2
2
11 1

• Differentiating with respect to the loads,
U   P   P  x
11 1
12 2
1
P1
U   P   P  x
12 1
22 2
2
P2
• Castigliano’s theorem: For an elastic structure subjected to n loads,
the deflection xj of the point of application of Pj can be expressed as
x j  U
Pj
and
 j  U
Mj
j  U
Tj
Deflections by Castigliano’s Theorem
• Application of Castigliano’s theorem is
simplified if the differentiation with respect to
the load Pj is performed before the integration
or summation to obtain the strain energy U.
• In the case of a beam,
L
2
M
U
dx
2
EI
0
L
x j  U   M M dx
Pj 0 EI Pj
• For a truss,
Fi 2 Li
U
i 1 2 AE
i
n
n FL F

U
xj 
 i i i
Pj i 1 AE
Pj
i
Example 11.3
SOLUTION:
• For application of Castigliano’s theorem,
introduce a dummy vertical load Q at C.
Find the reactions at A and B due to the
dummy load from a free-body diagram of
the entire truss.
• Apply the method of joints to determine the
axial force in each member due to Q.
Members of the truss shown consist
• Combine with the results of Sample Problem
of sections of aluminum pipe with
11.4 to evaluate the derivative with respect
the cross-sectional areas indicated.
to Q of the strain energy of the truss due to
Using E = 73 GPa, determine the
the loads P and Q.
vertical deflection of the joint C
caused by the load P.
• Setting Q = 0, evaluate the derivative which
is equivalent to the desired displacement at
C.
Example 11.3
• Find the reactions at A and B due to a dummy
load Q at C from a free-body diagram of the
entire truss.
Ax   34 Q
Ay  Q
B  34 Q
• Apply the method of joints to determine the
axial force in each member due to Q.
FCE  FDE  0
FAC  0; FCD  Q
FAB  0; FBD   34 Q
Example 11.3
• Combine with the results of Sample Problem 11.4 to evaluate the derivative
with respect to Q of the strain energy of the truss due to the loads P and Q.
 FL  F
yC    i i  i  1  4306P  4263Q
E
 AE
i  Q
• Setting Q = 0, evaluate the derivative which is equivalent to the desired
displacement at C.
yC 

4306 40  103 N
73  109Pa

yC  2.36 mm 