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ENERGY METHOD Conservation of Energy SOLID MECHANICS II BMCS 3333 Nadlene Razali
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ENERGY METHODConservation of Energy
SOLID MECHANICS IIBMCS 3333
Nadlene Razali
Conservation of Energy.Work and Energy Under Single Load.Deflection Under a Single Load.
Energy Method (continue...)
• A loading is applied slowly to a body, so that kinetic energy can be neglected.
• Physically, the external loads tend to deform the body as they do external work Ue as they are displaced.
• This external work is transformed into internal work or strain energy Ui, which is stored in the body.
• Thus, assuming material’s elastic limit not exceeded, conservation of energy for body is stated as,
ie UU
Conservation of Energy
• Consider a truss subjected to load P.
• When load P is applied gradually, then Ue = 1/2Px, where is vertical displacement of the truss.
• Assume that P develops an axial internal force F in a particular member, and strain energy stored , Ui = F2L/2AE.
• Summing strain energies for all members of the truss, we write axial strain energy as,
AE
LFxP
2.
2
1 2
Conservation of Energy
• Consider a beam subjected to load P. The External work done, Ue = 1/2P.
• Strain energy due to shear in beam can be neglected.
• Beam’s strain energy determined only by the moment M, thus
L
dxEI
MP
0
2
2.
2
1
Conservation of Energy
• Consider a beam loaded by a torsional couple T. A rotational displacement f is caused. External work done is
Ue = 1/2T.
• Thus equilibrium equation becomes
• Note that equilibrium equation is only applicable for a single external force or external couple moment acting on structure or member.
L
dxGJ
TT
0
2
22
1
Conservation of Energy
• Previously, we found the strain energy by integrating the energy density over the volume. For a uniform rod,
AE
LPdxA
E
AP
dVE
dVuU
L
22
2
21
0
21
2
• Strain energy may also be found from the work of the single load P1,
1
0
x
dxPU
• For an elastic deformation,
11212
121
00
11
xPxkdxkxdxPUxx
• Knowing the relationship between force and displacement,
AE
LP
AE
LPPU
AE
LPx
2
211
121
11
Work and Energy Under a Single Load
• Strain energy also may be found from the work of other types of single concentrated loads.
EI
LP
EI
LPP
yPdyPUy
63
321
31
121
1121
0
1
• Transverse load
EI
LM
EI
LMM
MdMU
2
211
121
1121
0
1
• Bending couple
JG
LT
JG
LTT
TdTU
2
211
121
1121
0
1
• Torsional couple
Maximum deflection, y1 and slope, can be obtained from Appendix D (text book)
Work and Energy Under a Single Load
Example:
The three-bar truss is subjected to a horizontal force of 20 kN.
If x-sectional area of each member is 100 mm2, determine the horizontal displacement at point B. (Given E = 200 GPa.)
Work and Energy Under a Single Load
Since only a single external force acts on the truss and required displacement is in same direction as the force, we use conservation of energy.
Also, the reactive force on truss do no work since they are not displaced.
Using method of joints (static), force in each member is determined as shown on free-body diagrams of pins at B and C.
Solution:
Work and Energy Under a Single Load
AE
LFP
22
1 2
AEAEAEhB 2
m732.1N1020
2
m2N10094.23
2
m1N10547.11N1020
2
1232323
3
AEhB
mN94640
Applying axial strain energy,
Substituting in numerical data for A and E and solving, we get
mm73.4m1073.4
N/mm10200mm1000/m1mm100
mN94640
3
2922
hB
Work and Energy Under a Single Load
• 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.
lLlL BDBC 8.06.0
From statics,PFPF BDBC 8.06.0
From the given geometry,
• Strain energy of the structure,
AE
lP
AE
lP
AE
LF
AE
LFU BDBDBCBC
2332
22
364.02
8.06.0
22
• Equating work and strain energy,
AE
Ply
yPAE
LPU
B
B
728.0
364.021
2
Deflection Under a Single Load
Example:
Members of the truss shown consist of sections of aluminum pipe with the cross-sectional areas indicated. Using E = 73 GPa, determine the vertical deflection of the point E caused by the load P.
Deflection Under a Single Load
Solution:
• Find the reactions at A and B from a free-body diagram of the entire truss.
821821 PBPAPA yx
• Apply the method of joints to determine the axial force in each member.
PF
PF
CE
DE
815
817
0
815
CD
AC
F
PF
PF
PF
BD
AD
821
45
0ABF
Deflection Under a Single Load
• Evaluate the strain energy of the truss due to the load P.
2
22
297002
1
2
1
2
PE
A
LF
EEA
LFU
i
ii
i
ii
• Equate the strain energy to the work by P and solve for the displacement.
9
33
2
21
1073
1040107.29
2
2970022
E
E
E
y
E
P
PP
Uy
UPy
mm27.16Ey
Deflection Under a Single Load
Example:
Using the method of work and energy, determine the deflection at point D caused by the load P.
Deflection Under a Single Load
Solution:
Deflection Under a Single Load
_________________________________________________________________________________________________________________
Kamarul Ariffin, FKM, UTeM BMCS 2333- 2008
Deflection Under a Single Load
