材料力学cai9

Therefore, the final strain energy for the second loading sequence is
dPi dδ i + U + dPiδ i (b) 2 Equating this expression to the earlier one (a), we have ∂U dPi dδ i U+ dPi = + U + dPiδ i ∂Pi 2 Neglecting the higher order infinitesimal amount, we then obtain
∂U ∂ M dx M ∂M = δi = ∫ 2 EI = ∫ EI ⋅ ∂Pi dx (8 - 21) ∂Pi ∂Pi 0 0 If we use the last equation and differentiate first, the calculations are much shorter and simple. For th 2 EI
l 2
[ M ( x)]2 dx dU = 2 EI ( x)
[ M ( x)] dx U =∫ 2 EI ( x) 0
5. Strain energy in the case of combined loading
If the materail is linear elastic and the deformation is small, we can get the strain energy in a member by superpersition
For a bar u=
U N u= = AL 2 EA2
2
Eδ u= 2 L2
2
2
σ
2
2E
Eε u= 2
3. Strain energy in a bar in Torsion Similarly, we can use equation (*) to calculate the strain energy in a torsional bar by substituting P with torque T and δ with angle of torsion ϕ, and get
Chapter 9
Energy Methods
9.1 Calculation of Strain Energy 1. Strain energy in the bars Consider a bar with a static load. The work done by P1 is
W = ∫ P dδ 1 1
1 1 1 dU = N ( x)dδ + M ( x)dθ + T ( x)dϕ 2 2 2
N 2 ( x)dx M 2 ( x)dx T 2 ( x)dx = + + 2 EA( x) 2 EI ( x) 2GI P ( x)
N 2 ( x)dx M 2 ( x)dx T 2 ( x)dx +∫ +∫ U =∫ 2 EA( x) 0 2 EI ( x) 0 2GI P ( x) 0
If the two loads are equal, then the last equation becomes
δ 12 = δ 21
The reciprocal - displaceme nt theorem
The displacement at point A due to the load acting at point B is equal to the displacement at point B due to the load acting at point A.
5 PL3 5PL3 δ cb = = δ bc = 48 EI 48EI
2. Reciprocal Theorems The work done by P1 and P2 together which equal to the total strain energy U of the beam is
l l l
9.2 Reciprocal Theorems 互换定理 1. Basic conditions a) the material must follow Hooke’s law; b) The displacement must be small enough that all calculations can be based upon the undeformed geometry of the structure.
For a nonprismatic bar with continuously varying axial force, its strain energy is
[ N（x）2 ] U =∫ dx 2 EA( x) 0
l
2. Strain energy density ----The strain energy per unit volume of material.
0
δ
which make the bar absorbs energy called strain energy.
U = W = ∫ P dδ 1 1
0
δ
PL Noting that δ= We have EA P2L EAδ 2 U= U= 2 EA 2L
Let us assume that the material of the bar follows Hooke’s law, then Pδ U =W = (*) 2
Tϕ U =W = 2
T L U= 2GI P
2
l
[T ( x)] dx dU = 2GI P ( x)
2
[T ( x)]2 dx U =∫ 2GI P ( x) 0
4. Strain energy in a beam Similarly, we can use equation (*) to calculate the strain energy in a beam by substituting P with bending moment M and δ with the relative angle of rotation of the two ends of the beam θ, and get
Comparing the equations (*) and (**), we get
P1δ 12 = P2δ 21 The reciprocal - work theor em
The work done by the forces in the first state of loading when they move through their corresponding displacements in the second state of loading is equal to the work done by the forces in the second state of loading when they move through their corresponding displacements in the first state of loading
P 2 L3 PM 0 L2 M 02 L = + + 6 EI 2 EI 2 EI
∂U PL M 0 L δA = = + ∂P 3EI 2 EI
3 2
∂U PL2 M 0 L θA = = + ∂M 0 2 EI EI
When calculating displacement with Castigliano’s theorem, we can differentiate the integral by differentiating under the integral sign following the rules of calculus.
When the second load is applied, an additional deflection results at B equal to δ22 ; hence, the second load dose work equal to 1 P2δ 22 2 Noting that the point A undergoes an additional deflection δ12 while the second load is being applied, the work done by P1 during this process is P1δ 12 In the way that one load is applied before the other, the total strain energy in the beam is 1 1 U = P1δ 11 + P2δ 22 + P1δ 12 (**) 2 2
∂U δi = ∂Pi (8 - 20) Castigliao’ s Theorems
Example9-1 Find the vertical displacement and angle of rotation at the free end of a cantilever with Castigliano’s theorem. Solution: M=-Px-M0 l l 2 M dx 1 U =∫ = (− Px − M 0 ) 2 dx 2 EI ( x) 2 EI ∫ 0 0
∂U dU = dPi ∂Pi Thus, the final strain energy of the beam is ∂U U + dU = U + dPi (a) ∂Pi
Now we let the load dPi is applied first, which produces its Corresponding displacement dδi. Thus, the strain energy due to the dPi is 1 dPi dδ i 2 When the loads P1 P2 ,…Pn are applied, they produce the same displacements as before (δ1, δ2, … ,δn) and do the same amount of work as before (U). However, during the application of these loads, the force dPi automatically moves through the displacement δi. Thus, dPi do additional work, equals to the additional strain energy, as following dPiδ i