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REVISION OF St. VENANT PROBLEM
The St. Venant's problem regards the definition of the stress and displacement field in an elastic beam w/ a rectilinear axis and subjected to actions applied only to the extreme sections
- Hypothesis:
- small strains & displacements
- linear elastic homogeneous material
- long beam w/ a constant cross section
- free body in the space (no kinematic constraints)
- no body forces acting on the body, F=0
- no surface forces on : f=0
- loads act only on the extreme sections
- Reference system: it is located in the centroid of one of the extreme section and it is characterized by having:
- Sx=aqn∫ y dA = 0
- Sy=aqn∫ x dA = 0
- Ix=aqn∫ y2 dA
- Iy=aqn∫ x2 dA
- it is a PRINCIPAL REFERENCE SYSTEM
- since we have Fo on V and fo on M the only acting force is the SURFACE FORCE P acting on the EXTERNAL BASES and it must be SELF-EQUILIBRATED (since there are no constraints)
- NOTE the distribution of P is not specified BTW the St Venant prob ensures that at a certain distance from the sections the solution depends only on the ACTIONS RESULTANTS (this is why the rod must be sufficiently long)
- The INTERNAL ACTIONS which defines the stress resultants are defined as:
- Nx: ∫A σx dA
- Tx: ∫A τxz dA, Ty: ∫A τyz dA
- Hx: ∫A σx y dA, Hy: ∫A σx x dA
- Mt: Tx y + Ty x = ∫A (τyz x - τxz y) dA
4) SHEAR
* The solution of the problem is found adopting a "SEMI-INVERSE APPROACH" that means making some preliminary assumption on the stress/displacements and verify it at the end of the computations
- the rigid displacement can be eliminated by imposing the following NORMALIZATION
∫A ψG(x,y)dA = 0
as a consequence
∫A ψ dA / A = Sz = 0 average displacement
- now the SOLUTION IS UNIQUE and the warping function ψG depends only on the CROSS SECTION SHAPE
- The TWIST B can be computed by exploiting the equilibrium on the basis
Mt = ∫A(τyzz - τxzy)dA
= ∫A[GB(∂ψG/∂y + x)z - GB(∂ψG/∂x + y)y]dA
= -GB ∫A(∂ψG/∂y z - ∂ψG/∂x y + x2 + y2)dA
J = [...]
J is the TORSIONAL INERTIA of the section and it depends only on the GEOMETRY OF THE CROSS SECTION
Mt = GBJ
-> β = Mt / GJ
➔ if we refer the rotation to the centre of torsion then it is possible to DECOUPLE the problem of torsion from the problem of shear and bending
➔ the CENTRE of TORSION coincides with the CENTRE of SHEAR (if we apply the shear force in this point then there is no torsional rotation)
check of the governing equation
- equilibrium:
- ∂xz/∂ + ∂yz/∂ = 0 in A
- + = 0
- compatibility:
- ∂yz/∂ - ∂xz/∂ = 2G
K = 2G
boundary condition
- xz: nx + yz ny = 0
- 0·1 + 2G·0·0 = 0 OK
- 0·0 + 2G·x·1 = 0 NO
- 0·(-1) + 2G·x·0 = 0 OK
- 0·0 + 2G·(-1) = 0 NO
! there is a violation of the boundary condition BUT since b q3 = q1 - q2 -> CONSTANT along the common branch
* so we have:
- q1 = const in loop 1 (not in the common branch)
- q2 = const in loop 2 ("")
- q3 = q1 - q2 in the common branch
{
∫J2q'*b ds - q1z∫J2dSverizb = -2L2GB
∫J2q'*dS - q1z∫J2dSverizb = -2L2GB
* We have 3 unknowns (q1, q2, β) W/ 3 equations
REMARKS:
- the position of the cuts are arbitrary, the only condition is 1 cut for loop
- q1, q2, q'* depends on the chosen cut BUT the final result q and β must be equal # choice
SHEAR CENTRE POSITION
- * Def. SHEAR CENTRE: is that specific point where if you apply the shear force there would be torsional moment
- * by imposing this definition we can re-write the governing eq
- a) we assume that the shear force is applied in the center of shear C = (xc, yc)
- Tyxc - Txyg = ∫0J2q'*b ds = 2L1q1 + 2L2q2
- the SHEAR FLUX q is connected to the TORSIONAL MOMENT
Mtwv = ∫0a q(s)·jc(s) ds
- recalling that the warping function is:
ψc = - ∫0s jc(s) ds + ψ0
- if we derive it:
dψc/ds = - jc(s)
- the TORSIONAL MOMENT becomes:
Mtwv = ∫0a q
+ EΘo ∫0a jψ' dψc/ds ds
- EΘo ∫0a jψ dψc/ds ds
- integrating by parts:
Mtwv = EΘo {[jψ'ψc]0a - ∫0a d2jψ/dx2 ψc ds}
1) jψ'ψc|0a = 0
Sψ(0) = ∫0a ψb ds = 0 & Sψ(a) = ∫0a ψb ds = 0
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