University:
Massachusetts Institute of Technology
Course:
16.01 | Unified Engineering I, II, III, & IV
Academic year:

2021
Views:

23

Pages:

5
Author:

sonalhava9qv

Lecture M4: Simple Beam Theory Reading: Crandall, Dahl and Lardner 7.2-7.6 We have looked at the statics of a beam, seen that loads are transmitted by internal forces: axial forces, shear forces and bending moments. Now look at how these forces imply stresses, strains, and deflections. Recall model assumptions: slenderness General, symmetric, cross section Geometry L >> h,b Loads Leads to assumptions on stresses † Load in x - z plane Æ † s yy = s xy = s yz = 0 Also L >> h,b implies s xx ,s xz >> s zz Consider moment equilibrium of a cross section of a beam loaded by some distributed stress szz, which is reacted by axial stresses on the cross section, sxx Â M X = 0 = s zz L - s xx h fi s zz h ª ª0 s xx L i.e. geometry of beam implies sxx >> szz † Assumptions on Deformations: The key to simple beam theory is the Bernoulli - Euler hypotheses (1750) "Plane sections remain plane and perpendicular to the mid-plane after deformation." It turns out that this is not really an assumption at all but a geometric necessity, at least for the case of pure bending. To see what the implications are of this, consider a beam element which undergoes transverse (bending) deformation. w= deflection of midplane/midline (function of x only) Obtain deflection in x-direction, displacement u of point k to K, defined by rotation f . Note, key assumption, "plane sections remain plane" † axial displacement, u, of an arbitrary point on the cross-section arises from rotation of cross, sections u= -z tanf Note, negative sign here due to use of consistent definitions of positive directions for w, x and dw/dx. If deformations/angles are small tan q ª q dw dw fi u = -z dx dx † field Hence obtain deformation fª † (1) dw dx v( x, y,z ) = 0 nothing happening in y direction u( x, y,z ) = -z w( x, y,z ) = w( x ) Deflection out of original plane i.e., cross sections remain rigid (2) Hence we can obtain distributions of strain. (compatible with deformation). In absence † of deformations in transverse directions partial derivatives can be replaced by regular derivatives, i.e. ∂_ d_ ≡ . Hence. ∂_ d_ ∂u d 2w = -z ∂x dx 2 † ∂v ∂w e yy = = 0, e zz = =0 ∂y ∂z Ê∂u ∂v ˆ Ê∂v ∂w ˆ g xy = 2e12 = Á + ˜ = 0, g yz = Á + ˜ = 0 Ë ∂y ∂x ¯ Ë∂z ∂y ¯ Ê∂u ∂w ˆ Ê dw dw ˆ g xz = Á + ˜ = Á+ ˜=0 Ë ∂z ∂x ¯ Ë dx dx ¯ e xx = (3) If no shear - consistent with B-E assumption of plane sections remaining plane. † for constant bending moment: dM =S=0 dx (will revisit for S ≠ 0) Next use stress - strain (assume orthotropic - for generality) e xx = s xx Ex (4) s xx Ex s e zz = -n xz xx Ex t xy g xy = G xy e yy = -n xy g yz = t yz Note inconsistency - G yz g xz = 0, t xz ≠ 0 (shear t g xz = xz G xz forces are non zero) The inconsistency on † the shear stress/strain arises from the plane/sections remain plane assumption. Does not strictly apply when there is varying bending moment (and hence † non-zero shear force). However, displacements due to g xz are very small compared to those due to e xx , and therefore negligible. Finally apply equilibrium: † † ∂s mn + fm = 0 ∂x m ∂s xx ∂s yz ∂s zx ∂s xx ∂szx + + = 0 fi + =0 ∂x ∂y ∂z ∂x ∂z ∂s xy ∂x + ∂s yy ∂y + ∂s zy ∂z = 0 fi 0 = 0 ∂s xz ∂s yz ∂s zz ∂s xz + + = 0 fi =0 ∂y ∂x ∂y ∂z 5 equations, 5 unknowns : w,u,e xx ,s xx ,s xz To be continued… † † (5)