CHAPTER 5. Beam Theory


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1 CHPTER 5. Beam Theory SangJoon Shin School of Mechanical and erospace Engineering Seoul National University ctive eroelasticity and Rotorcraft Lab.
2 5. The EulerBernoulli assumptions One of its dimensions much large than the other two Civil engineering structure assembly on grid of beams with crosssections having shapes such as T s on I s Machine parts beamlike structures lever arms, shafts, etc. eronautic structures wings, fuselages can be treated as thinwalled beams Beam theory important role, simple tool to analyze numerous structures valuable insight at a predesign stage EulerBernoulli beam theory simplest, must be useful ssumption Crosssection of the beam is infinitely rigid in its own plane inplane displacement field { rigid body translations rigid body rotation The crosssection is assumed to remain plane 3 The crosssection is assumed to normal to the deformed axis 
3 5. The EulerBernoulli assumptions Fig. 5. pure bending beam deforms into a curve of constant curvature a circle with center O, symmetric w.r.t. any plane perpendicular to its deformed axis Kinematic assumptions EulerBernoulli 3 Crosssection is infinitely rigid in its own plane Crosssection remains plane after deformation Crosssection remains normal to the deformed axis of the beam valid for long, slender beams made of isotropic materials with solid crosssections 
4 5. Implication of the EB assumption (,, ) (,, ) (,, ) u x x x u x x x u x x x displacement of an arbitrary point of the beam EB assumption Displacement field in the plane of crosssection consists solely of rigid body translations xial displacement field consists of rigid body translation 3 u ( x ), u ( x ) 3 (,, ) =, u x x x u x 3 Equality of the slope of the beam u x, x, x u ( x ) = (5.) u x { rigid body rotation Φ ( x ), Φ ( x ) 3 u x, x, x = u ( x ) + x Φ ( x ) x Φ ( x ) (5.) { the rotation of the section 3 du du Φ 3 = 3 Φ dx = dx (5.3) consequence of the sign convention
5 5. Implication of the EB assumption To eliminate the sectional rotation from the axial displacement field (5.4.a) Important simplification of EB : unknown displacements are functions of the spanwise coord,,alone Strain field ε = 0, ε = 0, γ = 0 ε 3 3 γ 3 du ( x ) du ( x ) u x x x u x x x = 0, γ = 0 3 (,, ) = 3 3 dx dx x u du ( x ) du( x) d u ( x ) 3 = = x3 x x dx dx dx (5.5.a) EB() (5.5.b) EB() (5.5.c) du( x) 3 ε ( x) =, κ ( x) =, κ 3 x = dx dx dx Sectional axial strain Sectional curvature about i axes, i3 du ( x) ε x, x, x = ε ( x ) + xκ ( x ) x κ ( x ) (5.7) du( x). (5.6) EB() 4 ssuming a strain field of the form Eqs (5.5.a),(5.5.b),(5.7) Math. Expression of the EB assumptions
6 5.3 Stress resultants 3D stress field described in terms of sectional stresses called stress resultants equipollent to specified components of the stress field N( x) { V x V3 x 3 force resultants axial force, M ( x ), M ( x ) transverse shearing forces moment resultants : bending moments 3 N ( x ) = σ x, x, x d 3 = τ 3 τ3 (,, 3 ) V ( x ) x, x, x d, 3 V x = x x x d M ( x ) = x σ x, x, x d 3 3 M ( x ) x σ x, x, x d = 3 3 (5.0a) (5.0b) (5.8) (5.9) 5 (+) equipollent bending moment about i (Fig 5.5) 3 bending moments computed about point p 3 M ( x ) = ( x x ) σ x, x, x d p 3 3p 3 p 3 ( ) = ( p ) σ,, 3 M x x x x x x d Px (, x ) p (5.a) (5.b) Fig Sign convention for the sectional stress resultans
7 5.4 Beams subjected to axial loads Distributed axial load p (x ) [N/m], concentrated axial load P [N] axial displacement field u( x) bar rather than beam 5.4. Kinematic description xial loads causes only axial displacement of the section Eq.(5.4) xial strain field u x, x, x = u ( x ) 3 u x, x, x = 0 3 u x, x, x = 0 ε 3 3 x, x, x = ε ( x ) 3 (5.a) (5.b) (5.c) (5.3) uniform over the xs (Fig. 5.7) 6
8 5.4 Beams subjected to axial loads 5.4. Sectional constitutive law σ << σ, σ 3 << σ transverse stress components 0, σ 0, σ 3 0 Generalized Hooke s law σ ( x, x, x ) = Eε ( x, x, x ) 3 3 (5.4) Inconsistency in EB beam theory Eq. (5.5a) Hooke s law if Eq.(5.3) (5.4) : xial force ε = 0, ε = 0 3 t the infinitesimal level ε = νσ / E, ε = νσ / E σ = σ3 = 0, then 3 (Poisson s effect) very small effect, and assumed to vanish σ ( x, x, x ) = Eε ( x, x, x ) 3 3 N ( x ) = σ x, x, x d 3 = Ed ε( x ) = S ε( x ) S xial stiffness for homogeneous material = E (5.5) (5.6) constitutive law for the axial behavior of the beam at the sectional level 7
9 5.4 Beams subjected to axial loads Equilibrium eqns Fig. 5.8 infinitesimal slice of the beam of length dx dn force equilibrium in axial dir. dx = p (5.8) Eq. (.4) Eq. (5.8) equilibrium condition for a differential element of a 3D solid equilibrium condition of a slice of the beam of differential length dx 8
10 5.4 Beams subjected to axial loads Governing eqns Eq (5.6) Eq. (5.8) and using Eq. (5.6) d du S = p( x) dx dx (5.9) 3 B.C Fixed(clamped) : Free (unloaded) : u = 0 du N = 0 0 dx = P N = P 3 Subjected to a concentrated load : S du dx = P The sectional axial stiffness Homogeneous material S = E (5.0) 9
11 5.4 Beams subjected to axial loads Rectangular section of width b made of layered material of different moduli(fig. 5.9) n n [] i [] i [] i [ i ] [] i [ i ] i= i= S = Ed = E d = E b( x x ) weighting factor thickness weighted average of the Young s modulus The axial stress distribution 0 Eliminating the axial strain form Eq.(5.5) and (5.6) E σ ( x, x, x3) = N( x) S (5.) Homogeneous material N( x) σ ( x, x, x3) = (5.) Uniformly distributed over the section Sections made of layers presenting different moduli [] i [] i N( x) σ ( x, x, x3) = E (5.3) S Stress in layer I is proportional to the modulus of the layer
12 5.4 Beams subjected to axial loads Eq (5.3) axial strain distribution is uniform over the section, Strength criterion i.e. each layer is equally strained (Fig. 5.0) E N S tens tens max σ allow, E N S comp max σ comp allow (5.4) in case compressive, buckling failure mode may occur Chap. 4 
13 5.5 Beams subjected to transverse loads Fig. 5.4 transverse direction distributed load, p( x) [N/m] concentrated load, [N] P bending moments, transverse shear forces, and axial stresses will be generated { transverse shearing 5.5. Kinematic description ssumption transverse loads only cause transverse displacement curvature of the section General displacement field (Eq.(5.6)) du( x) u( x, x, x3) = x dx (5.9a) u x, x, x = u ( x ) 3 u x, x, x = (5.9b) (5.9c) {  linear distribution of the axial displacement component over the xs
14 5.5 Beams subjected to transverse loads Only nonvanishing strain component ε x, x, x = x κ ( x ) (5.36) linear distribution of the axial strain Sectional constitutive law Linearly elastic material, axial stress distribution σ Sectional axial force by Eq. (5.8) x, x, x = Ex κ ( x ) 3 3 N ( x ) = σ x, x, x d 3 = Ex d κ3( x ) xial force =0 since subjected to transverse loads only κ3 0, then Exd = 0 S x = c Exd 0 S = S = (5.3) (5.3) (5.33) Location of the modulusweighted centroid of the xs 3
15 5.5 Beams subjected to transverse loads If homogeneous material x E x d = = x d = 0 E d c x is simply the area center of the section The axis system is located at the { (5.34) modulusweighted centroid area center if homogeneous material Center of mass x d nn = = = xc Bending moment by Eq.(5.3) x ρ ρ d x d d c M 3( x) = Ex d κ3( x) = H33κ3( x) center of mass 3 coincide (5.35) centroid bending stiffness about axis Constitutive law for the bending behavior of the beam bending moment i 3 the curvature 4 M x = H κ ( x ) c 33 3 momentcurvature relationship (5.37) Bending stiffness ( flexural rigidity )
16 5.5 Beams subjected to transverse loads Equilibrium eqns Fig. 5.6 infinitesimal slice of the beam of length dx M ( x ), V ( x ) acting at a face at location x 3 + evaluated using a Taylor series expansion, and H.O terms ignore x 5 equilibrium eqns vertical force { moment about O dm3 dx = p ( x ) dv dx = p x dm 3 V dx + = 0 (5.38a) (5.38b) (5.39)
17 5.5 Beams subjected to transverse loads Governing eqns Eq. (5.37) Eq. (5.39), and recalling Eq. (5.6) 4 B.C Clamped end 3 d dx Simply supported(pinned) Free(or unloaded) end = c du H33 p x dx u = 0, du dx du dx = 4 th order DE 0 u = 0, = 0, du dx = 0 d c du H33 0 = dx dx (5.40) 6 4 End subjected to a concentrated transverse load du dx = 0, P P dm = V = dx d = dx c du H33 P dx 3
18 5.5 Beams subjected to transverse loads 5 Rectilinear spring(fig. 5.7) d H ku ( L) = 0, dx dx = c du 33 x L V ( L) = ku ( L) sign convention du dx = 0 (+) when the spring is located at the left end 6 Rotational spring(fig. 5.8) H du c 33 k dx dx x = L du + = 0, () when at the left end M ( L) = kφ ( L) 3 3 d c du H33 0 = dx dx = x L 7
19 5.5 Beams subjected to transverse loads The sectional bending stiffness Homogeneous material I H = EI c c = x d c 33 (5.4) (5.4) : purely geometric quantity, the area second moment of the section computed about the area center Rectangular section of width b made of layered materials (Fig. 5.9) b H33 = Ex d = E x [ ] d = E ( x ) ( x ) i 3 n n c [] i [] i [] i [ i+ ] 3 [] i 3 i= i= (5.43) weighted average of the Young s moduli 8
20 5.5 Beams subjected to transverse loads The axial stress distribution Local axial stress eliminating the curvature from Eq.(5.3), (5.37) M3( x) σ ( x, x, x3) = Ex c H 33 (5.44) homogeneous material σ linearly distributed over the section, independent of Young s modulus various layer of materials σ 3 ( x, x, x ) = x 3 (,, ) x x x = E x M ( x ) I [] i [] i 3 3 c H33 axial STRIN distribution is linear over the section Eq.(5.30) axial stress distribution piecewise linear (Fig. 5.0) 33 M ( x ) (5.45) (5.46) 9
21 5.5 Beams subjected to transverse loads Strength criterion x max max max comp max tens EM3 σ allow, EM c 3 σ, c allow 33 H33 Maximum (+) bending moment in the beam H Layers of various material must be computed at the top locations of each ply { bottom x Rational design of beams under bending Neutral axis along axis i 3 which passes through the section s centroid Material located near the N. carries almost no stress Material located near the N. contributes little to the bending stiffness Rational design removal of the material located at and near the N. and relocation away from that axis 0
22 5.5 Beams subjected to transverse loads Fig. 5. rectangular section, same mass m = bhρ { ideal Ratio of bending stiffness Hideal For d / h= 0, 00 H For d / h= 0, H H a thin web would be used to keep the flanges ideal rect rect = = + 3 max σ rect 6( d / h) = 60 max σ ideal b( h / ) bh E + d d bh 4 h E Ratio of max. axial stress h d E MI + 3 ideal σ ideal 4 h = = σ h d rect IidealE d + M3 + 4 h ideal section can carry a 60 times larger bending moment  Ideal section I beam, but prone to instabilities of web and flange buckling
23 5.6 Beams subjected to combined and transverse loads Sec. 5.4, 5.5 convenient to locate the origin of the axes system at the centroid of the beam s xs 5.6. Kinematic description du( x) u( x, x, x3) = u( x) ( x xc ) dx u x, x, x = u ( x ) 3 u x, x, x = location of centroid (5.73a) (5.73b) (5.73c) Strain field ε x, x, x = ε ( x ) ( x x ) κ ( x ) 3 C 3 (5.74) 
24 5.6 Beams subjected to combined and transverse loads 5.6. Sectional constitutive law 3 xial stress distribution xial force Bending moment σ x, x, x = Eε ( x) Ex ( x ) κ ( x) 3 C 3 [ ε κ ] M = ( x x ) E ( x ) E( x x ) ( x ) d C 3 C C 3 [ ε κ ] N = E x E x x x d C 3 = Ed ε( x) + E( x xc ) d κ3( x) S = E( x xc) d ε( x) + E( x xc) d κ3( x) (5.75) = 0 c H 33 (bending stiffness) decoupled sectional constitutive law crucial steps Displacement field must be in the form of Eq.(5.73) { Bending moment must be evaluated w.r.t. the centroid Thus, centroid plays a crucial rule (axial stiffness) = Ex d x Ed = S Sx = C C 0 N M = Sε = H κ C c
25 5.6 Beams subjected to combined and transverse loads Equilibrium eqns Fig infinitesimal slice of the beam of length dx dn Force equilibrium in horizontal dir. dx = p dv Vertical equilibrium dx = p Equilibrium of moments about the centroid (5.8) (5.38a) dm 3 (5.77) V ( x a x C) p dx + = Moment arm of the axial load w.r.t the centroid 4
26 5.6 Beams subjected to combined and transverse loads Governing eqns d dx du S dx d c du d H33 = p( x ) + ( xa xc) p ( x ) dx dx dx = p( x) [ ] (5.78a) (5.78b) (5.9) almost similar to (5.9) concept { { u ( x ) u ( x ) decoupled eqns (5.78a) can be independently solved (5.78b) If axial loads are centroid, extension and bending are decoupled If axial loads are not centroid, extension and bending are coupled 5
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