S. Socrate 2013 K. Qian

Symmetric Slender Beams in Bending

Loading Conditions on each Section (x) Applied y-Forces & z-Moments The resultants at any section ┴ x are:

the bending moment M(x)

the shear force S(x) [ for slender beams stresses and deformation

due to the shear force S(x) are negligible]

y and z are section symmetry axes

Find • S(x)along the bar (shear force diagram) by cutting the bar at x and imposing force_y equilibrium. • M(x)along the bar (bending moment diagram) by cutting the bar at x and imposing moment_z equilibrium. For the example shown, equilibrium at x left of B gives: Σ Fy = 0 = – S(x) + FB – FC à S (x) = FB – FC

Σ (Mz )x= 0 = – M(x) + MD + LBFB – LCFC à M(x) = MD + LBFB – LCFC Note pivot for z-moment is at “x” to eliminate unknown S (x) from equation

Differential relationships:

For distributed loading qy (x), with qy (x) in N/m + along y obtain S(x) and M(x) by integration

)()( xdxxdx

dxxd SMqS

y −=−=)()(

S

M

x

Examples S(x) and M(x) diagrams

• start at x=L with S(x)=0 , M(x)=0 • “walk” the beam from right to left: (in –x direction) the vertical forces (applied and reactions at supports) point to where the S diagram is going: •  a concentrated load up (positive) means the S diagram has a step up •  a distributed load up (positive) means the S diagram is sloping up •  a concentrated load down (negative) means the S diagram has a step down •  a distributed load down (negative) means the S diagram is sloping down the S diagram and concentrated moments (applied and reactions at supports) point to where the M diagram is going: • a CCW concentrated moment (positive) means the M diagram has a step up •  an S diagram up (positive) means the M diagram is sloping up • a CW concentrated moment (negative) means the M diagram has a step down •  a S diagram down (negative) means the M diagram is sloping down

x

S

M

x

x

S

M

x

x

S

M

x

x

S

M

x

x

S

M

x

x

S

M

x

x

S

M

x

x

S

M

x

x

Section deformation Section at x has rotation θ (x) Section at x+dx has rotation θ (x+dx)= θ (x)+dθ

Local measure of deformation at section x

Kinematics constraint Cross sections ┴ x : stay flat, rotate around z by θ (x)

dxdθ

Section displacement in y-direction v(x): deflection

Section rotation = beam slope: dxdvx =)(θ

NEUTRAL AXIS: no change in x-length with deformation

)(1)( xdxddxdx

ρθ

θρ =→=⋅

Radius at neutral axis

curvature

Structural response

slope deflection with θ (x0)=θ0 and v(x0)=v0 determined by BC (e.g., θ0=0 and v0 =0 at wall) Note that for simply supported beams, must use instead two BCs on v(x) (zero at supports)

∫∫

∫∫

+=+=

+=+=

x

x

x

x

x

x

x

x

dxxvdxdxdxvxv

dxxdxdxdxx

00

00

)()()(

)(1)()(

00

00

θθ

ρθ

θθθ

sign convention

• sections rigidly rotateà strain varies linearly with y

• for positive curvature, top of section shortens, bottom elongates

• for cross sections with top-bottom symmetry the mid-plane is neutral no change in x-length on neutral mid-planeà set y=0 at midplane

section deformation ßà Strain

( ))()(

)()(),(xy

dxdxdyx

dxdxdsyxa ρθρ

θρθρε −=

⋅

⋅−⋅−=

−=

Note: εa is a function of y: larger strains away from the neutral axis

εa

ρε

2h

MAX =

Section equilibrium The Bending Moment M(x) at section x is obtained by integrating the contributions of each elemental area, dA at distance y from the axis, which carries a normal stress σn(x,y,z)

Constitutive Properties If the material is linear elastic and the Young’s modulus of elemental area dA is E the stress can be obtained as

Section Response

Constant over cross section

Effective Section Stiffness:

If only 1 material, E(x)à (EI)eff =E(x) I (x);

If 2 materials (E1, E2) à (EI)eff = E1 I1 + E2 I2

NOTE: I, I1 , I2 all calculated with respect to y=0

dAzyxy(xA

n∫ −= ),,() σM

),(),,(),,( yxzyxEzyx an εσ ⋅=

xA

eff

effinverteff

A

AAa

An

dAzyxEyxEI

xEI(xxxEIxdAzyxEyx(x

dAxzyxEydAyxzyxEydAzyxy(x

∫

∫

∫∫∫

=

=⎯⎯ →←==

⋅⋅=⋅−=−=

),,()()(

)()())(1)())((1),,()(1)

)(1),,(),(),,(),,()

2

2

2

MM

M

ρρρ

ρεσ

iAi

i dAyI ∫= 2

Caution: for simple superposition, all areas for I1 , I2 … must have centroid on y=0 if not à need to use parallel axis theorem

12

3bhI =4

4RI π=

( ) ( ) ( ) ( )

=

=

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ +−++⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ ++=++=

++= ∫∫∫

ci

ci

ici

cii

AAA

yI

AyII

btthbtthbtthbtIIII

dAydAydAyI

))((

212122122

23323

321

3

2

2

2

1

2

Moment of inertia of Ai with respect to its own centroid

Section moment of inertia dAyIA∫= 2

Obtain I for more complex sections by superposition: I=I1 + I2 +…

44

44IO RRI ππ

−=12

212

33 bhBHI −=12

212

33 bhBHI +=

Moment of inertia of rectangle and circle with respect to their centroid

y-coordinate of centroid of area Ai

ß Parallel axis theorem

Special case: homogeneous beam (modulus E); constant cross section with moment of inertia I

1ρ

( x )=M(x )EI

curvature along the beam (at neutral axis)

εa ( x , y )= − yρ( x )

= − y M(x )EI

axial strain

σ n ( x , y )= E εa ( x , y )= − y M(x )I

normal stress

Bending moment M (x) obtained from

ΣMz=0 à M (x) (may be NOT constant along shaft)

⎪⎪

⎩

⎪⎪

⎨

⎧

== ∫

12

4

3

4

2

bh

R

dAyIA

π

ρ(x) x