Beam Theory

Euler-Bernoulli Bending Theory (Pure Bending Moment) A

ψ

z

D dw C dx

M

neutral axis

M x

B

ux

uz = w(x) = vertical deflection of the neutral axis

z u x = − zψ ( x )

If the plane AB remains perpendicular to CD

ux = − z

dw dx

dw

dx

ψ

ψ=

dw dx

ux = − z

dw dx

∂u x d 2w = −z 2 ε xx = ∂x dx

If we assume

σ yy = σ zz = σ xy = σ yz = 0

The stress-strain relations give ε xx =

1 ⎡σ xx −ν (σ yy + σ zz ) ⎤ ⎦ E⎣

d 2w σ xx = − E z 2 dx

⎛ ∂u x ∂u z + ∂x ⎝ ∂z

σ xz = G ⎜

⎞ ⎛ dw dw ⎞ = G − ⎜ ⎟=0 ⎟ ⎝ dx dx ⎠ ⎠

z

d 2w σ xx = − E z 2 dx y

σ xx

M = − ∫ σ xx z dA

M

A

=E

xx

d w 2 z dA dx 2 ∫A

dA = 0

A

∫ σ xx y dA = 0 A

x

A

2

d 2w = EI 2 dx

∫σ

y

z

σ xx = −

Mz I

∫ zdA = 0

neutral axis is at centroid

A

∫ y z dA = 0 A

cross-section must be symmetric

Engineering Beam Theory

z

x

qz

M ( x) z σ = − Let xx I V ( x)Q ( z ) σ xz = − I t ( z)

M V

[ Note: we still have u x = − z

σ xz = 0

dM 2 = V ( x) d w dx M = EI 2 dx dV = qz ( x ) dx

dw ⎛ dw ⎞ = ψ ⎜ ⎟ dx ⎠ dx ⎝

(inconsistent) ]

d 4w EI 4 = qz ( x ) dx

qz … applied force/unit length on beam in zdirection

so that

dM = V ( x) dx dV = qz ( x ) dx

How are these internal force and bending moment equilibrium relations related to our local equilibrium equations?

∂σ xx ∂σ xy ∂σ xz + + =0 ∂x ∂y ∂z ∂σ xy ∂x

+

∂σ yy ∂y

+

∂σ yz ∂z

=0

∂σ xz ∂σ yz ∂σ zz + + =0 ∂x ∂y ∂z

∂σ xx ∂σ xy ∂σ xz + + =0 ∂x ∂y ∂z multiply by z and integrate over the cross-section, A

∫z A

⎛ ∂σ ∂σ xx ∂σ dA + ∫ z ⎜ xy + xz ∂x ∂y ∂z A ⎝

⎞ ⎟ dA = 0 ⎠

⎡∂ ⎤ d ∂ z σ dA + z σ + z σ ( ) ( ) xx ∫A ⎢⎣ ∂y xy ∂z xz ⎥⎦ dA − ∫A σ xz dA = 0 dx ∫A

or, equivalently

- V(x)

- M(x) z

∂f ∫A ∂y dA = vC∫ f ny ds

n nz ny

A

C

y

∂f ∫A ∂z dA = vC∫ f nz ds

Gauss’ theorem (2-D)

−

dM + ∫ z ( n yσ xy + nzσ xz ) ds + V ( x ) = 0 dx v C

Tx(

z

n)

n y

Tx( ) = 0 n

dM = V ( x) dx

Now, consider

∂σ xz ∂σ yz ∂σ zz + + =0 ∂x ∂y ∂z integrating over A

⎛ ∂σ yz ∂σ zz d σ xz dA + ∫ ⎜ + ∫ dx A ∂y ∂z A⎝ -V(x)

⎞ ⎟ dA = 0 ⎠

−

dV + ∫ (σ yz n y + σ zz nz ) ds = 0 dx v C Tz( n )

n

Tz( n )

z y

( ) T z v∫ ds = qz ( x ) n

C

applied force/unit length in zdirection

dV = qz ( x ) dx

Last remaining equilibrium equation is:

∂σ xy ∂x

+

∂σ yy ∂y

Integrating over A gives

+

∂σ yz ∂z

=0

⎛ ∂σ yy ∂σ yz d dA + + σ ⎜ xy ∫ ∫ dx A ∂y ∂z A⎝

⎞ ⎟ dA = 0 ⎠

Vy

dVy dx

+ v∫ (σ yy n y + σ yz nz ) ds = 0 C

Ty( n ) ( ) T y v∫ ds = q y ( x ) n

( n)

n

Ty

C

z y

Vy

applied force/unit length in ydirection

dVy dx

= −q y ( x )

which is identically satisfied if

Vy = 0, Ty( ) = 0 n

Timoshenko Beam Theory A

ψ

z

D dw C dx

M

neutral axis

M x

B

ψ

u x = − zψ ( x ) dψ dx ⎛ ∂u ∂u σ xz = G ⎜ x + z ∂x ⎝ ∂z

ψ ( x) ≠

dw dx

dw dx

σ xx = − E z

dw ⎞ ⎞ ⎛ = − ψ + G x ( ) ⎜ ⎟ ⎟ dx ⎠ ⎝ ⎠

better than Euler/Bernoulli but still a constant across the cross-section so introduce a form factor κ 2

σ xx = − E z

dψ dx

dw ⎞ ⎛ ⎟ dx ⎠ ⎝ = σ zz = σ xy = σ yz = 0

σ xz = κ 2G ⎜ −ψ ( x ) + σ yy

For bending moment and shear force

M = − ∫ zσ xx dA = E A

dψ dx

2 z ∫ dA = EI A

dψ dx

dw ⎞ ⎛ V = − ∫ σ xz dA = −κ 2G ⎜ −ψ + ⎟ ∫ dA dx ⎠ A ⎝ A dw ⎞ ⎛ = −κ 2G ⎜ −ψ + ⎟ A = −σ xz A dx ⎠ ⎝

Timoshenko Beam theory

dψ M = EI dx V ( x) dw −ψ ( x ) = − 2 dx κ GA

Example:

Euler-Bernoulli Theory

d 2w M = EI 2 dx dw ψ= dx P

P x

M = - Px

dψ = − Px dx Px 2 + C1 EIψ = − 2

x

L

V=-P

EI

ψ ( L) = 0 ⇒ ψ = This gives

σ xx =

Pxz I

P ⎡⎣ L2 − x 2 ⎤⎦

rotation of the beam cross-section

2 EI (same as ordinary beam theory)

dw V =− 2 +ψ slope of neutral axis κ GA dx P ( L2 − x 2 ) P = 2 + κ GA 2 EI Integrating

P ( L2 x − x3 / 3) Px + + C2 w= 2 κ GA 2 EI PL PL3 + + C2 = 0 w( L) = 0 ⇒ 2 κ GA 3EI

which gives

PL3 PL C2 = w(0) = − − 2 3EI κ GA

deflection due to:

− PL3 w ( 0) = 3EI

3E I ⎤ ⎡ + 1 ⎢⎣ κ 2G AL2 ⎥⎦

For a rectangular section of base b and height h

I 1 ⎛h⎞ = ⎜ ⎟ AL2 12 ⎝ L ⎠

2

bending

shear