FE Review Mechanics of Materials

FE Mechanics of Materials Review Stress F

V

M

N

N = internal normal force (or P) V = internal shear force M = internal moment

N P = Normal Stress = σ = A A Average Shear Stress =

τ=

V A

Double Shear F/2

F

F/2

V = F/2

F

V = F/2

τ=

F

2 A

FE Mechanics of Materials Review Strain Normal Strain

ε=

∆L L − L0 δ = = L0 L0 L0

Units of length/length

ε = normal strain ∆L = change in length = δ L0 = original length L = length after deformation (after axial load is applied) Percent Elongation =

∆L × 100 L0 Ai − A f

Percent Reduction in Area =

Ai

Ai = initial cross- sectional area Af = final cross-sectional area

× 100

FE Mechanics of Materials Review Strain Shear Strain = change in angle , usually expressed in radians y B

γ xy θ

B'

x

FE Mechanics of Materials Review

Stress-Strain Diagram for Normal Stress-Strain

FE Mechanics of Materials Review

FE Mechanics of Materials Review Hooke's Law (one-dimension)

σ = Eε σ = normal stress, force/length^2 E = modulus of elasticity, force/length^2 = normal strain, length/length

ε

τ = Gγ τ = shear stress, force/length^2

G = shear modulus of rigidity, force/length^2 = shear strain, radians

γ

FE Mechanics of Materials Review

E G= 2(1 + ν )

ν

= Poisson's ratio = -(lateral strain)/(longitudinal strain)

ε lat ν =− ε long ε lat =

δ' r

ε long =

δ L

change in radius over original radius

change in length over original length

FE Mechanics of Materials Review Axial Load If A (cross-sectional area), E (modulus of elasticity), and P (load) are constant in a member (and L is its length):

E=

σ P A = ⇒ ε δ L

δ=

PL AE

Change in length

If A, E, or P change from one region to the next:

PL δ =∑ AE

δ

Apply to each section where A, E, & P are constant

A / B = displacement of pt A relative to pt B

δA

= displacement of pt A relative to fixed end

FE Mechanics of Materials Review -Remember principle of superposition used for indeterminate structures - equilibrium/compatibility

FE Mechanics of Materials Review Thermal Deformations

δ t = α ( ∆T ) L = α (T − T0 ) L δt α

= change in length due to temperature change, units of length = coefficient of thermal expansion, units of 1/°

T

= final temperature, degrees

T0

= initial temperature, degrees

FE Mechanics of Materials Review Torsion Torque – a moment that tends to twist a member about its longitudinal axis

Shear stress, τ , and shear strain, γ , vary linearly from 0 at center to maximum at outside of shaft

FE Mechanics of Materials Review T

Tr τ= J

φ=

TL JG

τ

= shear stress, force/length^2

r

T = applied torque, force·length

r

= distance from center to point of interest in cross-section (maximum is the total radius dimension)

J

= polar moment of inertia (see table at end of STATICS section in FE review manual), length^4

φ

= angle of twist, radians

L

= length of shaft

G

= shear modulus of rigidity, force/length^2

τ φ z = Gγ φ z = Gr ( dφ / dz ) ( dφ / dz ) = twist per unit length, or rate of twist

FE Mechanics of Materials Review Bending Positive Bending Makes compression in top fibers and tension in bottom fibers

Negative Bending Makes tension in top fibers and compression in bottom fibers

FE Mechanics of Materials Review

dV = q( x ) Slope of shear diagram = negative of distributed loading value Î − dx dM Slope of moment diagram = shear value Î =V dx

FE Mechanics of Materials Review

x2

Change in shear between two points = neg. of area under V2 − V1 = [− q( x )]dx ∫ distributed loading diagram between those two points Î x1 x2 Change in moment between two points = area under M 2 − M1 = ∫ [V ( x )]dx shear diagram between those two points Î x1

FE Mechanics of Materials Review Stresses in Beams

My σ =− I

Mc σ max = ± I εx = − y ρ From

σ

= normal stress due to bending moment, force/length^2

y

= distance from neutral axis to the longitudinal fiber in question, length (y positive above NA, neg below)

I

= moment of inertia of cross-section, length^4

c

= maximum value of y; distance from neutral axis to extreme fiber

ρ = radius of curvature of deflected axis of the beam

σ = Eε = − E y ρ Î σ = − My I

and

1

ρ

=

M EI

FE Mechanics of Materials Review

S=I Then

c

S

= elastic section modulus of beam

Mc M σ max = ± =± I S

VQ Transverse Shear Stress: τ = It Transverse Shear Flow: VQ q= I Q = y ' A' t = thickness of cross-section at point of interest t = b here

FE Mechanics of Materials Review Thin-Walled Pressure Vessels (r/t >= 10) Cylindrical Vessels

σt =

pr = σ1 t

σ1 = hoop stress in circumferential direction p r t

= gage pressure, force/length^2 = inner radius = wall thickness

pr σa = = σ2 2t

= axial stress in longitudinal direction

See FE review manual for thick-walled pressure vessel formulas.

FE Mechanics of Materials Review 2-D State of Stress Stress Transformation σ x +σ y σ x −σ y σ x' = + cos 2θ + τ xy sin 2θ 2 2

σ y' =

σ x +σ y σ x −σ y −

2

τ x' y' = −

σx −σ y 2

2

cos 2θ − τ xy sin 2θ

sin 2θ + τ xy cos 2θ

Principal Stresses

σ 1, 2 =

σx +σy

tan 2θ p =

2

±

τ xy

⎛ σx −σy ⎜⎜ 2 ⎝

⎛ σx −σ y ⎞ ⎜ 2 ⎟⎠ ⎝

2

⎞ 2 ⎟⎟ + (τ xy ) ⎠

No shear stress acts on principal planes!

FE Mechanics of Materials Review Maximum In-plane Shear Stress

⎛ σx −σ y ⎞ 2 max ⎟⎟ + (τ xy ) τ in − plane = ⎜⎜ 2 ⎠ ⎝ 2

⎛ σ x −σ y ⎞ ⎟⎟ / τ xy tan 2θ s = −⎜⎜ 2 ⎠ ⎝

σ avg =

σx +σy 2

FE Mechanics of Materials Review Mohr's Circle – Stress, 2D

Center: Point C( σ avg =

σx +σy

,0)

2

−τ

σ, positive to the right tau, positive downward!

R = (σ x − σ avg ) + (τ xy ) 2

2

σ1 = σ avg + R = σ a σ 2 = σ avg − R = σ b

τ inmax − plane = R +τ A rotation of θ to the x’ axis on the element will correspond to a rotation of 2θ on Mohr’s circle!

FE Mechanics of Materials Review Beam Deflections

+ -

Fig. 12-2

Inflection point is where the elastic curve has zero curvature = zero moment

ε

σ

− My ⇒ and σ = =− Alsoε = E I ρ y

1

M = ρ EI 1

ρ

= radius of curvature of deflected axis of the beam

FE Mechanics of Materials Review 2 M d2y d y = = 2 ⇒ M ( x ) = EI ρ EI dx dx 2

1

from calculus, for very small curvatures

⎛ dM ( x ) ⎞ V =⎜ ⎟⇒ ⎝ dx ⎠

− w( x ) =

dV ( x ) ⇒ dx

V ( x ) = EI

− w( x ) = EI

d 3y dx

3

d 4v

for EI constant

= − q dx 4

for EI constant

Double integrate moment equation to get deflection; use boundary conditions from supports Î rollers and pins restrict displacement; fixed supports restrict displacements and rotations

FE Mechanics of Materials Review

M ( x ) = EI

d2y dx

2

⇒

[ ∫ M ( x )dx ]dx ∫ y= EI

• For each integration the “constant of integration” has to be defined, based on boundary conditions

FE Mechanics of Materials Review Column Buckling

Pcr =

π 2 EI A2

Pcr

I A r= r=

Euler Buckling Formula (for ideal column with pinned ends)

= critical axial loading (maximum axial load that a column can support just before it buckles) = the smallest moment of inertia of the cross-section = unbraced column length

I A

= radius of gyration, units of length

I ⇒ I = r2 A ⇒ A

A/r

Pcr π 2E = σ cr = A ( A / r )2

= slenderness ratio for the column

= critical buckling stress

FE Mechanics of Materials Review

Euler’s formula is only valid when When

σ cr > σ yield

σ cr ≤ σ yield

.

, then the section will simply yield.

For columns that have end conditions other than pinned-pinned:

Pcr =

π 2 EI

(KL )2

K = the effective length factor (see next page) KL = Le = the effective length

σ cr =

π 2E

(KL / r )2

KL/r = the effective slenderness ratio

FE Mechanics of Materials Review Effective Length Factors