Solid Mechanics Equation Sheet s4w3b

Solids

Ch01: Concept of Stress Axial Stress F

dF

P

σ=

=

A

where: FP = Normal Force

P

A = Cross Sectional Area

dA

Shear Stress F V τ= A Bearing Stress where: t= thickness d= diameter

F σ = b

P

td

Oblique Plane under Axial Loading F F P P 2 σ=  ( cos( θ) ) τ=  sin( θ)  cos( θ) A A 0

0

F τ

max

=

P

A

 sin( 45)  cos( 45)

0

Factor of Safety σ ultimate FS = σ allowable

Ch02: Stress and Strain - Axial Loading Normal Strain under Axial loading ε=

ΔL L

=

δ

=

L

dδ dL

Hooke's Law/Modulus of Elasticity (E) σ = E ε Determining Yield Strength using offset Method σ = E ( 0.002 ) Y

Deformation under Axial Loading  = δ= A E   P L

L

P A E

dx

0

Strain Due to Temperature Changes δ = α ΔT L T

where: α = coefficient of thermal expansion

Solids Page 1 of 9

Solids

Poisson's Ration (ν) ε lateral ν= ε axial

Multiaxial Loading ν σ σ ν σ x y z ε =   x E E E ν σ y

y



E

E

ν σ

x

z

z



E

ν σ ε =

ν σ

σ

x

ε =

y



E

E

σ 

z

E

Dilation/Bulk Modulus (K) ΔV e=ε ε ε = x y z V e=

K=

( 1  2ν) E



where: e = the change in volume per unit volume

 σ σ σ x

y

=

z

( 1  2ν) E

P

where: P = uniform hydrostatic pressure

E 3  ( 1  2ν)

Shear Strain and Modulus of Rigidity (G) τ = G γ τ γ

xy

τ

xy

=

γ

G

yz

=

yz

G

τ γ

zx

=

zx

G

E = 2  G ( 1  ν) Stress Concentrations σ max k= σ ave

Ch03: Torsion   T =  r dF =  r τ dA   r ϕ r γ= = γ max L r max

τ=

T r J

=

r r

τ

max

max Solids Page 2 of 9

Solids

Angle of Twist  = ϕ= J G   T L

L

T J G

dx

0

Transmission Shafts P = 2  π f  T P

T=

2  π f

Stress Concentrations τ = k

T r J

Thin-Walled Hollow Shafts where: T τ= Λ = area bound by the center line of the wall thickness 2  t Λ t = the thickness of the wall section Plastic Deformation   =  T 2 π   

rY

0

T

Y

r  τY   max 2  d  2  π  r  r r r  τ dr Y   rY     rY 2

 max  τY TP = 3  TY 4

3

= .5 π r

3 3   ϕ  1  rY   4 4 1  Y   T = T  1   = T  1     3 Y  4  ϕ   3 Y 4 r max   

1

ϕ

r

Y

r

max

=

Y

ϕ

=  4 



3 T  T

3



Y

Ch04: Pure Bending Deformations in a symmetric member L = p θ

where: p is the radius of deformation arc

L = ( p  y)  θ

δ=L  L

y

ε = x

±y p

=

±y c

ε

y

max

where: c = largest distance from the neutral surface (ymax) ± depends on direction of moment

Stresses and Deformations in the Elastic Range σ= 1 p

=

±M y I

=

±y c

σ

max

M E I Solids Page 3 of 9

Solids

Deformations in a Transverse Cross Section 1 ν Anticlastic_Curvature = = p p z

Bending of Disimilar Materials E 2 n= b = b n ( 1 2) 1 E 1

σ = 2

M y

σ = n 1

I

b2 b( 1 2)

M y I

b2 b( 1 2)

Stress Concentrations M c σ = k max I Eccentric Axial Loading F P ±M y σ =  x A I Unsymmetric Bending ±M  z ±M  y y z σ =  x I I y

z

I tan( ϕ) =

z

I

 tan( θ)

y

General Case Eccentric Axial Loading ±M  y F ±M  z P y z σ =   x I I A y

z

Bending in Curved A R=   1 dA  r  σ = x

±M y A e ( R  y )

=

M  ( r  R)

where: e = the distance from the centroid of the section to the neutral surface.

A e r

Plastic Bending 2   1  rY   M = M  1    2 Y 3 r   max  

3

M

Y

=

2 3

2

 b c  σ

Y

M = P

3 2

M

Y

Solids Page 4 of 9

Solids

Ch05: Beams dF

V

= ω dx

where: ω is distrubuted loads

dM = F  dx V

Ch06: Shearing Stresses in Beams and Thin-Walled Transverse Shear F Q ave

max

q=

I t 3

=

τ

where: Q is the first moment with respect to the neutral axis

V

=

τ

F

V



for a beam of rectangular cross section

2 A F Q

ΔH Δx

where: ∆H is horizontal shear q is the horizontal shear per unit length

V

=

I

where: A' is the area of the section y is the distance from the centeroid of the section to the neutral axis

Q = A' y

Unsymmetric Loading z

 2 F =  q ds Vz  z 1

1

h

F e=

y

 2 F =  q ds Vy  y

Vz

F

Vy

Ch07: Transformations of Stress and Strain Plane Stress σ σ σ =

x

x'

σ σ

y

2



σ σ σ =

x

y'

y

 cos( 2θ)  τ

sin( 2  θ)

 cos( 2θ)  τ

sin( 2  θ)

xy

2 σ σ

y

2

x



x

y

σ σ xy

2

σ

ave

=

x

y

2

σ σ τ

xy'

=τ

xy

cos( 2  θ) 

σ σ σ = p

x

y

2

x

y

2

 sin( 2θ)

2  σx  σy    τ 2 ±  xy 2  

 p

tan 2  θ

2τ =

xy

σ σ x

y Solids Page 5 of 9

Solids

2  σx  σy     τxy 2 τ =  max  2 

σ σ

 s

tan 2  θ

=

x

y

2 τ

xy

Plane Strain ε ε x

ε = x'

2 ε ε x

ε = y'

τ

x'y'

y

2



ε ε

p

max

ε ε

3

x

2

y

γ cos( 2  θ) 

xy

2 γ

cos( 2  θ) 

xy

2

ε ε

ν 1υ

x

ε

sin( 2  θ)

2 2  ε x  εy   γxy     R=   2   2 

 sin(2 θ)  γxy cos(2 θ)

y

ave

=

 p

tan 2  θ

εx  εy  γxy 2

y

sin( 2  θ)

±R

= 2 R =

ε =

y

2



x

ave

x



=ε ε

ε =ε

γ

y

2

γ =

xy

ε ε x

y

2

ε1  ε2

Cylindrical Pressure Vessels

σ = 1

P r

σ = 2

t

P r

τ

2 t

max

=σ = 2

P r 2 t

where: P is the pressure r is the radius t is the thickness

Spherical Pressure Vessels σ =σ = 1

2

P r

τ

2 t

max

=

1 2

σ = 1

P r 4 t

Ch08: Principal Stresses under a Given Loading Design 2 2 Shafts 2  ofMTransmission  M T y z   max J = τ c

    all

Ch09: Deflection of Beams M( x) E I

=

d

2

dx

2

y =

1

where: ρ is curvature

ρ

Solids Page 6 of 9

Solids

d

E I

4

dx d

E I

3

dx d

E I

y = F( x )

3

V

2

dx E I

y = ω( x )

4

y = M( x)

2

d y = E I θ( x ) dx

    1 1 3 2 E I y ( x ) =     ω( x ) dx dx dx dx   C  x   C  x  C  x  C 1 2 3 4 6 2     Ch10: Columns d

2

dx

2

M

y =

=

E I

P E I

y

L

efixed.free

= 2 L

L

epinned.pinned

=L

2

π  E I

P = cr

Le

L

2

efixed.pinned

= 0.7 L

L

efixed.fixed

= 0.5 L

Ch11: Strain Energy Strain Energy (U)    U =  F dx =  P     0 x1

σx

2

dV

2 E

Strain Energy Density (u)   u= = dV  

x1

dU

 dx =   A L  F

P

ε1

0

0

2

σ dε = x

x

σ

2 E

Axial Loading   U=  

L

0

FP

2

2  A E

FP  L dx = 2

2  A E

x

x

  1 2 U =  F dx =  k  x dx =  k  x P  2  0 0

Solids Page 7 of 9

Solids

Bending   U=  

L

M

2

θ

2

dx =

2  A E

 1 U =  M dθ =  M  θ  2 0

M L 2  A E

0

Shearing  u=  

γxy

τ

dγ

xy

xy

0

  U=  

τxy

2

dV

2 G

Torsion   U=  

L

T

2

2

2  G J

 U= 

T L

dx =

2  G J

ϕ

T dϕ =

0

0

1 2

 T ϕ

General State u=

u=

u =

1 2



 σ ε  σ ε  σ ε  τ γ x x

z z

xy

xy

 τ γ xy

xy

 τ γ xy



xy

 σ 2  σ 2  σ 2  2 ν σ  σ  σ  σ  σ  σ   1   τ 2  τ 2  τ 2  y  z  x y y z z x 2 G  xy  yz  zx  2  E  x 1

1  2 ν

v

6 E

u =

12 G 

d

y y



 σ σ σ a

b

 σ  σ

1

a



2

where u is the energy associated with a change in volume v

c

  σb  σc  σc  σa  2

2

2

b

where u is the energy associated with d

distortion.

Castigliano's Theorem x = j

δU j

θ = j

ϕ = j

 x = = j δP  

L

δU

δP

δU δM

j

δU δT

j

M δM  dx E I δP

 Fi Li δFi  x =  = j  Ai E δP  δP   i 1 n



j

δU δQ

j

j

0

δU

x =

where: x is deflection j

Q is a dummy/fictitious load j

Solids Page 8 of 9

Solids

Solids Page 9 of 9