summary of concepts and formulae

8/12/2019 Summary of Concepts and Formulae

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University of California at Berkeley CE 130Department of Civil and Environmental Engineering Section 2

J. Lubliner Spring 2007

Summary of Concepts and Formulas

I. General Principles

1. Statics

(a) External equilibrium (loads, reactions):

F= 0,

M= 0

(b) Internal forces (free-body diagram)

(c) Stress

i. stress tensor

x xy xzyx y yzzx zy z

ii. Moment equilibrium: xy =yx, etc.

iii. Mean stress0 = 13

(x+ y+ z); pressure p= 0iv. Equilibrium of thin shells of revolution: (cylindrical)z =pR/2t, =pR/t,r 0;

(spherical) ==pR/2t,r 0.

2. Geometry

(a) Displacement field: u(x,y,z), v(x,y,z), w(x,y,z)

(b) Strain

i. x=u

x (longitudinal),xy =yx =

u

y+

v

x(shear), etc.

ii. strain tensor

x12

xy12

xz12

yx y12

yz12

zx12

zy z

3. Constitutive Properties

(a) General: relation between stress, strain and temperature

(b) Linear elastic isotropic

i. general: x= 1

E[x y z] +T etc.,xy =

xyG

etc.;G= E

2(1 + )

ii. plane stress (inxy-plane): z =yz =zx = 0, x = E

1 2(x+ y) etc.

iii. plane strain: z =yz =zx = 0, x= E

(1 + )(1 2)[(1 )x+ y] etc.

(c) Elastic-perfectly plastic: stress-strain curve becomes horizontal after yield, unloading iselastic.

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(d) Yield stress: uniaxial yp, shear yp

(e) Yield criteria

i. Tresca: ||max yp= 12

yp

ii. Mises, 12

[(x y)2 + (y z)

2 + (z x)2] + 3(2xy+

2yz +

2zx)

2yp= 3

2yp

4. Work, energy

(a) Global work-energy relation: (elastic)W =U, W = U; (linear elastic) U= U

i. W =

F d or

M d,

ii. W =

dF or

dM

(b) Linear elastic

i. Local: 1

2(xx+ yy+ zz+ xyxy+ yzyz + zxzx) =Uo(x, . . .) = Uo(x, . . .);ii. Global: U=

VUodV(strain energy),

U=V UodV (complementary energy)

(c) Castiglianos theorems for elastic systems

i. 1st Theorem: Pi= U

i,Mi=

U

i

ii. 2nd Theorem: i= U

Pi, i=

U

Mi(d) Reciprocal relations (linear elastic)

i. Finite number of degrees of freedom: U= 12

ij

kijij ,kij =kji ([kij] = stiffness

matrix, {i} include displacements and rotations)

ii. Finite number of loads: U= 12

i

j

fijFiFj,fij =fji ([fij ] = flexibility matrix, {Fi}

include both forces and moments)

(e) Potential energy: =U+

i. =

(F +M ) or L0 qvdxetc.: potential energy of applied loads

ii. Minimum potential energy: = 0 for equilibrium (finite number of degrees offreedom: /i= 0, /j = 0)

5. Transformation of axes in two dimensions

(a) Vectors: Fx =ex F, etc.,

FxFy

=

cos sin sin cos

FxFy

(b) Tensors:

xx xyxy yy

=

cos sin sin cos

xx xyxy yy

cos sin sin cos

wherexx = x,xy =xy etc.

i. x = = 12(x+ y) + 1

2(x y)cos2+ xysin 2,y =+/2

ii. =xy = 12(x y)sin2+ xycos 2

iii. For strains usexx=x, xy = 12xy etc.

iv. Chain rule for derivatives: x

= xx

x

+ yx

y

, etc.

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v. Strain rosettes: i = 1

2[(x+ y) + (x y)cos2i+ xysin 2i],i= 1, 2, 3; solve for

x,y,xy

(c) Principal values

i. Definition: x =1 and y =2 whenxy = 0; alsodx

d = 0; convention: 1> 2

(note that 1 = 2 if and only ifx =y andxy = 0)

ii. Principal angles: tan 21,2= 2xy/(x y)

iii. Principal values: 1,2= 12(x+ y)

[ 12

(x y)]2 + 2xy

(d) Maximum shear

i. Definition: dxy

d = 0

ii. Maximum shear angle: tan 2s= (x y)/2xy

iii. Maximum shear: max=

[ 12

(x y)]2 + 2xy = 1

2(1 2)

iv. Relation to principal angle: s 1,2= /4

(e) Mohrs circle

i. To draw: in the - plane, mark the points (x, xy and (y, xy), and draw theline between them. The intersection of this line with the -axis, i.e. the point( 1

2[x+ y], 0), is the center of the circle.

ii. The radius isr= max.

iii. The circle intersects the-axis at (1, 0) and (2, 0).

iv. Rotation on the circle is twice the physical rotation and in the opposite direction.

6. Transformation of axes in three dimensions

(a) Principal values and principal directions

i. In order to solve the system

(x )nx+ xyny+ xznz = 0,xynx+ (y )ny+ yznz = 0,xznx+ yzny+ (z )nz = 0

, it is necessary

that

x xy xzxy y yz

xz yz z

= 0. This is a cubic equation in whose roots 1,

2,3 are the principal values (eigenvalues), and the vector n = inx+jny+ knz foreach of these values (eigenvector) gives the direction of the corresponding principalaxis.

ii. If one of the principal axes is known, two-dimensional analysis can be used to findthe other two.

iii. When all three principal values are known, Mohrs circles can be drawn. The maxi-mum shear is the radius of the largest Mohrs circle.

(b) Yield criteria in terms of principal stresses (no numbering convention in terms of value)

i. Tresca: max(|1

2|, |

2

3|, |

1

3|) = 2

ypii. Mises: (1 2)

2 + (2 3)2 + (1 3)

2 = 22yp

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II. Slender Bodies

1. Internal Forces(a) Axial force: P=

A x dA

(b) Torque (cylindrical shaft): T =A zr dA

(c) Bending moment: Mz = A xy dA,My =

A xz dA

(d) Shear force: Vy = A xydA, Vz =

A xzdA

2. Elastic energy

(a) Strain energy: U =A Uo dA,U=

L0 U

dx

(b) Complementary energy:U

=A

Uo dA,

U=

L0

U

dx

3. Axial loading

(a) General

i. Strain-displacement relation: = x=du

dxii. Elongation L= u(L) u(0) =

L0 dx

iii. Resultant forceP =A dA

iv. Equilibrium (differential equation): dP

dx+p= 0 (p= distributed axial load per unit

length)v. Work of virtual load: F =

L0 P dx (=

PL if uniform), where P = bar forcedue to virtual load F conjugate to

(b) Linear elastic

i. Hookes Law= /E

ii. Elongation L=L0 (P/EA) dx( =PL/EAif uniform)

iii. Strain energy: U = 12

(A E

2 dA)dx(= 12

EA2,U=EA(L)2/2L if uniform)

iv. Complementary energy: U =

A(2/E) dA (= P2/2EA if homogeneous, U =

P2L/2EA if uniform, EA/L= axial spring constant)

v. Work of virtual load: F =

L0 (PP/EA) dx(= PPL/EAif uniform)

4. Torsion

(a) General (cylindrical shaft)

i. Strain-displacement(twist) relation: =z =r ( =

d

dz)

ii. Total rotation = (L) (0) =L0

dz

iii. Resultant torqueT= 2cb r

2 dr

iv. Equilibrium (differential equation): dT

dz +t = 0 (t = distributed torque per unit

length)

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v. Work of virtual load: M or F =L0 T

dz

(b) Linear elastic (cylindrical shaft)

i. Hookes Law= /G

ii. TorqueT =GJ , J=A r

2 dA= (c4 b4)/2 (c= outer radius, b= inner radius)

iii. Stress=T r/J

iv. Total rotation =L0 (T/GJ) dz

v. Strain energy: U = 12

A G

2 dA (= 12

GJ2 if homogeneous, U = GJ()2/2L ifuniform, GJ/L= torsional spring constant)

vi. Complementary energy: U =

A(2/2G) dAdz (= T2/2GJ if homogeneous, U =

T2L/2GJ if uniform)

vii. Work of virtual load: M=L0 (TT/GJ) dz(=

TTL/GJif uniform)

(c) Elastic-perfectly plastic (cylindrical shaft)

i. Yield torque: Typ=ypJ/c

ii. Ultimate torque: Tu=ypA r dr= (2/3)(c

3 b3)

iii. Solid shaft: T =Tu[1 (1/4)(rp/c)3],Tu= (4/3)Typ

iv. In unloading strain change is linear, leading to residual stressres =init Tinitr/J,and residual twist res= (d/dz)init Tinit/GJ.

v. At elastic-plastic interface: =yp=rp

(d) Thin-walled closed tubes

i. Shear Flow q= t = constantii. EquilibriumT =

rq ds= 2q A(where A = area enclosed by mean curve of tube)

iii. Complementary elastic energy U = 12

(2/G)tdsdz= 1

2q2

(1/Gt)ds

iv. Twist = 12

[T /(2A)2]

(1/Gt)ds

v. Torsional stiffness: ifG = constant,J=T/G = (2A)2/

(1/t)ds

5. Bending

(a) General (loading in xy-plane)

i. Geometry: assume section symmetric about y-axis, origin ofyzaxes at the centroidof the section

ii. Kinematics

A. Strain-curvature relation: x =(y y0) (where y0 is the y-coordinate of theneutral axis)

B. Curvature-displacement relation: v,

C. Rotation-displacement relation: v

iii. Equilibrium (differential equations) dV

dx = q,

dM

dx = V (where V = Vy, M = Mz,

q= transverse load per unit length)

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iv. Work of virtual force: F =L

0 Mdx (M(x) = bending moment due to virtual

load F, which may be either a force or a moment, conjugate to the displacement

[rotation] )(b) Elastic pure bending about z-axis (V = 0, P = 0, My = 0)

i. Hookes Law=E

ii. Moment M=Mz =EI, (transformed sections) M=ErefI

iii. Second moment of area (moment of inertia): I=Iz =A y

2 dA(y measured fromcentroid); rectangle I = bh3/12, circle I = c4/4; for sections composed of simplesubsections use parallel-axis theoremI=Io+ Ad

2 (where Io is calculated about thecentroid of the subsection and d is the y-distance from there to the centroid of thewhole section)

iv. Neutral axis (

Fx = 0): y0 =

A Ey dA/A E dA (= 0 if homogeneous, i.e. E =constant)

v. Stress = E(y y0) (= My/Iif homogeneous)

vi. Strain energy: U = 12

A E

2 dA(= 12

EI 2 if homogeneous)

vii. Complementary energy: U =A(

2/2E) dA (=M2/2EIif homogeneous)

viii. Work of virtual load: F or M=L0 (

MM/EI) dx

(c) Elastic bending with shear

i. Shear flow: q=V Q/I,Q=Ay dA= yA

ii. Average shear stress: q/t; rectangular: (y) = (V /2I)[(h/2)2 y2], max =

3V /2A; I-beam: max V /Aweb(d) Elastic bending with axial loads

i. TensileP, use superposition for stresses (= P /A My/I)

ii. CompressiveP, use superposition for stresses; check for buckling

(e) Elastic-perfectly plastic pure bending about z-axis

i. Initial yield: Myp=ypI/ymaxii. Rectangular section: Myp = ypbh

2/6, M = Mu[1 (1/3)(yp/(h/2))2], Mu =

(3/2)Mypiii. Neutral axis: T = C, (ultimate state)AT

Typ dA =AC

Cyp dA; AC = AT =

1

2A if

Typ= Cyp= ypiv. Ultimate moment: T lT+ClC=Mu; if

Typ =

Cyp =yp, Mu =ypAd/2, where d is

the distance between the centroids of the half-areas

v. Unloading: strain change is linear, resulting in residual stressres=init+ Minity/Iand residual curvature res=init Minit/EI.

vi. At elastic-plastic interface: = yp= yp

(f) Singularity functions

i. Ramp function: =

x, x >00, x 0

ii. General (n >0): n =

(x a)n

, x > a0, x a

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iii. Special: HeavisideH(xa) = (d/dx), Dirac delta(xa) = (d/dx)H(xa),doublet (x a) = (d/dx)(x a).

iv. Integration rules:

(x a) dx = (x a) +C,

(x a) dx = H(x a) +C,H(x a) dx= + C,

n dx= n+1/(n+ 1) +C

(g) Deflections due to elastic bending

i. Differential equation: (EI v) =qor EI v =qfor constant E I

ii. Boundary conditions

A. Fixed (built-in): = 0, v= 0,

B. Pin or roller: M=EIv = 0, v= 0,

C. Free: M=EIv = 0, V =EIv = 0;

D. Intermediate roller support: v= 0 (v and v are continuous),

E. Intermediate hinge: M=EIv = 0 (v is continuous)

iii. Particular results: |v|max=

F L3/3EI endloaded cantileverF L3/48EI centerloaded simple beam

5wL4/384EI uniformly loaded simple beam

6. Elastic stability and buckling

(a) General

i. Potential energy: =U+ , where = P ( = displacement conjugate to P)

ii. Equilibrium (single-degree-of-freedom system): d

d = 0

iii. Equilibrium is stable when d2

d2 >0 and unstable when

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B. Clamped-free (cantilever): Pcr=PE/4, Le= 2L

C. Pinned-clamped: Pcr= 2.05PE, Le= 0.699L

D. Clamped-clamped (no sidesway): Pcr= 4PE,Le= 0.5LE. Clamped-clamped (with sidesway): Pcr=PE,Le= L

iv. Critical stresscr= Pcr/A= 2E

(Le/r)2, where r=

I/A

v. Elastic limit: cr=yp, so Le/r= (Le/r)cr=

E/yp

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summary of concepts and formulae

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