MAE 343 - Intermediate Mechanics of Materials

Tuesday, Sep. 21, 2004

Textbook Section 4.4

Strain Energy

Castigliano’s Theorem

Overview of Loads ON and IN Structures / Machines

B o d y fo rces & cou p les

C o nce n tra ted

D is tribu te d /P re ssu res

S u rface fo rce s& cou p les

Applied Loads-Forces &M om ents

F lo w L in e s fromA p p lie d to R e ac tion

F o rces

F B D o f E a chC o m p o ne n t fromF o rces th ro u gh

C o nta c t S u rfa ces

F ro m FB D o fe n tire s truc tu re

E q u ilib riu m E q s.(3 in 2D , 6 in 3 D )

Reaction Forces& M om ents (at supports)

R e su lta n ts onC u tting S urfa ce

F ro m FB D o fP a rt o f S truc tu re

E q u ilib riu m E q s.(3 in 2D , 6 in 3 D )

S h e ar F orce &B e n d in g M o m e nt

D ia g ra m s

D e pe n d o n L oca tio n&O rie n ta tion o fC u ttin g P la ne

Internal Forces& M om ents

Structural Analysis

Overview of Various Stress Patterns

U n ifo rm D is trib u tio nso f S tre sses

N o rm al S tre ss-S tra igh t B a rsS ig m a = F /A

D ire ct S h ea r in L a p Jo in tsT a u = P /A

Uniaxial Tension orCom pression ofStraight Bars or

Lap Joints

N e u tra l A x is a tC e n tro id o f C ro ss -se c tion

T h e R e su ltan t o fB e n d in g S tresses

is the B e n d in g M o m e ntin th e C ro ss -se c tion

M a xim um a t topo r bo tto m , a nd ze ro

a t n eu tra l a x is

L in e a r D is trib u tiono ve r fro m Ne u tra l A x is

S ig m a = (M y)/I(N o rm a l A x ia l S tre sse s)

Pure Bendingof Long Beams

T ra n sve rse S h ea r S tre ssesR e su lt (A d d-u p ) inS h e ar F orce in the

C ro ss -S e c tion

D is trib u tio n fromN e u tra l A xis De p e nd s onS h a pe o f C ross -sec tion

T a u (y)= (V *M o m e n tA re a )/(I*Z (y))

Z e ro a t topa n d bo ttom e dg es

M a xim u m m a y no t bea t n eu tra l a x is

T a u _m ax= (co e f.)T a u_ a vera ge

B e n d in g o f A sym m e tric B ea m s:IF P lan e o f T ran sverse Fo rces

p a sse s th ro u gh the S HE A RC E N T E R , N O T o rs io n o ccu rs

Beams Subjectedto Transverse Forces

M o m e n t o f d iffe ren tia lto rs io n a l s tress ab o ut ce n tro id re su lts in the

in te rna l to rq u e in c ross -se c tion

C ircu la r C ro ss -se c tio n :L in e ar d is trib u tion , w ithM a x. a t the ou ter e dg e :

T a u = (T r)/J

P rism a tic S h a fts -T o rs ionL e ad s to W a rp ing

M e m b ra n e A n a lo g y:T a u _ m ax= T /Q , T h e ta =(T L )/K G

P o w e r-T o rq u e -R P M R e la tio n s:h p = (Tn )/63 ,0 25kw = (T n )/9 5 49

Torsion of Roundor Prism atic Shafts

Stresses- from Distribution ofInternal Forces over G iven

Cross-Section

Problem No.2 in Exam No.1

yx

x

z

V

z

x

y

z

My x

xz

xz (z)

V

z

y

x

x

x(z)

z

Hy

x

z Mz

x

x(x)

x

H

x

y

zxy(y)

Superposition of stresses from applied loads “V” and “H”

y

A

xy dAzM )(

A

xzdAV

A

xz dAyM )(A

xydAH

z

x

z

x

z

x

y

Elastic Deformation for Different Types of Loadings (Stress Patterns)

• Straight uniform elastic bar loaded by centered axial force (Figs. 2.1, 2.2, 2.3)– Similar to linear spring in elastic range

– “Force-induced elastic deformation”, f must not exceed “design allowable”: failure is predicted to occur if (FIPTOI): f-max>f-allow

– “Spring Constant (Rate)” for elastic bar

– Normalize “Force-deflection” curve to obtain “Engineering Stress-Strain Diagram”

0, llykyF ff

fFyFinlbk //)/(

)'(:,

:,,

0

0

0

0

00

LawsHookeEwherel

EA

l

AFk

thatsolA

F

fff

ax

ff

Elastic Deformation for Different Types of Loadings (Stress Patterns)

• Torsional moment produces torsional shearing strain, according to Hooke’s Law: = G – Shear strain, , is change in initially right angle (radians)– Angular deflection (twist angle) for elastic members:

• Beam bending loads cause transverse deflections:– Deflection (elastic) curve obtained by integrating twice the governing

differential equation and using the boundary conditions:

– See Table 4.1 for deflection curves of several common cases

memberslcylindricaforJKwhereKG

TL :,

diagrammomentbendingfromMwhereEI

xM

dx

xyd

zz

z "",)()(

2

2

Stored Strain Energy (Potential Energy of Strain)

• From Work done by external forces or moments over corresponding displacements– Recovered by gradual unloading if elastic limit of the material

is not exceeded– Displacements (deformations) are LINEAR functions of

external loads if Hooke’s Law applies – Generalized forces include moments, and generalized

displacements include rotations (angular displacements)

• Strain energy per unit volume for differential cubic element (Fig. 4.11)

222zzyyxx

dxdydz

Uu

Total Strain Energy Formulas for Common Stress Patterns

• Members with uniform (constant) geometry material properties along the longitudinal axis– Tension and Direct Shear

– Torsion

– Pure bending

– Strain energy associated with transverse shearing stresses is complex function of cross-section and negligible in comparison to bending strain energy (except for short beams)

• Integrations are required if geometry or material properties vary along the member (Table 4.6, where “Q” denotes generalized displacement)

surfacecontactthetoreferLandAwhereAG

LP

AG

PLP

k

PPPU

areationalcrossisAwhereAE

LF

AE

FLF

k

FFFU

sheardirsavesheardir

axfavetens

,2

)(2

)(2

sec,2

)(2

)(2

2

2

KG

LT

KG

TLT

k

TTTU

torfavetor 2

)(2

)(2

2

EI

LM

EI

MLM

k

MMMU

bendyfavebend 2

)(2

)(2

2

Castigliano’s Theorem• Energy method for calculating displacements in a

deformed elastic body (Deflection equations - Table 4.6)– At ANY point where an external force is applied, the displacement

in the direction of that force is given by the partial derivative of the total strain energy with respect to that force.

– Example of simple tension in uniform prismatic bar:

– If no real force is applied at the point of interest, a “DUMMY” force is “applied” at that point, and then set equal to zero in the expression of the corresponding derivative of the total strain energy.

• Applicable also to calculating reactions at the supports of statically redundant (undetermined) structures.– Set partial derivative equal to zero since there is no displacement

fEA

FL

EA

LF

dF

d

dF

dU )2

(2

Summary of Example Problems

• Example 4.6 – Total strain energy in beam– Simply supported beam loaded by concentrated load, “P” at mid-

span and moment “M” at left support– Superposition of cases 1 and 4 in Table 4.1

• Example 4.7 – Beam deflections and slopes– Determine reactions and expressions of the bending moment on

both sides of mid-span (from equilibrium)– Use deflection equations in Table 4.6 for mid-span deflection and

angular displacement (slope of deflection curve) at left support– Apply dummy moment at mid-span to calculate angular deflection

(slope of deflection curve) at mid-span

Given: L1=10 in, L2=5 in, P=1000 lb, s=1.25 in, d=1.25 in

Find: Use Castigliano’s theorem to find deflection y0 under load “P”

Summary of Textbook Problems – Problem 4.24, Castigliano’s Theorem

• Select coordinate system and identify the elements of the total strain energy– Bending of square leg of support bracket– Torsion of square leg of support bracket– Bending of the round leg of support bracket

• Differentiate the expression of “U” with respect to load “P”, to find deflection y0

– Stiffness properties of steel: E=30x106psi, G=11.5x106psi– Use Case 3 in Table 4.4 to find the geometric rigidity

parameter, K, of the square leg, K1= 0.34– Substitute all numerical values to obtain: y0 = 0.13 inches