Overview of Loads ON and IN Structures / Machines.
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Transcript of Overview of Loads ON and IN Structures / Machines.
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