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Strength of Materials II
DETERMINATE BEAMS
Equations of pure bending:
M: Bending Moment [N*m]: normal stress [N/m2]E: Modulus of elasticity [N/m2]: adius of !ur"ature [m]y: #istance from neutral surface [m]$: Moment of inertia [m%]
Equation of elastic cur"e:
Boundary !onditions:
(c) !antile"er beam
I
y2
x2 = M
= ER
= y
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INDETERMINATE BEAMS
Macaulays Method (Singularity functions)
If positive then the brackets (< ! can
be rep"aced by parentheses# $ther%ise
the brackets %i"" be e&'a" to ER$#
& ' ( a ) n
d ' * +n , +
& ' ( a ) n , +
' ) a
- & ' & a& ' ( a ) n
* )
' ) a& ' ( a ) n
* . ' ( a /n
0
1 o
ab =
0
1 o
ab
1 o
*tep%ise processes in order to deter+ine the reactions, the s"ope and the def"ection#
Superposition Method
-e so"ve the prob"e+ separate"y
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STRAIN ENERGY
Strain Energy of a e!er. is the increase in ener/y associated %ith the
defor+ation of the +e+ber#
d '3 t r a i n E n e r g y * 4 *
Strain Energy "ensity of a aterial. *train Ener/y per 'nit 0o"'+e#
3 t r a i n E n e r g y # e n s i t y * u *d U
d 0 *
)
1 56d ' *
+2
2 7
E#ASTI$ STRAIN ENERGY %&R N&RMA# STRESS
*train ener/y 'nder 3xia" 4oadin/
4 *5 4
2
2 3 E
2 *train Ener/y in 6endin/
M 2
d '4 *2 E I
E#ASTI$ STRAIN ENERGY %&R S'EARING STRESS
7 8 9 d : 8 9u *7 8 9 . s h e a r i n / s t r e s s
: 78 9 8 9. s h e a r i n / s t r a i n c o r r e s p o n d i n / t o
u *d U
d 0U =
)
: 8 9
7 8 92
d 0
)
: 8 9
2 8U =
T 42
2 ; . E f f e c t i v e " e n / t h f a c t o r
5 =c r?
2E I
( > 4 !2
c r =?
2E
( > 4 r !@2
> 4 r@ . E f f e c t i v e A s " e n d e r n e s s r a t i o
G depends on tHe support:
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INE#ASTI$ EN"ING
Maximum Elastic Mm!"t
Issume tHat tHe applied moment M=MYis sufficient to produce yielding strains in tHe
top and bottom fibers of tHe beam as sHo1n in figure J+b 3ince tHe straindistribution is linearK 1e can determine tHe stress distribution by using tHe stressstrain diagram .figure J+c LHen tHese stresses are plotted at tHe measured pointsyH/2K yy+K yy2tHe stress distribution in figure J+d(e results
igure J+
Le must cHec; to see if equation .EqJ+ is satisfied $n otHer 1ords 1e must cHec;
to see if tHe resultant force is equal to EO
* -d II EqJ+
Po do so 1e 1ill first calculate tHe resultant force for eacH of tHe t1o portions of tHestress distribution sHo1n in figure J+e 8eometrically tHis is equi"alent to finding tHeVOLUMESunder tHe t1o triangular bloc;s
Le Ha"e:
3ince T#CK equation EqJ+ is satisfied and tHe neutral a'is passes tHrougH tHecentroid of tHe cross sectional area
PHe Maximum Elastic Mm!"tcan be calculated by multiplying tHe forces T$CbytHeir corresponding distances from tHe neutral a'is PHus:
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PHe same result can of course be obtained in a more direct manner by using tHe
fle'ure formulaK tHat isK . =Y YM c / I. I:Moment of inertiac: PHe perpendicular distancefrom tHe neutral to a pointfartHest a1ay 1Here tHe forceacts
PHe fle'ure formula tHenK gi"es: K or
Maximum %lastic Mm!"t
$f tHe internal moment M)MQK tHe material at tHe top and bottom of tHe beam 1illbegin to yieldK causing a redistribution of stress o"er tHe cross section until tHerequired internal moment M is de"eloped 4sing tHe stresses sHo1n in figure J2c tHestress distributions can result .figure J2d(e Rere tHe compression and tensionstress Sbloc;sT eacH consist of component rectangular and triangular bloc;s
PHeir &lum!sare:
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PHe Plastic Mm!"t can be calculated by multiplying tHe forces T$C by tHeircorresponding distances from tHe neutral a'is
Beams used in steel buildings are sometimes designed to resist a plastic momentLHen tHis is tHe caseK codes usually list a design property for a beam called tHe
s'a%! (act) PHes'a%! (act) *k
+ is design as tHe ratio of plastic moment .M"ersus tHe yielding moment .MQ:
k =M P
M Y