amem 314 theory

8/10/2019 AMEM 314 Theory

1/8

Debre Tabor University

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


2/8


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


3/8


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:


5/8


6/8


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:


7/8


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:


8/8


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

amem 314 theory

Documents