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EJERCICIO:ANALIZAR EL DIAGRAMA DE LA SIGUIENTE VIGA POR EL MTODO D

SOLUCIN:

1. Hallamos la !"!#$%& sa'!()#o *+( I , 1-

k1= I1/L1 k2= I2/L2

k1= 10/7 k2= 10/6

k1= 1,429 k2= 1,667

. Hallamos los /o(0/!()(s () los )+#os () R 2 S

Nudo R: QR= k1/(k1+k2) QR=0,462

RS= k2/(k2+k1) RS=0,538

Nudo S: RS= k2/(k2+k3) RS=0,5

ST= k3/(k3+k2) ST=0,5

. Hallamos (l Mom()o #( Em3oam!()o P(4(/o 5MEP6

MQ = (qL2)/12

MR1 = !(qL2)/12

MR2 = (("#($2))/(L2))+

MS1 = !(("(#2)$)/(L2))!(

7 3 2 1

%1 %2

Q R S

6T&/' 4T& 5T&

6T&/'

7MQ MR1

4T& 5T&

63 2 1

MR2 MS1

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MS2 = (qL2)/12

MT = !(qL2)/12

7. Hallamos los #!a"amas

= !24,5+3,694 = !20,806 'u*-#do

= !6,472+24 = 17,528

0462 0538 05

245 !245 3694 !6472

4806 916211194 5597

!5784 !11563

1336 26713111 1556

!0389 !0778

009 018 0209 0105

!0027 !0053

0006 00120015 0008

!0002 !0004

30738 0001 0001

!11605

!12024 12024

& '

6MS2 MT

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CROSS Y SACAR LOS MOMENTOS

k3= I3/L3

k3= 10/6

k3= 1,667

1

1

MQ = 245

MR1 = !24,5

("#($2))/(L2)) MR2 = 3,694

("(#2)$)/(L2)) MS1 = !6,472

6

%3

T

8T&/'

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MS2 = 24

MT = !24

20,8060,462 = 9,162

o. (!) = 20806

20,8060,538 = 11,194

05

24 !24

!23125

!11563 !5782

!0778 !0389

!0053 !0027

!0004 !0002

11602 !302

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