Resistencia de Materiales
Presentado por:
Javier Fuentes Fuentes
Juan Carlos Heredia Rojas
Presentado a:
Alfonso RodrΓguez PeΓ±a
Universidad del AtlΓ‘ntico
IngenierΓa MecΓ‘nica
11 Junio 2015
EJERCICIO (A)
Diagrama de cuerpo libre:
+βΊ ππ΅ = 0: β π΄π 4 + 1000 2 = 0
βΉ π΄π = 500
+βΊ ππ· = βπ΄π π + π1 = 0
βΉ π1 = 500 π
πΈπΌπ2π¦1
ππ₯2 = 500 π
πΈπΌππ¦1
ππ₯= πΈπΌπ1 = 250 π2 + πΆ1 (1)
πΈπΌπ¦1 = 83,333 π3 + πΆ1 π + πΆ2 (2)
2m
A B
1000 N
2m
2m
A B
1000 N
AX
AY BY 2m
A
AY
M1
V1
D
X
+βΊ ππΈ = βπ΄π π + 1000 π β 2 + π2 = 0
βΉ π2 = π΄π π β 1000 π + 2000
π2 = β500 π + 2000
πΈπΌπ2π¦2
ππ₯2= β500 π + 2000
πΈπΌππ¦2
ππ₯= πΈπΌπ2 = β250 π2 + 2000 π + πΆ3 (3)
πΈπΌπ¦2 = β83,333 π3 + 1000 π2 + πΆ3 π + πΆ4 (4)
X-2
A
1000 N
AY
M2
V2
E
X
(X=0, Y1=0)
(X=2, Y1= Y2) (X=2, Ο΄1= Ο΄2)
(X=4, Y2=0)
Y
X
2m 2m
A B
1000 N
π = 0, π1 = 0 ππ πππ’πππΓ³π (2)
πΈπΌ π1 = 83,333 π3 + πΆ1 π + πΆ2
πΆ2 = 0 (5)
π = 2, π1 = π2 π π πππ’ππππ πππ’πππππππ 1 π¦ 3 :
250 22 + πΆ1 = β250 22 + 2000 2 + πΆ3
πΆ1 = πΆ3 + 2000 (6)
π = 2, π1 = π2 π π πππ’ππππ πππ’πππππππ 2 π¦ 4 :
83,333 23 + πΆ1 2 + πΆ2 = β83,333 23 + 1000 22 + πΆ3 2 + πΆ4
666,664 + 2πΆ1 + πΆ2 = 3333,336 + 2πΆ3 + πΆ4 (7)
π = 4, π2 = 0 ππ πππ’πππΓ³π (4)
πΈπΌ 0 = β83,333 43 + 1000 42 + πΆ3 4 + πΆ4
πΆ4 = β10666,688 β 4πΆ3 (8)
ππππππππ§ππππ πππ ππ. 5 , 6 π¦ 8 ππ ππ. 7 :
666,664 + 2 πΆ3 + 2000 + 0 = 3333,336 + 2πΆ3 + (β10666,688 β 4πΆ3)
πΆ3 = β12000,016 (9)
ππππππππ§ππππ ππ ππ. 9 ππ 6 :
πΆ1 = (β12000,016) + 2000
πΆ1 = β10000,016 (10)
ππππππππ§ππππ ππ ππ. 9 ππ 8 :
πΆ4 = β10666,688 β 4(β12000,016)
πΆ4 = 37333,376 (11)
πΈππ’πππΓ³π ππ ππ πΆπ’ππ£π πΈπΓ‘π π‘πππ
ππππ 0 β€ π β€ 2
πΈπΌπ1 = 250 π2 β 10000,016
πΈπΌπ¦1 = 83,333 π3 β 10000,016 π
ππππ 2 β€ π β€ 4
πΈπΌπ2 = β250 π2 + 2000 π β 12000,016
πΈπΌπ¦2 = β83,333 π3 + 1000 π2 β 12000,016 π + 37333,376
MΓ©todo de singularidad
Haciendo momento en B tenemos que:
ππ΅ = 0 β π΄ππ + π π β 2 1 + π = 0
π(π₯) = π΄ππ + π π β 2 1
πΈπΌπ =π΄ππ
2
2β
π π β 2 2
2+ πΆ1 (1)
πΈπΌπ¦ =π΄ππ
3
6β
π π β 2 3
6+ πΆ1π + πΆ2 (2)
Evaluando en la ecuaciΓ³n (2) X=0; y=0
0 = 0 + 0 + πΆ2
πΆ2 = 0
Evaluando la ecuaciΓ³n 2 cuando X=4; y=0
Nos queda que:
0 =500 4 3
6β
1000 2 3
6+ 4πΆ1
πΆ1 = β1000
πΈπΌπ =π2
2π΄π β
π
2 π β 2 2 + 1000
La ecuaciΓ³n de la curva elΓ‘stica por el mΓ©todo de singularidad es:
πΈπΌπ¦ = 83,333π3 β 166,666 π β 2 3 + 1000π
EJERCICIO (B)
πππ π ππππ‘πΓπ: π΄π¦ = π΅π¦ = 3 ππ
+βΊ ππΆ = βπ΄π π₯ + π1 = 0
βΉ π1 = 3000 π₯
πΈπΌπ2π¦1
ππ₯2 = 3000 π₯
πΈπΌππ¦1
ππ₯= πΈπΌπ1 = 1500 π₯2 + πΆ1 (1)
πΈπΌπ¦1 = 500 π₯3 + πΆ1 π₯ + πΆ2 (2)
2m 1m 1m
A B
30 kN/m
2 m 1 m 1 m
A B
3 kN/m
6 kN
AX
AY BY
2 m
A
AY
M1
V1
C
X
+βΊ ππ· = βπ΄π π₯ +
3000 π₯ β 1 (π₯β1
2) + π2 = 0
βΉ π2 = 1500(π₯2) + 1500
πΈπΌπ2π¦2
ππ₯2 = 1500 π₯2 + 1500
πΈπΌππ¦2
ππ₯= πΈπΌπ2 = 500 π₯3 +
1500 π₯ + πΆ3 (3)
πΈπΌπ¦2 = 125 π₯4 + 750 π₯2 +
πΆ3 π₯ + πΆ4 (4)
+βΊ ππΈ = βπ΄π π₯ + 6000 π₯ β 2 + π3 = 0
βΉ π3 = 3000 π₯ β 6000 π₯ + 12000
βΉ π3 = β3000 π₯ + 12000
πΈπΌπ2π¦3
ππ₯2= β3000 π₯ + 12000
πΈπΌππ¦3
ππ₯= πΈπΌπ3 = β1500 π₯2 + 12000(π₯) + πΆ5 (5)
πΈπΌπ¦3 = β500 π₯3 + 6000(π₯2) + πΆ5(π₯) + πΆ6 (6)
M2
A
AY
V2
D
X
X-1
(X-1)/2
3000(X-1)
X
X-2
A E
AY (X-3)
M3
V3
1 m
3 kN/m
6 kN
π = 0, π1 = 0 ππ πππ’πππΓ³π (2)
πΈπΌ(0) = 500 0 + πΆ1 0 + πΆ2
πΆ2 = 0 (7)
π = 1, π1 = π2 π π πππ’ππππ πππ’πππππππ 1 π¦ 3 :
1500 22 + πΆ1 = 500 23 + 1500 2 + πΆ3
πΆ1 = πΆ3 + 1000 (8)
π = 1, π1 = π2 π π πππ’ππππ πππ’πππππππ 2 π¦ 4 :
500 23 + πΆ1 2 + πΆ2 = 125 24 + 750 22 + πΆ3 2 + πΆ4
4000 + 2πΆ1 = 5000 + 2πΆ3 + πΆ4 (9)
π = 3, π2 = π3 π π πππ’ππππ πππ’πππππππ 3 π¦ 5 :
500 33 + 1500 3 + πΆ3 = β1500 32 + 12000(3) + πΆ5
18000 + πΆ3 = 22500 + πΆ5 (10)
2m 1m 1m
A B
30 kN/m
(X=0, Y1=0) (X=1, Y1= Y2) (X=1, Ο΄1= Ο΄2)
(X=3, Y2= Y3) (X=3, Ο΄2= Ο΄3)
(X=4, Y3=0)
Y
X
π = 3, π2 = π3 π π πππ’ππππ πππ’πππππππ 4 π¦ 6 :
125 34 + 750 32 + πΆ3 3 + πΆ4 = β500 33 + 6000(32) + πΆ5(3) + πΆ6
16875 + 3πΆ3 + πΆ4 = 40500 + 3πΆ5 + πΆ6 (11)
π = 4, π3 = 0 ππ πππ’πππΓ³π 6 :
πΈπΌ(0) = β500 43 + 6000(42) + πΆ5(4) + πΆ6
0 = 64000 + 4πΆ5 + πΆ6 (12)
ππ ππ ππ. 10 : πΆ3 = πΆ5 + 4500
ππππππππ§ππππ πππ ππ. 10 π¦ 8 ππ ππ. 9 :
4000 + 2(πΆ3 + 1000) = 5000 + 2πΆ3 + πΆ4
4000 + 2πΆ3 + 2000 = 5000 + 2πΆ3 + πΆ4
4000 + 2(πΆ5 + 4500) + 2000 = 5000 + 2(πΆ5 + 4500) + πΆ4
4000 + 2πΆ5 + 9000 + 2000 = 5000 + 2πΆ5 + 9000 + πΆ4
πΆ4 = 1000 (13)
ππ ππ. 12 : πΆ6 = β64000 β 4πΆ5
ππππππππ§ππππ 10 , 12 π¦ (13) ππ 11 :
16875 + 3 πΆ5 + 4500 + (1000) = 40500 + 3πΆ5 + (β64000 β 4πΆ5)
πΆ5 = β13718,75 (14)
ππππππππ§ππππ 14 ππ 12 :
πΆ6 = β64000 β 4(β13718,75)
πΆ6 = β9125 (15)
ππππππππ§ππππ 14 ππ 10 :
πΆ3 = (β13718,75) + 4500
πΆ3 = β9218,75 (16)
ππππππππ§ππππ 16 ππ 8 :
πΆ1 = (β9218,75) + 1000
πΆ1 = β8218,75 (17)
πΈππ’πππΓ³π ππ ππ πΆπ’ππ£π πΈπΓ‘π π‘πππ
ππππ 0 β€ π β€ 1
πΈπΌπ1 = 1500 π₯2 β 8218,75
πΈπΌπ¦1 = 500 π₯3 β 8218,75 π₯
ππππ 1 β€ π β€ 3
πΈπΌπ2 = 500 π₯3 + 1500 π₯ β 9218,75
πΈπΌπ¦2 = 125 π₯4 + 750 π₯2 β 9218,75 π₯ + 1000
ππππ 3 β€ π β€ 4
πΈπΌπ3 = β1500 π₯2 + 12000(π₯) β 13718,75
πΈπΌπ¦3 = β500 π₯3 + 6000(π₯2) β 13718,75(π₯) β 9125
EJERCICIO (C)
Hallar EcuaciΓ³n de la curva elΓ‘stica
Diagrama de Cuerpo libre:
+βΊ ππ΅ = 0: β π΄π 4 + 1000 2 + 30 β 103 0,5 = 0
βΉ π΄π = 4250
+βΊ ππΈ = βπ΄π π + π1 = 0
βΉ π1 = 4250 π
πΈπΌπ2π¦1
ππ₯2 = 4250 π
πΈπΌππ¦1
ππ₯= πΈπΌπ1 = 2125 π2 + πΆ1 (1)
πΈπΌπ¦1 = 708,333 π3 + πΆ1 π + πΆ2 (2)
2m 1m 1m
A B
1000 N 30 kN/m
2m 1m 1m
A B
1000 N
30 kN
AX
AY BY
0,5m
A
AY
M1
V1
E
X
+βΊ ππΉ = βπ΄π π + 1000 π β 2 + π2 = 0
βΉ π2 = π΄π π β 1000 π + 2000
π2 = 3250 π + 2000
πΈπΌπ2π¦2
ππ₯2= 3250 π + 2000
πΈπΌππ¦2
ππ₯= πΈπΌπ2 = 1625 π2 + 2000 π + πΆ3 (3)
πΈπΌπ¦2 = 541,667 π3 + 1000 π2 + πΆ3 π + πΆ4 (4)
+βΊ ππΊ = βπ΄π π + 1000 π β 2 + 30 β 103 π β 3 (π β 3)
2+ π3 = 0
X-2
A
1000 N
AY
M2
V2
F
X
X
X-2
A G
1000 N
30(X-3)
AY
(X-3)/2
(X-3)
M3
V3
βΉ π3 = 4250 π β 1000 π β 2 β 30 β 103
2 π β 3 2
βΉ π3 = 4250 π β 1000 π + 2000 β 15 β 103 π 2 + 90 β 103 π + (135
β 103)
βΉ π3 = β 15 β 103 π 2 + (93,250 β 103) π + (137 β 103)
πΈπΌπ2π¦3
ππ₯2= β 15 β 103 π 2 + (93,250 β 103) π + (137 β 103)
πΈπΌππ¦3
ππ₯= πΈπΌπ3 = β 5 β 103 π 3 + 46,625 β 103 π 2 + 137 β 103 π + πΆ5 (5)
πΈπΌπ¦3 = β 1,25 β 103 π 4 + 15,542 β 103 π 3 + 68,5 β 103 π 2 + πΆ5 π + πΆ6 (6)
π = 0, π1 = 0 ππ πππ’πππΓ³π (2)
πΈπΌ π1 = 708,333 π3 + πΆ1 π + πΆ2
πΆ2 = 0 (7)
π = 4, π3 = 0 ππ πππ’πππΓ³π (6)
πΈπΌ π3 = β 1,25 β 103 π4 + 15,542 β 103 π3 + 68,5 β 103 π2 + πΆ5 π + πΆ6
0 = 1770688 + 4πΆ5 + πΆ6 (8)
2m 1m 1m
A B
1000 N 30 kN/m
(X=0, Y1=0)
(X=2, Y1= Y2) (X=2, Ο΄1= Ο΄2)
(X=3, Y2= Y3) (X=3, Ο΄2= Ο΄3)
(X=4, Y3=0)
Y
X
π = 2, π1 = π2 π π πππ’ππππ πππ’πππππππ 1 π¦ 3 :
2125 22 + πΆ1 = 1625 22 + 2000 2 + πΆ3
πΆ1 = πΆ3 + 2000 (9)
π = 2, π1 = π2 π π πππ’ππππ πππ’πππππππ 2 π¦ 4 :
708,333 23 + πΆ1 2 + πΆ2 = 541,667 23 + 1000 π2 + πΆ3 π + πΆ4
πΆ4 = 2πΆ1 β 2666,672 β 2πΆ3 (10)
π = 3, π2 = π3 π π πππ’ππππ πππ’πππππππ 3 π¦ 5 :
1625 32 + 2000 3 + πΆ3 = β5000 33 + 46625 32 + 137000 3 + πΆ5
πΆ3 = 675000 + πΆ5 (11)
π = 3, π2 = π3 π π πππ’ππππ πππ’πππππππ 4 π¦ 6 :
541,667 33 + 1000 32 + πΆ3 3 + πΆ4
= β1250 3 4 + 15542 3 3 + 68500 3 2 + πΆ5 3 + πΆ6
23625,009 + 3πΆ3 + πΆ4 = 934884 + 3πΆ5 + πΆ6 (12)
ππππππππ§ππππ ππ ππ. 9 ππ 10 :
πΆ4 = 2 πΆ3 + 2000 β 2666,672 β 2πΆ3
πΆ4 = 1333,328 (13)
ππ ππ ππ. 8 π‘ππππππ : πΆ6 = β4πΆ5 β 1770688
ππππππππ§ππππ πππ ππ. 13 , 11 π¦ 8 ππ 12 :
23625,009 + 3 675000 + πΆ5 + 1333,328 = 934884 + 3πΆ5 + (β4πΆ5 β 1770688)
πΆ5 = β721440,584 (14)
ππππππππ§ππππ ππ ππ. 14 ππ (11)
πΆ3 = 675000 + (β721440,584)
πΆ3 = β46440,584 (15)
ππππππππ§ππππ ππ ππ. 14 ππ ππ. 8 :
0 = 1770688 + 4πΆ5 + πΆ6
πΆ6 = 1115074,336 (16)
ππππππππ§ππππ ππ ππ. 15 ππ ππ ππ. 9 :
πΆ1 = β46440,584 + 2000
πΆ1 = β44440,584 (17)
πΈππ’πππΓ³π ππ ππ ππ’ππ£π ππΓ‘π π‘πππ:
ππππ 0 β€ π β€ 2
πΈπΌ π1 = 2125 π2 β 44440,584
πΈπΌ π1 = 708,333 π3 β 44440,584(π)
ππππ 2 β€ π β€ 3
πΈπΌ π2 = 1625 π2 + 2000(π) β 46440,584
πΈπΌ π2 = 541,667 π3 + 1000 π2 β 46440,584 π + 1333,328
ππππ 3 β€ π β€ 4
πΈπΌ π3 = β5000 π3 + 46625 π2 + 137000(π) β 721440,584
πΈπΌ π3 = β1250 π4 + 15542(π3) + 68500 π2 β 721440,584 π + 1115074,337
MΓ©todo de singularidad:
Haciendo momento en B encontramos la funciΓ³n de momento, la cual la integraremos
dos veces
ππ΅ = 0 β π΄π¦ 4 + 1000 2 + 3000 0.5 = 0
π΄π = 4250
π π₯ = π΄ππ β π π β 2 1 βπ0
2 π β 0 2 +
π0
2 π β 1 2
πΈπΌπ =π΄ππ2
2β
π
2 π β 2 2 β
π0
6 π β 0 3 +
π0
6 π β 1 3 + πΆ1 (1)
πΈπΌπ¦ =π΄ππ3
6β
π
6 π β 2 3 β
π0
24 π β 0 4 +
π0
24 π β 1 4 + πΆ1π + πΆ2 (2)
Evaluamos la ecuaciΓ³n (2) cuando X=0; y=0
πΆ2 = 0
Y evaluamos la ecuaciΓ³n (1) cuando X=4; y=0
0 =32
3π΄π β 4π β π0
8
3 +
9
8π0 + πΆ1(4)
πΆ1 = 1229,17
La ecuaciΓ³n de la curva elΓ‘stica por el mΓ©todo de singularidad es:
πΈπΌπ¦ = 708,333π3 β 166,666 π β 2 3 β 125 π β 0 4 + 125 π β 1 4 + 1129,17π
πΈπ½πΈπ πΆπΌπΆπΌπ (π·)
+β πΉπ¦ = 0: π΄π¦ β 1000 + π΅π¦ = 0 (A)
+βΊ ππ΄ = 0: β 1000 2 + π΅π¦ 4 + ππ΅ = 0
βΉ π΅π¦ = 500 β 1
4 ππ΅ (B)
2m
A B
1000 N
2m
2m
A B
1000 N
AX
AY BY 2m
BX
MB
+βΊ ππΆ = 0: π΅π¦ π₯ + ππ΅ β π1 = 0
βΉ π1 = π΅π¦ π₯ + ππ΅
πΈπΌπ2π¦1
ππ₯2 = π΅π¦ π₯ + ππ΅
πΈπΌπ1 =ππ¦1
ππ₯=
1
2π΅π¦ π₯
2 + ππ΅ π₯ + πΆ1 (1)
πΈπΌπ¦1 =1
6π΅π¦ π₯
3 +1
2ππ΅(π₯2) + πΆ1 π₯ + πΆ2 (2)
+βΊ ππ· = 0: π΅π¦ π₯ + ππ΅ β π2 β
1000(π₯ β 2) = 0
βΉ π2 = π΅π¦ π₯ + ππ΅ β 1000 π₯ + 2000
πΈπΌπ2π¦2
ππ₯2= π΅π¦ π₯ + ππ΅ β 1000 π₯ + 2000
πΈπΌπ2 =ππ¦2
ππ₯=
1
2π΅π¦ π₯
2 + ππ΅ π₯ β 500 π₯2 +
2000(π₯) + πΆ3 (3)
πΈπΌπ¦2 =1
6π΅π¦ π₯
3 +1
2ππ΅ π₯
2 β 166,667 π₯3 + 1000 π₯2 + πΆ3 π₯ + πΆ4 (4)
C
By
MB
V1
X
M1
X-2
X
D
B
1000 N
BY
MB V2
M2
2m
A B
1000 N
2m
(X=2, Y1= Y2) (X=2, Ο΄1= Ο΄2)
(X=4, Y2=0)
Y
X
(X=0, Y1= 0) (X=0, Ο΄1= 0)
π₯ = 0, π¦1 = 0 ππ πππ’πππΓ³π 2 :
πΈπΌ(0) =1
6π΅π¦ 0 +
1
2ππ΅(0) + πΆ1 0 + πΆ2
πΆ2 = 0 (5)
π₯ = 0, π1 = 0 ππ πππ’πππΓ³π 1 :
πΈπΌ(0) =1
2π΅π¦ 0 + ππ΅ 0 + πΆ1
πΆ1 = 0 (6)
π₯ = 2, π¦1 = π¦2 π π πππ’ππππ πππ’πππππππ 2 π¦ 4 :
1
6π΅π¦ 23 +
1
2ππ΅(22) + πΆ1 2 + πΆ2
=1
6π΅π¦ 23 +
1
2ππ΅ 22 β 166,667 23 + 1000 22 + πΆ3 2 + πΆ4
β2666,664 β 2πΆ3 = πΆ4 (7)
π₯ = 2, π1 = π2 π π πππ’ππππ πππ’πππππππ 1 π¦ 3 :
1
2π΅π¦ 22 + ππ΅ 2 + πΆ1 =
1
2π΅π¦ 22 + ππ΅ 2 β 500 22 + 2000(2) + πΆ3
πΆ3 = β2000 (8)
ππππππππ§ππππ ππ πππ’πππΓ³π 8 ππ ππ πππ’πππΓ³π (7)
β2666,664 β 2(β2000) = πΆ4
πΆ4 = 1333,336 (9)
π₯ = 4, π¦2 = 0 ππ πππ’πππΓ³π (4)
πΈπΌ(0) =1
6π΅π¦ 43 +
1
2ππ΅ 42 β 166,667 43 + 1000 42 + πΆ3 4 + πΆ4
0 = π΅π¦ 10,667 + ππ΅ 8 + 5333,312 + (β2000) 4 + (1333,336)
ππ΅ = 166,669 β π΅π¦(1,333) (10)
ππππππππ§ππππ ππ ππ. 10 ππ πππ’πππΓ³π (B)
π΅π¦ = 500 β 1
4 166,669 β π΅π¦(1,333)
π΅π¦ = 687,157 (11)
βΉ ππ΅ = 166,669 β 687,157 (1,333)
ππ΅ = β749,311 βΉ ππ΅ = 749,311 β» (12)
ππππππππ§ππππ π΅π¦ ππ ππ. 11 ππ ππ. (π΄)
π΄π¦ β 1000 + π΅π¦ = 0
π΄π¦ = 312,843 (13)
πΈππ’πππΓ³π ππ ππ πΆπ’ππ£π πΈπΓ‘π π‘πππ:
ππππ 0 β€ π β€ 2
πΈπΌπ1 = 343,578 π₯2 β 748,628 π₯
πΈπΌπ¦1 = 114,526 π₯3 β 374,314 π₯2
ππππ 2 β€ π β€ 4
πΈπΌπ2 = β156,421 π₯2 + 1251,372 π₯ β 2000
πΈπΌπ¦2 = β52,140 π₯3 + 625,686 π₯2 β 2000 π₯ + 1333,336
MΓ©todo de singularidad
Haciendo momento en B encontramos la funcion de momento:
ππ΅ = 0 β π π₯ β π΄π+π π β 2 1 = 0
π π₯ = π΄πβπ π β 2 1
πΈπΌπ =π΄π¦π
2
2β
π π β 2
2
2
+ πΆ1 (1)
πΈπΌπ¦ =π΄π¦π
3
6β
π π β 2
6
3
+ πΆ1π + πΆ2 (2)
Evaluando la ecuacion (2) en π = 0; π¦ = 0
Tenemos que:
πΆ2 = 0
Y evaluando la ecuaciΓ³n (2) en π = 4 π = 0
0 =32π΄π¦
3β
4000
3+ 4πΆ1
πΆ1 = 333,333 β 2,667π΄π¦
Ahora evaluamos la ecuacion (1) en
π = 0 π = 0
0 = 8π΄π¦ β 2000 + 333,333 β 2,667π΄π¦
0 = 5,333π΄π¦ β 1666,667
π΄π¦ =1666,667
5,333
π΄ = 312,519
πΆ1 = β500
La ecuaciΓ³n de la curva elΓ‘stica queda de la siguiente forma:
πΈπΌπ¦ = 52,086π3 β 166,667 < π β 2 >3β 500π
πΈπ½πΈπ πΆπΌπΆπΌπ (πΈ)
+β πΉπ¦ = 0: π΄π¦ β 6000 + π΅π¦ = 0 (A)
+βΊ ππ΄ = 0: β 6000 2 + π΅π¦ 4 β ππ΅ = 0
βΉ π΅π¦ = 3000 + 1
4 ππ΅ (B)
+β πΉπ¦ = 0: βπ΄π₯ + π΅π₯ = 0
2m 1m 1m
30 kN/m
A B
2 m 1 m 1 m
3 kN/m
6 kN
2 m
A B AX
AY BY
BX
MB
+βΊ ππΆ = 0: π΅π¦ π₯ β ππ΅ β π1 = 0
βΉ π1 = π΅π¦ π₯ β ππ΅
πΈπΌπ2π¦1
ππ₯2 = π΅π¦ π₯ β ππ΅
πΈπΌπ1 =ππ¦1
ππ₯=
1
2π΅π¦ π₯
2 β ππ΅ π₯ + πΆ1 (1)
πΈπΌπ¦1 =1
6π΅π¦ π₯
3 β1
2ππ΅(π₯2) + πΆ1 π₯ + πΆ2 (2)
+βΊ ππ· = 0: π΅π¦ π₯ β ππ΅ β π2 β
3000 π₯ β 1 π₯β1
2 = 0
βΉ π2 = π΅π¦ π₯ β ππ΅ β 1500 π₯2 +
3000 π₯ β 1500
πΈπΌπ2π¦2
ππ₯2
= π΅π¦ π₯ β ππ΅ β 1500 π₯2
+ 3000 π₯ β 1500
πΈπΌπ2 =ππ¦2
ππ₯=
1
2π΅π¦ π₯
2 β ππ΅ π₯ β 500 π₯3 + 1500 π₯2 β 1500(π₯) + πΆ3 (3)
πΈπΌπ¦2 =1
6π΅π¦ π₯
3 β1
2ππ΅ π₯
2 β 125 π₯4 + 500 π₯3 β 750 π₯2 + πΆ3 π₯ + πΆ4 (4)
C
By
MB
V1
X
M1
MB
(X-1)/2
X
D B
BY
V2 M2
X-1
3000(X-1)
X
X-2
3 kN/m
6 kN
E V3
M3
BY
MB
+βΊ ππΈ = 0: β π3 β 6000 π₯ β 2 + π΅π¦ π₯ β ππ΅ = 0
βΉ π3 = β6000 π₯ + 12000 + π΅π¦ π₯ β ππ΅
πΈπΌπ2π¦3
ππ₯2= β6000 π₯ + 12000 + π΅π¦ π₯ β ππ΅
πΈπΌπ3 =ππ¦3
ππ₯= β3000 π₯2 + 12000 π₯ +
1
2π΅π¦ π₯
2 β ππ΅ π₯ + πΆ5 (5)
πΈπΌπ¦3 = β1000 π₯3 +1
6π΅π¦ π₯
3 + 6000 π₯2 β1
2ππ΅ π₯
2 + πΆ5(π₯) + πΆ6 (6)
π₯ = 0, π1 = 0 ππππππππ§ππππ ππ πππ’πππΓ³π 1 :
πΈπΌ 0 =1
2π΅π¦ 0 β ππ΅ 0 + πΆ1
πΆ1 = 0 (7)
π₯ = 0, π¦1 = 0 ππππππππ§ππππ ππ πππ’πππΓ³π 2 :
πΈπΌ 0 =1
6π΅π¦ 0 β
1
2ππ΅(0) + πΆ1 0 + πΆ2
πΆ2 = 0 (8)
(X=1, Y1= Y2) (X=1, Ο΄1= Ο΄2)
(X=4, Y3=0)
Y
X
(X=0, Y1= 0) (X=0, Ο΄1= 0)
2m 1m 1m
30 kN/m
A B
(X=3, Y2= Y3) (X=3, Ο΄2= Ο΄3)
π₯ = 1, π1 = π2 π π πππ’ππππ πππ’πππππππ 1 π¦ 3 :
1
2π΅π¦ 12 β ππ΅ 1 + πΆ1 =
1
2π΅π¦ 12 β ππ΅ 1 β 500 13 + 1500 12 β 1500(1) + πΆ3
πΆ3 = 500 (9)
π₯ = 1, π¦1 = π¦2 π π πππ’ππππ πππ’πππππππ 2 π¦ 4 :
1
6π΅π¦ 13 β
1
2ππ΅(12) + πΆ1 1 + πΆ2
=1
6π΅π¦ 13 β
1
2ππ΅ 12 β 125 14 + 500 13 β 750 12 + πΆ3 1 + πΆ4
πΆ4 = 125 (10)
π₯ = 3, π2 = π3 π π πππ’ππππ πππ’πππππππ 3 π¦ 5 :
1
2π΅π¦ π₯
2 β ππ΅ π₯ β 500 π₯3 + 1500 π₯2 β 1500 π₯ + πΆ3
= β3000 π₯2 + 12000 π₯ +1
2π΅π¦ π₯
2 β ππ΅ π₯ + πΆ5
πΆ5 = β13000 (11)
π₯ = 3, π¦2 = π¦3 π π πππ’ππππ πππ’πππππππ 4 π¦ 6 :
1
6π΅π¦ π₯
3 β1
2ππ΅ π₯
2 β 125 π₯4 + 500 π₯3 β 750 π₯2 + πΆ3 π₯ + πΆ4
= β1000 π₯3 +1
6π΅π¦ π₯
3 + 6000 π₯2 β1
2ππ΅ π₯
2 + πΆ5(π₯) + πΆ6
πΆ6 = 10250 (12)
π₯ = 4, π¦3 = 0 ππππππππ§ππππ ππ πππ’πππΓ³π 6 :
πΈπΌπ¦3 = β1000 π₯3 +1
6π΅π¦ π₯
3 + 6000 π₯2 β1
2ππ΅ π₯
2 + πΆ5(π₯) + πΆ6
ππ΅ = β1218,75 +4
3π΅π¦ (13)
ππππππππ§ππππ ππ πππ’πππΓ³π 13 ππ ππ πππ’πππΓ³π π΅ :
βΉ π΅π¦ = 3000 + 1
4 β1218,75 +
4
3π΅π¦
π΅π¦ = 4042,969 (14)
ππππππππ§ππππ ππ πππ’πππΓ³π 14 ππ ππ πππ’πππΓ³π 13 :
ππ΅ = β1218,75 +4
3(4042,969)
ππ΅ = 4171,875 (15)
ππππππππ§ππππ ππ πππ’πππΓ³π 14 ππ ππ πππ’πππΓ³π π΄ :
π΄π¦ β 6000 + (4042,969) = 0
π΄π¦ = 1957,030
πΈππ’πππΓ³π ππ ππ πΆπ’ππ£π πΈπΓ‘π π‘πππ:
ππππ 0 β€ π β€ 1
πΈπΌπ1 = 2021,484 π₯2 β 4171,875 π₯
πΈπΌπ¦1 = 673,828 π₯3 + 2085,937 π₯2
ππππ 1 β€ π β€ 3
πΈπΌπ2 = β500 π₯3 + 3521,484 π₯2 β 5671,875 π₯ + 500
πΈπΌπ¦2 = β125 π₯4 + 1173,828 π₯3 β 2835,9375 π₯2 + 500 π₯ + 125
ππππ 1 β€ π β€ 3
πΈπΌπ3 = β978,515 π₯2 + 7828,125 π₯ β 13000
πΈπΌπ¦2 = β326,172 π₯3 + 3914,062 π₯2 β 13000 π₯ + 10250
MΓ©todo de singularidad
Encontramos la funciΓ³n de momento haciendo momento en B
ππ΅ = 0 β β π π₯ β 15000 π β 1 2 + 15000 π β 3 2 + π΄π¦π = 0
π π₯ = 15000 π β 1 2 + 15000 π β 3 2 + π΄π¦π
πΈπΌπ = β5000 π β 1 3 + 5000 π β 3 3 +π΄π¦π
2
2+ πΆ1 (1)
πΈπΌπ¦ = β1250 π β 1 4 + 1250 π β 3 4 +π΄π¦π
3
6+ πΆ1π + πΆ2 (2)
Evaluamos la ecuaciΓ³n (2) en: π = 0; y= 0 πΆ2 = 0
Evaluamos la ecuaciΓ³n (1) cuando π = 4; π = 0
0 = β5000 3 3 + 5000 + 8π΄π¦ + πΆ1
0 = β130000 + 8π΄π¦ + πΆ1
πΆ1 = 130000 β 8π΄π¦ (3)
Evaluamos la ecuacion (2) cuandoπ = 4; π¦ = 0
Para encontrar el valor de la fuerza π΄π¦
0 = β1250 3 4 + 1250 + 10,667π΄π¦ β 32π΄π¦ + 520000
π΄π¦ = 19687,521
Ahora evaluamos el valor de la fuerza π΄π¦ en la ecuacion (3)
πΆ1 = 130000 β 8(19687,521)
πΆ1 = β27500,168
Entonces la ecuacion de la curva elastica queda de la siguiente forma:
πΈπΌπ¦ = β1250 π β 1 4 + 1250 π β 3 4 + 3281,25π3 β 27500,168π
πΈπ½πΈπ πΆπΌπΆπΌπ (πΉ)
ππππ = 8 ππ π
β + ππ₯ = π cos 30 7 β 500 cos 20 5 β 600 cos 20 5 = 0
π = 500 cos 20 5 + 600 cos 20 5
7(cos 30)
y
z
x
30Β°
P A
B
C
D
E
R = 7in 500 lb
20Β°
R = 5in
R = 5in
20Β° 600 lb 8 in
8 in
8 in
8 in
Y
Z
A π (cos 30)7
30Β°
By
Bz 500 lb
500 cos 20 5
20Β°
Dz
Dy 20Β° 600 lb
600 cos 20 5
B
C
D
E
P
π = 852.550 ππ
Plano XY
ππ΅ = 0:
β π sin 30 8 + 500 sin 20 8 + π·π 16 β 600 sin 20 24
= 0
π·π = 9,176 ππ
+β πΉπ = 0: β π sin 30 + π΅π + 500 sin 20 + π·π β 600 sin 20 = 0
π΅π = 451,301 ππ
Y
8 in 8 in 8 in 8 in
500 sin 20
600 sin 20 π sin 30
A C E D B
Dy By
X
+
-426,275
25,026
196,036 205,212
-500
-400
-300
-200
-100
0
100
200
300
0 8 16 24 32
V (
lb)
X (in)
Diagrama de Fuerza cortante XY
0
-3.410,200-3.209,992
-1.641,704
0
-4000
-3500
-3000
-2500
-2000
-1500
-1000
-500
0
0 8 16 24 32
M (
lb*i
n)
X (in)
Diagrama de Momento flector XY
Plano XZ
ππ΅ = 0:
852,550 cos 30 8 + 500 cos 20 8 β π·π 16
+ 600 cos 20 24 = 0
π«π = ππππ, πππ ππ
+β πΉπ = 0:
852,550 cos 30 + π΅π β 500 cos 20 + 1449,810 β 600 cos 20
= 0
π©π = βππππ, πππ π³π
Z
8 in 8 in 8 in 8 in
500 cos 20 600 cos 20
π cos 30
A C E D B
Dz Bz
X
+
738,33
-416,149
-885,995
563,816
-1000
-800
-600
-400
-200
0
200
400
600
800
1000
0 8 16 24 32
V (
lb)
X (in)
Diagrama de Fuerza Cortante XZ
0
5906,64
2577,448
-4510,512
0
-6000
-4000
-2000
0
2000
4000
6000
8000
0 8 16 24 32
M (
lb*i
n)
X (in)
Diagrama de Momento Flector XZ
SecciΓ³n critica C
Elemento critico 1
ππ£
πππ
ππ‘ πππ
πππ
πππ
ππ‘ ππ£
ππ£
ππ£
ππ‘
ππ‘
1
2
3
4
πππ
ππ£
ππ‘
1
ππΆ = (500)(cos 20)(5)
π»πͺ = ππππ, πππ π³π β ππ
Esfuerzo cortante por carga transversal:
πππππ₯ =4π
3π΄ ; sabemos que Γ‘rea de una circunferencia es π΄ =
π
4π2 si
reemplazamos,
πππππ₯ =4π
3(π
4π2)
πππππ₯ =16π
3ππ2
πππππ₯ =16 500 cos 20
3ππ2
ππ½πππ = π, πππ β πππ
ππ πππ
Esfuerzo por flexion;
ππππ₯ =32π
ππ3
ππππ₯ =32 3.209,976
ππ3
ππππ = ππ, πππ β πππ
ππ πππ
Esfuerzo cortante por torsiΓ³n:
ππ =16π
ππ3 ; π =
ππ =16(2349,231)
ππ3
ππ» = ππ, πππ β πππ
ππ πππ
Como el esfuerzo por carga transversal es muy pequeΓ±o con respecto a los
esfuerzos de flexion y torsiΓ³n, podemos despreciarlos.
Entonces tenemos esfuerzo de torsiΓ³n en X y esfuerzo de flexion en Y.
ππ = 32,696 β 103
π3 ππ π
ππ» = ππ, πππ β πππ
ππ πππ
Ahora calculamos el centro y el radio del circulo de Mohr:
πΆ =ππ₯ + ππ¦
2
πΆ =0 + 32,696 β 103
2
πΆ =16,348 β 103
π3
π = ππ₯ + ππ¦
2
2
+ ππ₯π¦ 2
π = 0 + 32,696 β 103
2
2
+ 11,964 β 103 2
π =20,258 β 103
π3
Dibujo
Como πππππππππππ = π. π β ππππππ, y el radio del circulo es ππππ, entonces
igualo el radio con el πππππππππππ y asi obtengo el diΓ‘metro del eje AE.
πππππππππππ = πΉ
8.5 β 103 =20,258 β 103
π3
π = 20,258β103
8,5β103
3
π = π, πππ ππ
ππππ₯ = πΆ + π
ππππ₯ = 16,348 + 20,258
ππππ₯ = 36,606
Como esta en funciΓ³n del diΓ‘metro, el ππππ₯ queda de la siguiente manera:
ππππ₯ =36,606 β 103
π3
Como π = π, πππ ππ entonces:
ππππ₯ =36,606 β 103
1,336 3
ππππ = ππ, πππ β ππππππ
ππππ = πΆ β π
ππππ = 16,348 β 20,258
ππππ = β3,910
ππππ =β3,910 β 103
π3
ππππ =β3,91 β 103
1,395 3
ππππ = βπ, πππ β ππππππ
El angulo principal es:
2ππ = tanβ1 16,348
11,964 = 54,423Β°
π½π = ππ, πππΒ°
Curva Elastica
Plano (X,Y)
π1 = 426.2804πΏπ
π΅π¦ = 451.309586πΏπ
π2 = 171.01πΏπ
π·π¦ = 9.1729πΏπ
π3 = 205.21208πΏπ
Hacemos corte de A a B (0<X<8)
π1
F +βΊ π
π1
X
+βΊ π1 = π
426.2804 π₯ + π1 = 0
π1 = β426.2804π
πΈπΌ =π2π¦1
ππ₯2 = β426.2804π
Integrando
πΈπΌ =π2π¦1
ππ₯2 = β213.1402π₯2 + π1
Integrando
πΈπΌπ¦1 = β71.04673π₯3 + π1π₯ + π2
Hacemos corte de B a c (8<X<16)
π1 F
+βΊ π
By V
X
βΊ + ππ = 0
426.2804 π₯ β 451.3095 π₯ β 8 + π2 = 0
π2 = β426.2804 π₯ + 451.3095 π₯ β 3610.476
π2 = 25.0292 π₯ β 3610.476
πΈπΌπ2π¦2
ππ₯2 = π2 (π₯)
πΈπΌπ2π¦2
ππ₯2 = 25.0292π₯ β 3610.476
πΈπΌππ¦2
ππ₯2= 12.5146π₯2 β 3610.476π₯ + π3
πΈπΌπ¦2 = 4.17533π₯3 β 1805.238π₯2 + π3π₯ + π4
Hacemos corte de C a D (16<X<24)
P1 P2
F
π +βΊ π
By
βΊ + ππ = 0
426.2804 π₯ β 451.3095 π₯ β 8 β 171.01 π₯ β 16 + π3 = 0
π3 = β426.2804π₯ + 451.309586π₯ β 3610.476688 + 171.01π₯ β 2736.16
π3 = 196.039186π₯ β 6346.636688
πΈπΌπ2π¦3
ππ₯2 = π3 (π₯)
πΈπΌπ2π¦3
ππ₯2 = 196.039186π₯ β 6346.636688
πΈπΌπ2π¦3
ππ₯2 = 98.019593π₯2 β 6346.636688π₯ + π5
πΈπΌπ¦3 = 32.673197π₯3 β 3173.318344π₯2 + π5π₯ + π6
Hacemos corte de D a E (24<x<32)
P1 P2
F
V +βΊ π
By Dy
βΊ + ππ = 0
426.2804 π₯ β 451.3095 π₯ β 8 β 171.01 π₯ β 16 Β± 901729 π₯ β 249
+ π4 = 0
π4 = β426.2804π₯ + 451.309586π₯ β 3610.476688 + 171.01π₯ β 2736.16
+ 9.1279π₯ β 220.1496
π4 = 205.212086π₯ β 6566.786288
πΈπΌπ2π¦4
ππ₯2 = π4(π₯)
πΈπΌπ2π¦4
ππ₯2 = 205.212086π₯ β 6566.786288
πΈπΌπ2π¦4
ππ₯2 = 102.606043π₯2 + 6566.786288π₯ + π7
πΈπΌπ¦3 = 34.202014π₯3 β 3288.393144π₯2 + π7π₯ + π8
Plano (X,Z) la viga E,D,C,B,A
π3 = 852.5608πΏπ cos 30 = 738.3393πΏπ
π1 = 500πΏπ cos 20 = 469.8463014πΏπ
π2 = 600πΏπ cos 20 = 563.8155725πΏπ΅
π·π§ = 1449.81605πΏπ
π΅π§ = β1154.49355πΏπ
Hacemos corte de A a D (0<X<8)
E F
V +βΊ π
P2
βΊ + ππ = 0
βπ2 + π1 = 0
π1 = π2 = 563.8155725π₯
π2π¦1
ππ₯2 =ππ₯
πΈπΌ
πΈπΌπ2π¦1
ππ₯2 = 563.8155725π
Integrando
πΈπΌβ¦΅1 =π2π¦1
ππ₯2 = 281.9077863π₯2 + π1
Integrando
πΈπΌπ¦1 = 93.96926208π₯3 + π1π₯ + π2
Hacemos corte desde D a C (8<x<16)
E F
V +βΊ π
P2 Dz
βΊ + ππ = 0
βπ2π₯ β π·π§(π₯ β 8) + π2 = 0
π2 = 563.855725π₯ + 1449.81605π₯ β 8(1449.81605)
π2 = 2013.631623π₯ β 11598.5284
π2π¦2
ππ₯2 =π2π₯
πΈπΌ
πΈπΌπ2π¦2
ππ₯2= 2013.631623π₯ β 11598.
Integrando
πΈπΌ =π2π¦2
ππ₯2 = 1006.815812π₯2 β 11598.5284π₯ + π3 = πΈπΌβ¦΅2
Integrando
πΈπΌπ¦2 = 335.6052705π₯3 β 579.2642π₯2 + π3π₯ + π4
Hacemos desde C a B (16<x<24)
E D P1 F
P2 Dz V3 +βΊ π
βΊ + ππ = 0
βπ2π₯ β π·π§ π₯ β 8 + π1 π₯ β 16 + π3 = 0
π3 = 563.8155725π₯ + 1449.81605π₯ β 11598.5284 β 469.84104π₯
+ 7517.540966
π3 = 1543.785312π₯ β 4080.987434
π2π¦3
ππ₯2 =π3π₯
πΈπΌ
πΈπΌπ2π¦3
ππ₯2 = 1543.785312π₯ β 4080.987434
Integrando
πΈπΌ =π2π¦3
ππ₯2 = 771.892656π₯2 β 4080.987π₯ + π5
Integrando
πΈπΌπ¦3 = 257.297552π₯3 β 2040.493717π₯2 + π5π₯ + π6
Hacemos de B a A (24<x<32)
E D P1 F
P2 Dz Bz V3 +βΊ π
βΊ + ππ = 0
βπ2π₯ β π·π§ π₯ β 8 + π1 π₯ β 16 β π΅π§ π₯ β 24 + π4 = 0
π4 = 563.8155725π₯ + 1449.81605π₯ β 11598.5284 β 469.84104π₯
+ 7517.540966 β 1154.49355π₯ + 27707.8452
π4 = 389.2917621π₯ + 23626.85777
π2π¦4
ππ₯2=
π4π₯
πΈπΌ
πΈπΌπ2π¦4
ππ₯2 = π4π₯
πΈπΌπ2π¦4
ππ₯2 = 389.2917621π₯ + 23626.85777
Integrando
πΈπΌ =π π¦4
ππ₯2= 194.6458811π₯2 + 23626.8577π₯ + π7
Integrando
πΈπΌπ¦4 = 64.88196037π₯3 β 11813.42889π₯2 + π7π₯ + π8
Encontrando las condiciones de frontera
X=8 π1=0
0 = 48110.6985 + 8π1 + π2
π1 =β48110.6985 β π2
8 (1)
X=24 π3=0
0 = 3556881.359 β 1175329.381 + 24π5 + π6
π5 =β23815556.978 + π6
24 (2)
X=16 π2 = π3
72168.3947 + π3 = 132308.22 + π5
π3 = 60140.32753 + π5 (3)
48110.6985 + 8π1 + π2 = 199323.0103 + 8π3 + π4
8π1 + π2 = 247433.7088 + 8π3 + π4 (4)
X=24 π3 = π4
2381556.978 + 24π5 + π6 = 7701463.261 + 24π7 + π8
24π5 + π6 = 5319906.283 + 24π7 + π8 (5)
X=8 π2 = 0
0 = β199323.0103 + 8π3 + π4
π3 =199323.0103 + π4
8 (6)
X=24 π4 = 0
0 = 7701463.261 + 24π7 + π8
π7 =β7701463.261 β π8
24 (7)
X=16 π2 = π3
109972.4472 + 16π3 + π4 = 531524.3814 + 16π5 + π6
16π3 + π4 = 641496.8286 + 16π5 + π6 (8)
X=16 π2 = 0
0 = 72168. 3934 + π3
π3 = 72168. 3934 (9)
X=16 π3 = 0
0 = 132308.721 + π5
π5 = β132308.721 (10)
Reemplazando π3en (6) para hallar π4
π4 = 776670.1581
Reemplazando π5en (2) para hallar π6
π6 = 793852.326
Reemplazando π5, π6 en (5)
2381556.978 = 5319906.283 + 24π7 + π8
π8 = β7701463.261 β 24π7
Reemplazando π8 en (7)
π7 =7701463.261 + 7701463.261 + 24π7
24
π7 β π7 = 0
π7 = π7
Reemplazando π1, π3, π4 en (4)
β48110.6985 + π2 = β48110.6985
π2 = 0
Reemplazando π2 en (1)
π1 = β6013.8371313
Teniendo los valores de las constantes de integraciΓ³n podemos hallar
la deflexiΓ³n y pendiente de la viga en cualquier punto del Plano XZ
X=8 π1=0
0 = β36375.92576 + 8π1 + π2
π1 =36375.92576 β π2
8 (1)
X=24 π3= 0
0 = β1376157.098 + 24π5 + π6
π5 =1376157.098 + π6
24 (2)
X=16 π2 = π3
54563.8784 + π3 = β76453.1712 + π5
π3 = β21889.28928 + π5 (3)
X=8 π1 = π2
36375.92576 + 8π1 + π2 = 113399.4271 + 8π3 + π4
8π1 + π2 = β77023.48134 + 8π3 + π4 (4)
X=24 π3 = π4
β1376157.091 + 24π5 + π6 = β1421305.809 + 24π7 + π8
24π5 + π6 = β45148.71841 + 24π7 + π8 (5)
X=8 π2 = 0
0 = β113399.407 + 8π3 + π4
π3 =113399.407 + π4
8 (6)
X=24 π4 = 0
0 = β1421305.809 + 24π7 + π8
π7 =1421305.809 β π8
24 (7)
X=16 π2 = π3
44505.3288 + 16π3 + π4 = β678540.0812 + 16π5 + π6
16π3 + π4 = β233485.7524 + 16π5 + π6 (8)
X=16 π2 = 0
0 = β54563.3784 + π3
π3 = 54563.3784 (9)
X=16 π3 = 0
0 = β76453.1712 + π5
π5 = 76453.1712 (10)
Reemplazando π3en (6) para hallar π4
π4 = β323111.6201
Reemplazando π5en (2) para hallar π6
π6 = β458718.0178
Reemplazando π5, π6 en (5)
1376157.091 = β45148.71841 + 24π7 + π8
π7 = 1421305.809 + π8
24
Reemplazando π7 en (7)
=1421305 .809βπ8
24 =
1421305 .809βπ8
24
π8 = π8
Reemplazando π1, π3, π4 en (4)
β36375.92576 β π2 + π2 = 1023759258
π2 = 0
Reemplazando π2 en (1)
π1 = β4546.99072
Teniendo los valores de las constantes de integraciΓ³n podemos hallar
la deflexiΓ³n y pendiente de la viga en cualquier punto del Plano XY