VIGAS-UNSAAC

1. ELEMENTO 1
E= 3*109 𝑘𝑔/𝑚2
A= 30cm*30cm = 0.09 𝑚2
I= 1.9*10−4 𝑚4
L= 3m
𝑆 𝑜= 900*105 𝑘𝑔/𝑚 A= 900*105 𝑘𝑔/𝑚
𝑆1= 7.6*105 𝑘𝑔𝑚 A’= 2.53*105 𝑘𝑔/𝑚
𝑆2= 3.8*105 𝑘𝑔 B= C’=0
𝑆3= 2.53*105 𝑘𝑔/𝑚 C= 3.8*105 𝑘𝑔
ʎ 𝑥 = 1 ʎ 𝑦 = 0
5 6 3 1 4 2
900 0 -3.8 -900 0 -3.8 5
0 2.53 3.8 0 -2.53 3.8 6
K'1= -3.8 3.8 7.6 0 -3.8 3.8 3
-900 0 0 900 0 0 1
0 -2.53 -3.8 0 2.53 -3.8 4
-3.8 3.8 3.8 0 -3.8 7.6 2
ELEMENTO 2
E= 3*109 𝑘𝑔/𝑚2
A= 30cm*30cm = 0.09 𝑚2
I= 1.9*10−4 𝑚4
L= 5m
𝑆 𝑜= 540*105 𝑘𝑔/𝑚 A= 540*105 𝑘𝑔/𝑚
𝑆1= 4.6*105 𝑘𝑔𝑚 A’= 0.55*105 𝑘𝑔/𝑚
𝑆2= 1.4*105 𝑘𝑔 B= C’=0
𝑆3= 0.55*105 𝑘𝑔/𝑚 C= 1.4*105 𝑘𝑔
ʎ 𝑥 = 1 ʎ 𝑦 = 0
1 4 2 7 8 9
540 0 -1.37 -540 0 -1.37 1
0 0.55 1.368 0 -0.55 1.368 4
K'2= -1.37 1.368 4.56 0 -1.37 2.28 2
-540 0 0 540 0 0 7
0 -0.55 -1.37 0 0.55 -1.37 8
-1.37 1.368 2.28 0 -1.37 4.56 9

2. ENSAMBLANDO LA MATRIZ K
HALLANDO LA MATRIZ “K11”
1 2 3 4 5 6 7 8 9
1440 -1.37 0 0 -900 0 -540 0 -1.37 1
-1.37 12.16 3.8 -2.43 -3.80 3.8 0 -1.37 2.28 2
0 3.8 7.6 -3.8 -3.8 3.8 0 0 0 3
0 -2.43 -3.8 3.08 0 -2.53 0 -0.55 1.368 4
K= -900 -3.80 -3.8 0 900 0 0 0 0 5
0 3.8 3.8 -2.53 0 2.53 0 0 0 6
-540 0 0 0 0 0 540 0 0 7
0 -1.37 0 -0.55 0 0 0 0.55 -1.37 8
-1.37 2.28 0 1.368 0 0 0 -1.37 4.56 9
0 0 -0
inv(K11)= 0 0.1 -0
-0 -0 0.16
HALLANDO LA MATRIZ “K” Y “T” DE CADA ELEMENTO
900 0 0 -900 0 0 540 0 0 540 0 0
0 2.53 3.8 0 -2.53 -3.8 0 0.55 1.4 0 -0.55 -1.4
K1= 0 3.8 7.6 0 -3.8 3.8 K2= 0 1.4 4.6 0 -1.4 2.3
-900 0 0 900 0 0 540 0 0 540 0 0
0 -2.53 -3.8 0 2.53 3.8 0 -0.55 -1.4 0 0.55 1.4
0 -3.8 3.8 0 3.8 7.6 0 -1.4 2.3 0 1.4 4.6
1 0 0 0 0 0
0 1 0 0 0 0
T1=T2= 0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1

3. HALLANDO LAS DISTANCIAS DESCONOCIDAS (Dd)
{Dd} = [𝑲𝟏𝟏]−𝟏 ∗{Qc}
0 0 D1
Qc= -3.06 ENTONCES, Dd= -0.2 D2
-2.7 -0.3 D3
HALLANDO LAS FUERZAS INTERNAS DE CADA ELEMENTO
{Q} = [ 𝑲] ∗ [𝑻] ∗{D}
qxi1 900 0 0 -900 0 0 0 0
qyi1 0 2.53 3.8 0 -2.5 -3.8 0 -0.38
qzi1 = 0 3.8 7.6 0 -3.8 3.8 * -0.3 = -3.04
qxj1 -900 0 0 900 0 0 0 0
qyj1 0 -2.5 -3.8 0 2.53 3.8 0 0.38
qzj1 0 -3.8 3.8 0 3.8 7.6 -0.2 -2.66
qxi2 540 0 0 540 0 0 0 0
qyi2 0 0.55 1.4 0 -0.6 -1.4 0 -0.28
qzi2 = 0 1.4 4.6 0 -1.4 2.3 * -0.2 = -0.92
qxj2 540 0 0 540 0 0 0 0
qyj2 0 -0.6 -1.4 0 0.55 1.4 0 0.28
qzj2 0 -1.4 2.3 0 1.4 4.6 0 -0.46
HALLANDO LAS FUERZAS DESCONOCIDAS
Q4 = 8.54 tnf
Q5 = 0
Q6 = 6.38 tnf
Q7 = 0
Q8 = 3.88 tnf
Q9 = 3.34 tnf-m

4. SOLUCIÓN DEL PORTICO

6. SIMPLIFICACIÓN DE LA VIGA
ELEMENTO 1
E= 3*109 𝑘𝑔/𝑚2
A= 30cm*30cm = 0.09 𝑚2
I= 1.9*10−4 𝑚4
L= 3m
1 2
K=K'1= 7.6 3.8 1
3.8 7.6 2
ELEMENTO 2
E= 3*109 𝑘𝑔/𝑚2
A= 30cm*30cm = 0.09 𝑚2
I= 1.9*10−4 𝑚4
L= 5m
2 3
K=K'2= 4.56 2.28 2
2.28 4.56 3
ENSAMBLANDO LA MATRIZ K
HALLANDO LA MATRIZ “K11”
1 2 3
7.6 3.8 0 1
K= 3.8 12.16 2.28 2
0 2.28 4.56 3
inv(K11)= 0.156 -0.05
-0.05 0.097
HALLANDO LAS DISTANCIAS DESCONOCIDAS (Dd)
{Dd} = [𝑲𝟏𝟏]−𝟏 ∗{Qc}
Qc=
-2.7
ENTONCES,
Dd=
-0.25 D1
-3.6 -0.22 D2

7. HALLANDO LAS FUERZAS INTERNAS DE CADA ELEMENTO
{Q} = [ 𝑲] ∗ [𝑻] ∗{D}
qzi1
=
7.6 3.8
*
-0.25
=
-3.04
qzj1 3.8 7.6 -0.22 -2.66
qzi2
=
4.56 2.28
*
-0.22
=
-0.92
qzj2 2.28 4.56 0 -0.46
HALLANDO LAS FUERZAS DESCONOCIDAS
Q3 = -0.5+3.84=3.34 tnf-m
HALLANDO LAS FUERZAS CORTANTES
V1 = (3.04+2.66)/3=2.85 tnf ; V2 = (0.92+0.46)/5=0.276 tnf
SOLUCIÓN DEL PORTICO

10. TERCERA SOLUCION DE LA VIGA
1.26 0.63
*
Ɵ1
=
2.7
0.63 2.02 Ɵ2 3.06