UNIVERSIDAD POLITÉCNICA SALESIANA
FACULTAD: INGENIERÍAS
CARRERA: MECÁNICA
LABORATORIO DE INGENIERÍA MECÁNICA
TEMA: ENSAYO DE COMPRESIÓN EN MADERA
(Fime, 2006) 5TO “A” GRPO “!”
BRAYAN MONTERO
0"#!2#20!$
QUITO-ECUADOR
ÍNDICE
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Di&t$"(i$ De C$#7$ A(!#t$2ie"t F8P9A ! 57)6 5226
ESFUER;O <8F=A 5>7)=(2,6
DEFORMACION UNITARIA ?8@L=L!
M./1%! E%$&ti(i/$/ 5'&i6
5!0
0,02
92;,2;
2",!$
!,00E0
,!E06
!500
0,0$
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;5,6;
!,50E0
;,!$E06
/e
200
0,0
$$00
0,09
200
0,!!
5200
0,!6
6200
$29,2
!$9,0"
2,00E0
",95E06
9!2,$"
!;;,5
$,50E0
9,66E06
5""9,6
2$","
5,50E0
6,20E06
925,5;
2"9,02
;,00E0
5,2;E06
0,22
;;5$,5;
$5,!
!!,00E0
,2;E06
9200
0,$0
!02;!,6
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!5,00E0
$,"0E06
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P#e&i." PSI
Di&t$"(i$ De C$#7$ A(!#t$2ie"t F8P9A ! 57)6 5226
ESFUER;O <8P=A 5>7)=(2,6
DEFORMACION UNITARIA ?8@L=L!
M./1%! E%$&ti(i/$/ 5PSI6
500
0,!0
9!$,"";
2;,56
5,00E0
0,;!$E06
!500
0,20
2!!,""
;5,6;
!0,00E0
!,22E06
2500
0,$5
$56",""
!2,;
!9,50E0
!,!6E06
$500
0,50
""9,""
!"","2
25,00E0
!,!E06
500
0,6"
625,";
259,0
$,50E0
!,06E06
5200
0,"$
925,5;
2"9,02
6,50E0
0,"0"E06
6000
!,!9
;569,";
$2,92
5;,50E0
0,;$$E06
/e
P.-e+/ M/&e./ H$ Pi"!
P#e&i." 5PSI6
Di&t$"(i$ De C$#7$ A(!#t$2ie"t F8P9A ! 57)6 5226
ESFUER;O <8P=A 5>7)=(2,6
DEFORMACION UNITARIA ?8@L=L!
M./1%! E%$&ti(i/$/ 5PSI6
550
0,!2
$!,2
6,00E0
0,90E06
9;5,
/e
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0,2
22!$,$"
;;,5
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!,05E06
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0,$5
$56",""
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!,!6E06
$250
0,!9
60,""
!;5,6
22,00E0
!,20E06
000
0,!;
59!!,""
22;,;
26,00E0
!,25E06
950
0,!"
69;2,";
29!,$2
29,00E0
!,$E06
C0%(1%!& CALCULAR: EUCALIPTO A&: As =5 x 5 cm
As =25 cm
2
2
2
2
Ac = π ∗r = π ∗(1 ) =π ¿
F = P∗ A
F 1=
F 2=
F 6 =
2
:
510∗ π 2,2
=728,28 kgf
1500∗π 2,2
=2141,99 kgf
5200∗ π 2,2
=7425,58 kgf
F 5=
4200∗ π 2,2
=5997,6 Kgf
F 3 =
F 7 =
F 4 =
F 9 =
σ =
2400∗ π 2,2
=3427,2 kgf
6200∗ π
=8853,58 kgf
2.2
3300∗π 2,2
8000∗ π
F As
2,2
= 4712,39 kgf
=11424 kgf
σ 1=
728,28
σ 5=
5997,6
σ 2=
2141,99
σ 6 =
7425,58
25
25
=29,13 ( Kgf / cm 2 )
=239,90 ( Kgf / cm 2)
25
25
=85,68 ( Kgf / cm 2)
=297,02 ( Kgf / cm 2)
F 8 =
7200∗ π 2.2
=10281,6 kgf
σ 3=
3427,2
σ 7 =
8853,6
σ 4 =
4712,39
σ 8 =
10281,6
σ 9 =
11424
25
25
=137,09 ( Kgf / cm 2 )
=354,14 ( Kgf / cm 2 )
25
25
25
δ ε= L
=188,5 ( Kgf / cm 2)
= 411,26 ( Kgf / cm 2)
= 456,96 ( Kgf / cm 2 )
:
ε 1=
0,02
ε 2=
0,03
ε 3=
0,04
ε 4=
0,07
200
200
200
200
0,11
=1∗10−4
ε 5=
=1.5∗10−4
ε 6=
200
0,16 200
0,22
=2∗10−4
ε 7=
=3,5∗10−4
ε 8=
=
200
−
5,5∗10
4
=8∗10−4
=11∗10− 4
0,30 200
=15∗10−4
ε 9=
Е=
0,42
=21∗10−4
200
σ ε
:
29,13 −4
Е 1=
1∗10
= 4,14∗10 psi
0.0703
−4
1,5∗10
2∗10
188,5 −4
3,5∗10
−4
E=
=7,66∗106 psi
=3,1∗106 psi
297,02 −239,9 −4
−5,5∗10−4
Е 8=
=
=4,58∗106 psi
0.0703
15∗10
6
5,28∗10 psi
∗1
−4
Δσ σ 6 −σ 5 = ΔЄ Є 6 −Є 5
8∗10
1
∗
411,26
M./1%! /e e%$&ti(i/$/ t!t$% 5E6: E=
Е 7=
=6,20∗10 6 psi
0.0703
11∗10
∗1
∗1
0.0703
4
−4
0.0703
456,96
Е 9=
Е 6=
8∗10
354,14
= 9,75∗10 6 psi
0.0703
21∗10
−
= 8,13∗106 psi
∗1
0.0703
297,02
∗1
−4
Е 4 =
Е 5=
5,5∗10
∗1
0.0703
137,09
Е 3=
−4
6
85,68
Е 2=
239,9
∗1
∗1
0.0703
=3,9∗10 6 psi
E= 228480 psi
CALCULAR: MALVA A&: As =5 x 5 cm
As =25 cm
2
2
Ac = π ∗r = π ∗(1 ) =π ¿ 2
F = P∗ A
2,2
=713,998 kgf
4500∗ π
F 2=
F 6 =
2
:
500∗ π
F 1=
F 5=
2
2,2
1500∗π 2,2
=2141,99 kgf
5200∗ π 2,2
=6425,98 Kgf
=7425,58 kgf
F 3 =
F 7 =
F 4 =
σ =
2500∗ π 2,2
=3569,99 kgf
6000∗ π 2.2
=8567,98 kgf
3500∗π
F As
2,2
= 4997,99 kgf
σ 1=
713,998
σ 5=
6425,98
σ 2=
2141,99
σ 6 =
7425,58
σ 3=
3569,99
σ 7 =
8567,98
25
25
25
25
25
25
=28,56 ( Kgf / cm 2)
=257,04 ( Kgf / cm 2 )
=85,68 ( Kgf / cm 2)
=297,02 ( Kgf / cm 2)
=142,8 ( Kgf / cm 2)
=342,72 ( Kgf / cm 2)
4997,99
σ 4 =
25
δ ε= L
:
ε 1=
0,10
ε 2=
0,20
ε 3=
0,35
ε 4=
0,50
Е=
=199,92 ( Kgf / cm 2)
200
200
200
200
σ ε
ε 5=
=10∗10−4
ε 6=
=17,5∗10−4
ε 7=
0,93 200
=
1.17 200
−
46,5∗10
4
= 58,5∗10−4
:
257,04
∗1 −4
5∗10
0.0703
−4
= 0.813∗106 psi
Е 5=
85,68 10∗10
0.0703
34,5∗10
−4
=1,22∗106 psi
Е 6=
∗1 =1,06∗10 6 psi
0.0703
297,02
∗1 −4
Е 2=
200
=34,5∗10−4
=25∗10−4
28,56
Е 1=
0,69
=5∗10−4
46,5∗10
0.0703
∗1 =0,909∗106 psi
142,8 −4
Е 3=
17,5 ∗10
−4
Е 4 =
25∗10
−4
=1,16∗10 6 psi
0.0703
199,92
342,72
∗1 Е 7=
58,5∗10
∗1
0.0703
=0,833∗10 6 psi
∗1
0.0703
=1,14∗106 psi
M./1%! /e e%$&ti(i/$/ t!t$% 5E6: E=
E=
Δσ σ 5 −σ 4 = ΔЄ Є 5 −Є 4
257,04 −199,92 −4
34,5∗10
−25∗10−4
E= 60126,3 psi
CALCULAR: PINO A&: As =5 x 5 cm
As =25 cm
2
2
2
2
Ac = π ∗r = π ∗(1 ) =π ¿
F = P∗ A
F 1=
:
550∗ π 2,2
2
=785,4 kgf
F 4 =
3250∗π 2,2
= 4640,99 kgf
1550∗π
F 2=
F 5=
2,2
4000∗ π 2,2
F 3 =
F 6 =
σ =
=2213,39 kgf
=5711,99 Kgf
2500∗ π 2,2
=3569,99 kgf
4750∗ π 2,2
F As
=6782,98 kgf
σ 1=
785,4
σ 4 =
4640,99
σ 2=
2213,39
σ 5=
5711,99
25
=31,42 ( Kgf / cm 2)
25
25
25
=185,64 ( Kgf / cm 2 )
=88,54 ( Kgf / cm 2 )
= 228,48 ( Kgf / cm 2 )
σ 3=
3569,99
σ 6 =
6782,98
25
25
δ ε= L
=271,32 ( Kgf / cm 2)
:
ε 1=
0,12
ε 2=
0,24
ε 3=
0,35
Е=
=142,8 ( Kgf / cm 2)
200
200
200
σ ε
=6∗10−4
=12∗10−4
=17,5∗10−4
ε 4=
ε 5=
ε 6=
0,52 200
200
=22∗10−4
=26∗10−4
=27∗10−4
:
∗1 −4
6∗10
0.0703
185,64
=0,7∗106 psi
−4
Е 4 =
88,54
∗1 −4
Е 2=
200
0,54
31,42
Е 1=
0,44
12∗10
0.0703
22∗10
0.0703
228,48
=1,05∗106 psi
−4
Е 5=
∗1
26∗10
=1,20∗106 psi
∗1
0.0703
=1,25∗10 6 psi
142,8 −4
Е 3=
17,5 ∗10
271,32
∗1
0.0703
−4
=1,16∗10 6 psi
Е 6=
27∗ 10
∗1
0.0703
=1,43∗106 psi
M./1%! /e e%$&ti(i/$/ t!t$% 5E6: E=
E=
Δσ σ 4 −σ 3 = ΔЄ Є 4 −Є 3
185,64 −142,8 −4
22∗10
−17,5∗10−4
E= 95200 psi
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