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7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
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CONTOH 4.3
Properties penampang
Batang 1 :
b = 0.30 m
h = 0.50 m
A = 0.15 m2
I = 0.003125 m4
L1 = 2.5 m
Batang 2 :
b = 0.30 m
h = 0.50 m
A = 0.15 m2I = 0.003125 m4
L2 = 5 m
Batang 3 :
b = 0.30 m
h = 0.50 m
A = 0.15 m2
I = 0.003125 m4
L3 = 2.5 m
Bahan / material :E = 23500000 kN/m2
Hitung matriks kekakuan batang 1 (SM1)
Akibat disla!ement arah 1 ""# $1 %translasi arah sb l&kal 1'
() 1.1 = E.A / L = 1410000
() 2.1 = 0 = 0
() 3.1 = 0 = 0
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() 1.2 = 0 = 0
() 2.2 = 12.EI/L*3 = 5+400
() 3.2 = +.EI / L*2 = ,0500
Akibat disla!ement arah 3 ""# $3 %r&tasi arah sb l&kal 3'
() 1.3 = 0 = 0
() 2.3 = +.EI / L*2 = ,0500
() 3.3 = 4 EI / L = 11,500
Akibat disla!ement arah 4 ""# $4 %translasi arah sb l&kal 4'
() 1.4 = " E.A / L = "1410000
() 2.4 = 0 = 0
() 3.4 = 0 = 0
Akibat disla!ement arah 5 ""# $5 %translasi arah sb l&kal 5'
() 1.5 = 0 = 0
() 2.5 = "12.EI/L*3 = "5+400
() 3.5 = "+.EI / L*2 = ",0500
Akibat disla!ement arah + ""# $+ %r&tasi arah sb l&kal +'
() 1.+ = 0 = 0
() 2.+ = +.EI / L*2 = ,0500
() 3.+ = 2 EI / L = 5-,50
(snan matriks kekakan batang 1 %()1' :
()11 ()12 ()13
()21 ()22 ()23
()1 = ()31 ()32 ()33
()41 ()42 ()43
()51 ()52 ()53
()+1 ()+2 ()+3
(dt batang 1 : =1
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0
= 0
0 0 1
0 0 0
0 0 0
0 0 0
rans&se matrik &tasi
1.00 0 0
= 0 1.00 0
0 0 1
0 0 0
0 0 0
0 0 0
)atriks kekakan strktr menadi :
1.00 0.00 0.00
0.00 1.00 0.00
0.00 0.00 1.00
()( 1 = 0.00 0.00 0.00
0.00 0.00 0.000.00 0.00 0.00
1410000 0 0
0 5+400 ,0500
0 ,0500 11,500
"1410000 0 0
0 "5+400 ",0500
0 ,0500 5-,50
Hitung matriks kekakuan batang 2 (SM2)
Akibat disla!ement arah 1 ""# $1 %translasi arah sb l&kal 1'
() 1.1 = E.A / L = ,05000
() 2.1 = 0 = 0
() 3.1 = 0 = 0
Akibat disla!ement arah 2 ""# $2 %translasi arah sb l&kal 2'
!&s 1
sin 1
" sin 1
!&s 1
()( = . () .
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() 1.2 = 0 = 0
() 2.2 = 12.EI/L*3 = ,050
() 3.2 = +.EI / L*2 = 1,+25
Akibat disla!ement arah 3 ""# $3 %r&tasi arah sb l&kal 3'
() 1.3 = 0 = 0
() 2.3 = +.EI / L*2 = 1,+25
() 3.3 = 4 EI / L = 5-,50
Akibat disla!ement arah 4 ""# $4 %translasi arah sb l&kal 4'
() 1.4 = " E.A / L = ",05000
() 2.4 = 0 = 0
() 3.4 = 0 = 0
Akibat disla!ement arah 5 ""# $5 %translasi arah sb l&kal 5'
() 1.5 = 0 = 0
() 2.5 = "12.EI/L*3 = ",050
() 3.5 = "+.EI / L*2 = "1,+25
Akibat disla!ement arah + ""# $+ %r&tasi arah sb l&kal +'
() 1.+ = 0 = 0
() 2.+ = +.EI / L*2 = 1,+25
() 3.+ = 2 EI / L = 23,5
(snan matriks kekakan batang 2 %()2' :
()11 ()12 ()13
()21 ()22 ()23
()2 = ()31 ()32 ()33
()41 ()42 ()43
()51 ()52 ()53
()+1 ()+2 ()+3
(dt batang 2 : =
ata =
)atriks trans&rmasi r&tasi :
2
2
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0
= 0
0 0 1
0 0 0
0 0 0
0 0 0
rans&se matrik &tasi
0.80 -0.6 0
= 0.6 0.80 0
0 0 1
0 0 0
0 0 0
0 0 0
)atriks kekakan strktr menadi :
0.-0 "0.+0 0.00
0.+0 0.-0 0.00
0.00 0.00 1.00
()( 2 = 0.00 0.00 0.00
0.00 0.00 0.000.00 0.00 0.00
5+4000 "4230 "105,5
423000 5+40 14100
0 1,+25 5-,50
"5+4000 4230 105,5
"423000 "5+40 "14100
0 1,+25 23,5
Hitung matriks kekakuan batang 3 (SM3)
Akibat disla!ement arah 1 ""# $1 %translasi arah sb l&kal 1'
() 1.1 = E.A / L = 1410000
() 2.1 = 0 = 0
() 3.1 = 0 = 0
Akibat disla!ement arah 2 ""# $2 %translasi arah sb l&kal 2'
() 1 2 = 0 = 0
!&s 1
sin 1
" sin 1
!&s 1
()( = . () .
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() 2.2 = 12.EI/L*3 = 5+400
() 3.2 = +.EI / L*2 = ,0500
Akibat disla!ement arah 3 ""# $3 %r&tasi arah sb l&kal 3'
() 1.3 = 0 = 0
() 2.3 = +.EI / L*2 = ,0500
() 3.3 = 4 EI / L = 11,500
Akibat disla!ement arah 4 ""# $4 %translasi arah sb l&kal 4'
() 1.4 = " E.A / L = "1410000
() 2.4 = 0 = 0
() 3.4 = 0 = 0
Akibat disla!ement arah 5 ""# $5 %translasi arah sb l&kal 5'
() 1.5 = 0 = 0
() 2.5 = "12.EI/L*3 = "5+400
() 3.5 = "+.EI / L*2 = ",0500
Akibat disla!ement arah + ""# $+ %r&tasi arah sb l&kal +'
() 1.+ = 0 = 0
() 2.+ = +.EI / L*2 = ,0500
() 3.+ = 2 EI / L = 5-,50
(snan matriks kekakan batang 3 %()3' :
()11 ()12 ()13
()21 ()22 ()23
()3 = ()31 ()32 ()33
()41 ()42 ()43
()51 ()52 ()53
()+1 ()+2 ()+3
(dt batang 3 : =
ata =)atriks trans&rmasi r&tasi :
3
3
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0
= 0
0 0 1
0 0 0
0 0 0
0 0 0
rans&se matrik &tasi
1.00 0 0
= 0 1.00 0
0 0 1
0 0 0
0 0 0
0 0 0
)atriks kekakan strktr menadi :
1.00 0.00 0.00
0.00 1.00 0.00
0.00 0.00 1.00
()( 3 = 0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
1410000 0 0
0 5+400 ,0500
0 ,0500 11,500
"1410000 0 0
0 "5+400 ",0500
0 ,0500 5-,50
Susun MAT!"S "#"A"$AN ST$"T$
0 0 0
1 2 3
164106000 0 0
0 5+6400 ,06500
()( 1 = 0 ,06500 11,6500
"164106000 0 0
0 "5+6400 ",06500
0 ,06500 5-6,50
$1 0 $2
4 5 +
!&s 1
sin 1
" sin 1
!&s 1
()( = . () .
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4536,3- 335601+ "1065,5
335601+ 25-6312 146100
()( 2 = "1065,5 146100 5-6,50
"4536,3- "335601+ 1065,5
"335601+ "25-6312 "146100
"1065,5 146100 263,5
$3 0 $4
, -
164106000 0 0
0 5+6400 ,06500
()( 3 = 0 ,06500 11,6500
"164106000 0 0
0 "5+6400 ",06500
0 ,06500 5-6,50
MAT!"S "#"A"$AN ST$"T$
$1 $2 $3
4 + ,
16-+36,3- "1065,5 "4536,3-
()( = "1065,5 1,+6250 1065,5
"4536,3- 1065,5 16-+36,3-
"1065,5 263,5 1065,5
Batang 1
Gbr AM 1 =
Batang 2
Gbr AM 2 =
Hitung rekasi di ujung batang / freebody (AML) ---> Beban dimasukkan kecuai beban di joi
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Batang !
Gbr AM 3 =
"ransformasi dari sumbu oka ke sumbu goba struktur
Batang 1
0 1 1
30 2 2
AM 1 = 12.5 3 3
0 4 4
30 5 5
-12.5 6 6
SB LOKAL SB GLOBAL
1.00 0.00 0.00
0.00 1.00 0.00
= 0.00 0.00 1.00
0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
0 1 1
30 2 2
= 12.53
3
0 4 4
30 5 5
-12.5 + 6
SB LOKAL SB GLOBAL
Batang 2
22.5 1
25.3125 2
AM 2 = 35.15625 3
37.5 4
RT
AMS
1
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54.6875 5
-58.59375 6
SB LOKAL
0.80 -0.60 0.000.60 0.80 0.00
= 0.00 0.00 1.00
0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
2.8125 4
33.75 5
= 35.15625 6
-2.8125 7
66.25 8
-58.59375 9
SB GLOBAL
Batang !
0 1
60 2AM 3 = 25 3
0 4
60 5
-25 6
SB LOKAL
1.00 0.00 0.00
0.00 1.00 0.00
= 0.00 0.00 1.00
0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
0 7
60 8
= 25 9
0 10
60 11
RT
AMS2
RT
AMS3
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SB GLOBAL
BEBAN EKIVALEN GABUNGAN AE!
B7 1
0 1 0
0 2 30
0 3 12.5
$1 4 0
0 5 30
"AE = $2 + "12.5 8
$3 , 0
0 - 0
$4 0
0 10 0
0 11 0
0 12 0
SB GLOBAL
B"b#$ %# '()$* +#$,$, A'!
0 1 0
0 2 0
0 3 0
$1 4 0
A = 0 5 0$2 + 0
$3 , 0
0 - 0
$4 0
0 10 0
0 11 0
0 12 0
SB GLOBAL
BEBAN /OINT GABUNGAN A!
A = A' AE
1 0 0
2 0 "30
3 0 "12.5
4 0 "2.-125
A! = 5 0 8 "+3.,5
+ 0 "22.+5+25
, 0 2.-125- 0 "12+.25
0 33.53,5
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10 0 0
11 0 "+0
12 0 25
$1 "2.-125
$2 "22.+5+25
$3 2.-125
$4 33.53,5
#$%&'MA "'*+"*'
9itng erindahan
5.,05+5E"00, 2.22125E"00- 1.3-+5524+4-E"00,
= 2.22125E"00- 5.-3-15E"00+ "2.22124-1+3-E"00-
1.3-+55E"00, "2.22125E"00- 5.,05+4+11++,E"00,
2.22125E"00- ".,03+0E"00, "2.22124-1+3-E"00-
,AA-,AA #ALAM *.*, $L$M$
Batang 1
"(r#) *r*r +# #r# SUMBU GLOBAL &) b#*#$, 1 bb
0 1 0
0 2 0
0 0 3 0
-0.000001 4 1
0.000000 5 0
-0.000165 6 2
"(r#) *r*r +# #r# SUMBU LOKAL &) b#*#$, 1 bb
1 0 0
0 1 0
0 0 0 1
0 0 0
0 0 0
0 0 0
G## '$, b#*#$,
B"b#$ '()$* #$, "+#r# O: A:;
!
A%
0
:= S
::-1.A
:;S
::-1= SMS -1
:
#M1
M 1 = R . MS1
M 1
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0 1410000
30 0
AM 1 0 12.5 0
0 -141000030 0
-12.5 0
Batang 2
"(r#) *r*r +# #r# SUMBU GLOBAL &) b#*#$, 2 bb
-9.7180E-07 4 1
0 5 0
0 -1.6499E-04 6 29.7180E-07 7 3
0 8 0
2.1799E-04 9 4
"(r#) *r*r +# #r# SUMBU LOKAL &) b#*#$, 2 bb
0.8 0.6 0-0.6 0.8 0
0 0 0 1
0 0 0
0 0 0
0 0 0
G## '$, b#*#$,
22.5 705000
25.3125 0
AM 2 0 35.15625 0
37.5 -705000
54.6875 0
-58.59375 0
Batang !"(r#) *r*r +# #r# SUMBU GLOBAL &) b#*#$, 3 bb
AM 1 = AML1 SM 1 . M 1
#M2
M 2 = R . MS2
M 2
AM 2 = AML2 SM 2 . M 2
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9.7180E-07 7 3
0 8 0
0 2.1799E-04 9 4
0 10 0
0 11 0
0 12 0
"(r#) *r*r +# #r# SUMBU LOKAL &) b#*#$, 3 bb
1.0 0.0 0.0
0.0 1.0 0.0
0 0.0 0.0 1.0
0.0 0.0 0.00.0 0.0 0.0
0.0 0.0 0.0
G## '$, b#*#$,
0 1410000
60 0
AM 3 0 25 0
0 -1410000
60 0
-25 0
#M!
M 3 = R . MS3
M !
AM 3 = AML3 SM 3 . M 3
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%ata beban
= 100 kN
;1 = 24 kN/m
;2 = 4- kN/m
#&emen 'a&am matriks kekakuan batang
Batang EA / L 4.EI / L 2.EI / L +.EI / L*2
1 1410000 11,500 5-,50 ,0500
2 ,05000 5-,50 23,5 1,+25
3 1410000 11,500 5-,50 ,0500
9bngan antar batang
N(. B#*#$, U'$, ) U'$, '
1 1 2
2 2 3
3 3 4
(AL $AN (
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() 4.2 = 0 =
() 5.2 = "12.EI/L*3 =
() +.2 = +.EI / L*2 =
() 4.3 = 0 =
() 5.3 = "+.EI / L*2 =
() +.3 = 2 EI / L =
() 4.4 = E.A / L =
() 5.4 = 0 =
() +.4 = 0 =
() 4.5 = 0 =
() 5.5 = 12.EI/L*3 =
() +.5 = "+.EI / L*2 =
() 4.+ = 0 =
() 5.+ = "+.EI / L*2 =
() +.+ = 4 EI / L =
""# !atatan matriks kekakan batang dalam arah smb l&kal %smb l&
1 2
()14 ()15 ()1+ 1410000 0
()24 ()25 ()2+ 0 5+400
()34 ()35 ()3+ = 0 ,0500
()44 ()45 ()4+ "1410000 0
()54 ()55 ()5+ 0 "5+400
()+4 ()+5 ()++ 0 ,0500
0deg
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0 0 0
0 0 0 =
0 0 0
0
0
0 0 1
0 0 0
0 0 0
0 0 0
1.00 0 0
0 1.00 0
0 0 1
0.00 0.00 0.00 1410000 0
0.00 0.00 0.00 0 5+400
0.00 0.00 0.00 ? 0 ,0500
1.00 0.00 0.00 "1410000 0
0.00 1.00 0.00 0 "5+4000.00 0.00 1.00 0 ,0500
"1410000 0 0 1 0
0 "5+400 ,0500 0 1
0 ",0500 5-,50 ? 0 0
1410000 0 0 0 0
0 5+400 ",0500 0 0
0 ",0500 11,500 0 0
(AL $AN (
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() 4.2 = 0 =
() 5.2 = "12.EI/L*3 =
() +.2 = +.EI / L*2 =
() 4.3 = 0 =
() 5.3 = "+.EI / L*2 =
() +.3 = 2 EI / L =
() 4.4 = E.A / L =
() 5.4 = 0 =
() +.4 = 0 =
() 4.5 = 0 =
() 5.5 = 12.EI/L*3 =
() +.5 = "+.EI / L*2 =
() 4.+ = 0 =
() 5.+ = "+.EI / L*2 =
() +.+ = 4 EI / L =
""# !atatan matriks kekakan batang dalam arah smb l&kal %smb l&
1 2
()14 ()15 ()1+ ,05000 0
()24 ()25 ()2+ 0 ,050
()34 ()35 ()3+ = 0 1,+25
()44 ()45 ()4+ ",05000 0
()54 ()55 ()5+ 0 ",050
()+4 ()+5 ()++ 0 1,+25
3+.-+-,+4+ deg 0.-
3+.-+-,+4+ deg 0.+
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0 0 0
0 0 0 =
0 0 0
0
0
0 0 1
0 0 0
0 0 0
0 0 0
0.80 -0.6 0
0.6 0.80 0
0 0 1
0.00 0.00 0.00 ,05000 0
0.00 0.00 0.00 0 ,050
0.00 0.00 0.00 ? 0 1,+25
0.-0 "0.+0 0.00 ",05000 0
0.+0 0.-0 0.00 0 ",0500.00 0.00 1.00 0 1,+25
"5+4000 4230 "105,5 1 1
"423000 "5+40 14100 "1 1
0 "1,+25 23,5 ? 0 0
5+4000 "4230 105,5 0 0
423000 5+40 "14100 0 0
0 "1,+25 5-,50 0 0
(AL $AN (
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() 5.2 = "12.EI/L*3 =
() +.2 = +.EI / L*2 =
() 4.3 = 0 =
() 5.3 = "+.EI / L*2 =
() +.3 = 2 EI / L =
() 4.4 = E.A / L =
() 5.4 = 0 =
() +.4 = 0 =
() 4.5 = 0 =
() 5.5 = 12.EI/L*3 =
() +.5 = "+.EI / L*2 =
() 4.+ = 0 =
() 5.+ = "+.EI / L*2 =
() +.+ = 4 EI / L =
""# !atatan matriks kekakan batang dalam arah smb l&kal %smb l&
1 2
()14 ()15 ()1+ 1410000 0
()24 ()25 ()2+ 0 5+400
()34 ()35 ()3+ = 0 ,0500
()44 ()45 ()4+ "1410000 0
()54 ()55 ()5+ 0 "5+400
()+4 ()+5 ()++ 0 ,0500
0 deg 1
0 deg 0
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0 0 0
0 0 0 =
0 0 0
0
0
0 0 1
0 0 0
0 0 0
0 0 0
1.00 0 0
0 1.00 0
0 0 1
0.00 0.00 0.00 1410000 0
0.00 0.00 0.00 0 5+400
0.00 0.00 0.00 ? 0 ,0500
1.00 0.00 0.00 "1410000 0
0.00 1.00 0.00 0 "5+400
0.00 0.00 1.00 0 ,0500
"1410000 0 0 1 0
0 "5+400 ,0500 0 1
0 ",0500 5-,50 ? 0 0
1410000 0 0 0 0
0 5+400 ",0500 0 0
0 ",0500 11,500 0 0
$1 0 $2
4 5 + ==# EN)AN (B 7LBAL
"164106000 0 0 1 0
0 "5+6400 ,06500 2 0
0 ",06500 5-6,50 3 0
164106000 0 0 4 $1
0 5+6400 ",06500 5 0
0 ",06500 11,6500 + $2
$3 0 $4
, - ==# EN)AN (B 7LBAL
!&s 1
sin 1
" sin 1
!&s 1
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"4536,3- "335601+ "1065,5 4 $1
"335601+ "25-6312 146100 5 0
1065,5 "146100 263,5 + $2
4536,3- 335601+ 1065,5 , $3
335601+ 25-6312 "146100 - 0
1065,5 "146100 5-6,50 $4
0 0 0
10 11 12 ==# EN)AN (B 7LBAL
"164106000 0 0 , $3
0 "5+6400 ,06500 - 0
0 ",06500 5-6,50 $4
164106000 0 0 10 0
0 5+6400 ",06500 11 0
0 ",06500 11,6500 12 0
$4
"1065,5 4 $1
263,5 + $2 =
1065,5 , $3
1,+6250 $4
0 1
30 2 kN
= 12.5 3 kN.m
0 4
30 5 kN
"12.5 + kN.m
SB LOKAL
22.5 1
25.3125 2
= 35.15+25 3
3,.5 4 a =
54.+-,5 5 b =
"5-.53,5 + L =
()("1
t
AML1
AML2
AML3
AML4
AML5
AML6
AML1 .!&s =
AML2 .sin =
AML3
AML4
AML5
AML6
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SB LOKAL
0 1
+0 2 kN
= 25 3 kN.m
0 4
+0 5 kN
"25 + kN.m
SB LOKAL
0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
1.00 0.00 0.00
0.00 1.00 0.00
0.00 0.00 1.00
0
0
0
1
0
2
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0.00 0.00 0.000.00 0.00 0.00
0.00 0.00 0.00
0.80 -0.60 0.00
0.60 0.80 0.00
0.00 0.00 1.00
1
0
2
3
0
4
0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
1.00 0.00 0.00
0.00 1.00 0.00
0.00 0.00 1.00
3
0
4
0
0
AMS2 = R
T
. AML2
AMS2
AMS3 = RT. AM
L3
AMS3
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B7 2 B7 3
0 0 0
0 0 30
0 0 12.5
2.-125 0 2.-125
33.,5 0 = +3.,5
35.15+25 8 0 22.+5+25
"2.-125 0 "2.-125
++.25 +0 12+.25
"5-.53,5 25 "33.53,5
0 0 0
0 +0 +0
0 "25 "25
0 0
"30 0
"12.5 0
"2.-125 $1
= "+3.,5 0
"22.+5+25 $2
2.-125 $3"12+.25 0
33.53,5 $4
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0 0
"+0 0
25 0
2.22125E"00- "2.-125 ".,1-0E"0, $1
".,03+E"00, ? "22.+5+25 = "1.+4E"04 $2
"2.2212E"00- 2.-125 .,1-0E"0, $3
5.-3-15E"00+ 33.53,5 2.1,E"04 $4
0 0 0 0
0 0 0 0
0 0 0 3 0 0
1 0 0 -9.7179607E-007
0 1 0 0
0 0 1 -0.0001649936
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0 0 -1410000 0 0
56400 70500 0 -56400 70500
70500 117500 0 -70500 58750 3
0 0 1410000 0 0
-56400 -70500 0 56400 -70500
70500 58750 0 -70500 117500
0 0 0 -9.7179607E-0070 0 0 0
0 0 0 3 -0.0001649936 0
0.8 0.6 0 9.71796072E-007
-0.6 0.8 0 0
0 0 1 0.0002179852
0 0 -705000 0 0
7050 17625 0 -7050 17625
17625 58750 0 -17625 29375 3
0 0 705000 0 0
-7050 -17625 0 7050 -17625
17625 29375 0 -17625 58750
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0.0 0.0 0.0 9.71796072E-007
0.0 0.0 0.0 0
0.0 0.0 0.0 3 0.0002179852 0
1.0 0.0 0.0 00.0 1.0 0.0 0
0.0 0.0 1.0 0
0 0 -1410000 0 0
56400 70500 0 -56400 70500
70500 117500 0 -70500 58750 3
0 0 1410000 0 0
-56400 -70500 0 56400 -70500
70500 58750 0 -70500 117500
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
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12.EI / L*3
5+400
,050
5+400
)AN $AN AN$A %8' %"'
"1410000
0
0
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0
"5+400
,0500
0
",0500
5-,50
1410000
0
0
0
5+400
",0500
0
",0500
11,500
al ata gl&bal menentkan en&m&ran dan tanda %8' %"'
3 4 5 + sb l&kal
0 "1410000 0 0 1
,0500 0 "5+400 ,0500 2
11,500 0 ",0500 5-,50 3
0 1410000 0 0 4
",0500 0 5+400 ",0500 5
5-,50 0 ",0500 11,500 +
sb l&kal
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1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 "1410000 0 0 1
,0500 0 "5+400 ,0500 0
11,500 0 ",0500 5-,50 ? 0
0 1410000 0 0 0
",0500 0 5+400 ",0500 05-,50 0 ",0500 11,500 0
1
0 0 0 0 164106000
0 0 0 0 0
1 0 0 0 = 0
0 1 0 0 "164106000
0 0 1 0 0
0 0 0 1 0
)AN $AN AN$A %8' %"'
",05000
0
0
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0
",050
1,+25
0
"1,+25
23,5
,05000
0
0
0
,050
"1,+25
0
"1,+25
5-,50
al ata gl&bal menentkan en&m&ran dan tanda %8' %"'
3 4 5 + sb l&kal
0 ",05000 0 0 1
1,+25 0 ",050 1,+25 2
5-,50 0 "1,+25 23,5 3
0 ,05000 0 0 4
"1,+25 0 ,050 "1,+25 5
23,5 0 "1,+25 5-,50 +
sb l&kal
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
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0.8 0.6 0 0 0 0
-0.6 0.8 0 0 0 0
0 0 1 0 0 0
0 0 0 0.8 0.6 0
0 0 0 -0.6 0.8 0
0 0 0 0 0 1
0 ",05000 0 0 1
1,+25 0 ",050 1,+25 "1
5-,50 0 "1,+25 23,5 ? 0
0 ,05000 0 0 0
"1,+25 0 ,050 "1,+25 023,5 0 "1,+25 5-,50 0
4
0 0 0 0 4536,3-
0 0 0 0 335601+
1 0 0 0 = "1065,5
0 1 1 0 "4536,3-
0 "1 1 0 "335601+
0 0 0 1 "1065,5
)AN $AN AN$A %8' %"'
"1410000
0
0
0
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"5+400
,0500
0
",0500
5-,50
1410000
0
0
0
5+400
",0500
0
",0500
11,500
al ata gl&bal menentkan en&m&ran dan tanda %8' %"'
3 4 5 + sb l&kal
0 "1410000 0 0 1
,0500 0 "5+400 ,0500 2
11,500 0 ",0500 5-,50 3
0 1410000 0 0 4
",0500 0 5+400 ",0500 5
5-,50 0 ",0500 11,500 +
sb l&kal
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1.0 0.0 0 0 0 0
0.0 1.0 0 0 0 0
0 0 1 0 0 0
0 0 0 1.0 0.0 0
0 0 0 0.0 1.0 0
0 0 0 0 0 1
0 "1410000 0 0 1
,0500 0 "5+400 ,0500 0
11,500 0 ",0500 5-,50 ? 0
0 1410000 0 0 0
",0500 0 5+400 ",0500 0
5-,50 0 ",0500 11,500 0
,
0 0 0 0 164106000
0 0 0 0 0
1 0 0 0 = 0
0 1 0 0 "164106000
0 0 1 0 0
0 0 0 1 0
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5.,05+E"00, 2.22124-E"00- 1.3-++E"00, 2.2212E"00-
2.2212E"00- 5.-3-151E"00+ "2.221E"00- ".,04E"00,
1.3-++E"00, "2.22125E"00- 5.,05+E"00, "2.221E"00-
2.2212E"00- ".,03+0E"00, "2.221E"00- 5.-3-2E"00+
-0 kN = b/L =
+0 kN =
=
3.125 m = a/L =
1.-,5 m =
5 m =
AML1
AML2 b2%3a8b'/L3=
AML3 ab2/L2=
AML4
AML5 a2%a83b'/L3=
AML6 "a2b/L2=
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1.00 0.00 0.00 0.00 0.00
0.00 1.00 0.00 0.00 0.00
0.00 0.00 1.00 0.00 0.00
= 0.00 0.00 0.00 1.00 0.00
0.00 0.00 0.00 0.00 1.00
0.00 0.00 0.00 0.00 0.00
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0.80 -0.60 0.00 0.00 0.000.60 0.80 0.00 0.00 0.00
0.00 0.00 1.00 0.00 0.00
= 0.00 0.00 0.00 0.80 -0.60
0.00 0.00 0.00 0.60 0.80
0.00 0.00 0.00 0.00 0.00
1.00 0.00 0.00 0.00 0.00
0.00 1.00 0.00 0.00 0.00
0.00 0.00 1.00 0.00 0.00
= 0.00 0.00 0.00 1.00 0.00
0.00 0.00 0.00 0.00 1.00
0.00 0.00 0.00 0.00 0.00
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0
"30
"12.5
"2.-125
AE = "+3.,5
"22.+5+25
2.-125
"12+.25
33.53,5
0
"+0
25
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
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-9.71796E-07
-1.64994E-04
9.71796E-07
2.17985E-04
0 1
0 2
0 3
-9.717961E-007 4
0 5
-0.0001649936 6
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
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0 0 1.370
0 30 -11.632
0 0 12.5 -9.693 0
-9.717961E-007 0 -1.3700 30 11.632
-0.0001649936 -12.5 -19.387
-7.774369E-007 4 15.830776E-007 5 0
-0.0001649936 6 2
7.774369E-007 7 3
-5.830776E-007 8 0
0.0002179852 9 4
-7.774369E-007 22.5 -1.096
5.830776E-007 25.3125 0.942
0.000 0 35.15625 -3.270 0
0.000 37.5 1.096
0.000 54.6875 -0.942
0.000 -58.59375 7.980
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
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9.717961E-007 7 3
0 8 0
0.0002179852 9 4
0 10 00 11 0
0 12 0
9.717961E-007 0 1.370
0 60 15.368
0.0002179852 0 25 25.613 0
0 0 -1.370
0 60 -15.368
0 -25 12.807
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
43/101
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44/101
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45/101
0 0 0 0 0
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 00 0 0 0 1
2 3 4 5 + sb gl&bal
0 0 "164106000 0 0 1
5+6400 ,06500 0 "5+6400 ,06500 2
,06500 11,6500 0 ",06500 5-6,50 3
0 0 164106000 0 0 4
"5+6400 ",06500 0 5+6400 ",06500 5
,06500 5-6,50 0 ",06500 11,6500 +
sb gl&bal
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
46/101
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47/101
1 0 0 0 0
1 0 0 0 0
0 1 0 0 0
0 0 1 1 0
0 0 "1 1 00 0 0 0 1
5 + , - sb gl&bal
335601+ "1065,5 "4536,3- "335601+ "1065,5 4
25-6312 146100 "335601+ "25-6312 146100 5
146100 5-6,50 1065,5 "146100 263,5 +
"335601+ 1065,5 4536,3- 335601+ 1065,5 ,
"25-6312 "146100 335601+ 25-6312 "146100 -
146100 263,5 1065,5 "146100 5-6,50
sb gl&bal
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
48/101
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49/101
0 0 0 0 0
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
- 10 11 12 sb gl&bal
0 0 "164106000 0 0 ,
5+6400 ,06500 0 "5+6400 ,06500 -
,06500 11,6500 0 ",06500 5-6,50
0 0 164106000 0 0 10
"5+6400 ",06500 0 5+6400 ",06500 11
,06500 5-6,50 0 ",06500 11,6500 12
sb gl&bal
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22.5 kN
25.3125kN
35.15+25kN.m
3,.5 kN
54.+-,5kN
"5-.53,5kN.m
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0.00 0
0.00 30
0.00 > 12.5
0.00 0
0.00 30
1.00 -12.5
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0.00 22.50.00 25.3125
0.00 > 35.15625
0.00 37.5
0.00 54.6875
1.00 -58.59375
0.00 0
0.00 60
0.00 > 25
0.00 0
0.00 60
1.00 -25
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
53/101
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
54/101
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55/101
1.37 1 1.3,02 kN
18.37 2 1-.3+-0 kN
2.81 3 2.-0++ kN.m
-1.37 4 "1.3,02 kN41.63 5 41.+320 kN
-31.89 6 "31.--+,kN.m
21.40 1 1 21.403- kN
26.25 2 0 2+.254, kN
31.89 3 2 31.--+, kN.m
38.60 4 3 3-.5+2 kN
53.75 5 0 53.,453 kN
-50.61 6 4 "50.+133kN.m
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1.37 1 3 1.3,02 kN
75.37 2 0 ,5.3+-0 kN
50.61 3 4 50.+133 kN.m
-1.37 4 0 "1.3,02 kN
44.63 5 0 44.+320 kN
-12.19 6 0 "12.134kN.m
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
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0
0
0
$1
0
$2
@E>
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
60/101
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61/101
$1
0
$2
$3
0
$4
@E>
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
62/101
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63/101
$3
0
$4
0
0
0
@E>
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
64/101
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
65/101
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
66/101
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
67/101
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
68/101
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
69/101
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
70/101
7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.186)
71/101
Properties penampang
Bahan / material :
E = 2.35E80, kN/m2
Batang 1 :
0.30m
0.50m
1.50E"01m2
E = 2.35E80, kN/m2
3.13E"03m4
2.5m
0 degrees
#&emen 'a&am matriks kekakuan batang
EA / L 4.EI / L 2.EI / L +.EI / L*2
1.41E80+ 1.1-E805 5.--E804 ,.05E804
ekap #&emen 'a&am Matriks "ekakuan atangBatang EA / L 4.EI / L 2.EI / L
1 1.41E80+ 1.1-E805 5.--E804
2 ,.05E805 5.--E804 2.4E804
3 1.41E80+ 1.1-E805 5.--E804
1 2 3 4
* su'a+ %A,AM
Batang = 1
0 degrees
EA / L 4.EI / L 2.EI / L +.EI / L*21.41E80+ 1.1-E805 5.--E804 ,.05E804
b1=
h1=
A1=
I1=
L1=
theta1=
Hitung matriks kekakuan batang 1 /SM10
theta1=
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? = !&s theta = 1
C = sin theta = 0
1 D1 hi1
1 1410000 0 0
D1 0 5+400 ,.05E804hi1 0 ,.05E804 1.1-E805
2 "1410000 0 0.00E800
D2 0 "5+400 ",0500
hi2 0 ,.05E804 5.--E804
* su'a+ %A,AM
Batang = 2
3+.-+-,+5 degrees
EA / L 4.EI / L 2.EI / L +.EI / L*2
,.05E805 5.--E804 2.4E804 1.,+E804
? = !&s theta = 0.-
C = sin theta = 0.+
2 D2 hi2
2 453,3- 33501+ "105,5
D2 33501+ 25-312 1.41E804
hi2 "105,5 1.41E804 5.--E804
3 "453,3- "33501+ 1.0+E804D3 "33501+ "25-312 "14100
hi3 "105,5 1.41E804 2.4E804
* su'a+ %A,AM
Batang = 3
0 degrees
EA / L 4.EI / L 2.EI / L +.EI / L*2
1.41E80+ 1.1-E805 5.--E804 ,.05E804
? = !&s theta = 1
C = sin theta = 0
3 D3 hi3
3 1410000 0 0
D3 0 5+400 ,.05E804
hi3 0 ,.05E804 1.1-E805
4 "1410000 0 0.00E800
D4 0 "5+400 ",0500
hi4 0 ,.05E804 5.--E804
()1 =
Hitung matriks kekakuan batang 2 /SM20
theta2=
()2 =
Hitung matriks kekakuan batang 3 /SM30
theta3=
()3 =
MAT!"S "#"A"$AN ST$"T$ /SM 0
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BAAN7 3
&int 3 : , 0 reaksi ng titik 1 ""#
- +0 reaksi ng titik 1 ""#
25 reaksi ng titik 1 ""#
&int 4 : 10 0 reaksi ng titik 2 ""#
11 +0 reaksi ng titik 2 ""#12 "25 reaksi ng titik 2 ""#
Beban ekiDalen
&int (b 7l&bal Element 1
1 1 0
1 2 30
1 3 12.5
"Ae = 2 4 0
2 5 30
2 + "12.5
3 , 0
3 - 0
3 0
4 10 0
4 11 0
4 12 0
eban oint akibat beban &uar 'i oint
&int (b 7l&bal
1 1 01 2 0
A = 1 3 0
2 4 0
2 5 0
2 + 0
3 , 0
3 - 0
3 0
4 10 0
4 11 04 12 0
eban kombinasi atau gabungan (Ae 5 A)
A! = Ae 8 A
1 0
A! = 2 "30
3 "12.5
4 "2.-125
5 "+3.,5+ "22.+5+25 8
, 2 -125
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- "12+.25
33.53,5
10 0
11 "+0
12 25
)atri? e;ati&n:ea!ti&n Alied
0 "2.-125
0 "22.+5+25
0 2.-125
0 33.53,5
91 0
@1 "30 =
)1 "12.5
@2 "+3.,5
@3 "12+.25
94 0
@4 "+0
)4 25
$isla!ements:
2 = ".,1,+0,1+5E"00,
hi2 = "0.0001+43+
3 = .,1,+0,1+5E"00,
hi3 = 0.00021,-52
ea!ti&ns:91 = 0
@1 = "30
)1 = "12.5
@2 = "+3.,5
@3 = "12+.25
94 = 0
@4 = "+0
)4 = 25
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atatan a = N/m2
Batang 2 :
0.30
0.50
1.50E"01
E = 2.35E80,
3.13E"03
5
3+.-+-,+4+
#&emen 'a&am matriks kekakuan batan
12.EI / L*3 EA / L 4.EI / L
5.+4E804 ,.05E805 5.--E804
+.EI / L*2 12.EI / L*3 theta
,.05E804 5.+4E804 0 deg
1.,+E804 ,.05E803 3+.-+-,+ deg
,.05E804 5.+4E804 0 deg
5 +
$M$ ST$"T$-,OA,
12.EI / L*35.+4E804
b2=
h2=
A2=
I2=
L2=
theta2=
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2 D2 hi2 !ek :
"1410000 0 0 164106000
0 "5+400 ,.05E804 00.00E800 ",0500 5.--E804 0
1.41E80+ 0 0.00E800 "164106000
0 5.+4E804 ",0500 0
0.00E800 ",0500 1.1-E805 0
$M$ ST$"T$-,OA,
12.EI / L*3
,.05E803
3 D3 hi3 ek:
"453,3- "33501+ "105,5 4536,3-
"33501+ "25-312 1.41E804 335601+
1.0+E804 "14100 2.4E804 "1065,5
4.54E805 33501+ 1.0+E804 "4536,3-33501+ 2.5-E805 "14100 "335601+
1.0+E804 "14100 5.--E804 "1065,5
$M$ ST$"T$-,OA,
12.EI / L*3
5.+4E804
4 D4 hi4 ek:
"1410000 0 0 164106000
0 "5+400 ,.05E804 0
0.00E800 ",0500 5.--E804 0
1.41E80+ 0 0.00E800 "164106000
0 5.+4E804 ",0500 0
0.00E800 ",0500 1.1-E805 0
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hi3 1 D1 hi1 D2 D3
"105,5 "1410000 0 0 33501+ "33501+
23,5 0 ,0500 5-,50 "5+400 "14100
105,5 0 0 0 "33501+ 33501+
1,+250 0 0 0 14100 5+4000 1410000 0 0 0 0
0 0 5+400 ,0500 "5+400 0
0 0 ,0500 11,500 ",0500 0
14100 0 "5+400 ",0500 314,12 "25-312
5+400 0 0 0 "25-312 314,12
0 0 0 0 0 0
",0500 0 0 0 0 "5+400
5-,50 0 0 0 0 ,0500
> ? FdG
kekang (keua&i beban &uar 'i oint)
gaCa h&riH&ntal / arah ? %sb strktr'
gaCa Dertikal /arah C %sb strktr'
m&men arah H %sb strktr'
gaCa h&riH&ntal / arah ? %sb strktr'
gaCa Dertikal /arah C %sb strktr'
m&men arah H %sb strktr'
gaCa h&riH&ntal / arah ? %sb strktr'
gaCa Dertikal /arah C %sb strktr'
m&men arah H %sb strktr'
gaCa h&riH&ntal / arah ? %sb strktr'
gaCa Dertikal /arah C %sb strktr'
m&men arah H %sb strktr'
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gaCa h&riH&ntal / arah ? %sb strktr'
gaCa Dertikal /arah C %sb strktr'
m&men arah H %sb strktr'
gaCa h&riH&ntal / arah ? %sb strktr'
gaCa Dertikal /arah C %sb strktr'm&men arah H %sb strktr'
Element 2 Element 3
0 0 0
0 0 30
0 0 12.5
2.-125 0 2.-125
33.,5 0 +3.,5
8 35.15+25 8 0 = 22.+5+25
"2.-125 0 "2.-125
++.25 +0 12+.25
"5-.53,5 25 "33.53,5
0 0 0
0 +0 +0
0 "25 "25
$
d1
d2
d3
d4
$
0 1 0
0 D1 "30
0 hi1 "12.5
0 2 "2.-125 d1
0 D2 "+3.,50 = hi2 "22.+5+25 d2
0 3 2 -125 d3
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0 D3 "12+.25
0 hi3 33.53,5 d4
0 4 0
0 D4 "+0
0 hi4 25
2 hi2 3 hi3 1
2 1-+3,3- "105,5 "453,3- "105,5 "1410000
hi2 "105,5 1,+250 105,5 23,5 0
3 "453,3- 105,5 1-+3,3- 105,5 0
hi3 "105,5 23,5 105,5 1,+250 0
1 "1410000 0 0 0 1410000
D1 0 ,0500 0 0 0
hi1 0 5-,50 0 0 0
D2 33501+ "5+400 "33501+ 14100 0
D3 "33501+ "14100 33501+ 5+400 0
4 0 0 "1410000 0 0
D4 0 0 0 ",0500 0
hi4 0 0 0 5-,50 0
ek:
".,1,+0,1+5E"00, "
"0.0001+43+ "
.,1,+0,1+5E"00, "
0.00021,-52 "
ek:8 1.3,02324+1 = 1.3,02324+ 1.3,02324+1
8 "11.+3204+0- = "41.+3204+ 1-.3+,530
8 ".+33,2424- = "22.133,24
8 11.,2-03+-04 = "52.0210+3
8 15.2,10+31+ = "110.,-04
8 "1.3,02324+1 = "1.3,02324+
8 "15.3+,53002 = ",5.3+,531
8 12.-0++2,5,52 = 3,.-0++2,+
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%ata beban
= 100 kN
;1 = 24 kN/m;2 = 4- kN/m
Batang 3 :
m
m
m2
kN/m2 E =
m4
m tan theta 0.,5 3+.-+-,+
degrees 0.,5
#&emen 'a&am matriks kekak
2.EI / L +.EI / L*2 12.EI / L*3 EA / L
2.4E804 1.,+E804 ,.05E803 1.41E80+
b3=
h3=
A3=
I3=
L3=
theta3=
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0 0 "164106000 0 0 "
5+6400 ,06500 0 "5+6400 ,06500 ",06500 11,6500 0 ",06500 5-6,50 "
0 0 164106000 0 0 "
"5+6400 ",06500 0 5+6400 ",06500 "
,06500 5-6,50 0 ",06500 11,6500 "
335601+ "1065,5 "4536,3- "335601+ "1065,5 "
25-6312 146100 "335601+ "25-6312 146100 "
146100 5-6,50 1065,5 "146100 263,5 "
"335601+ 1065,5 4536,3- 335601+ 1065,5 ""25-6312 "146100 335601+ 25-6312 "146100 "
146100 263,5 1065,5 "146100 5-6,50 "
0 0 "164106000 0 0 "
5+6400 ,06500 0 "5+6400 ,06500 "
,06500 11,6500 0 ",06500 5-6,50 "
0 0 164106000 0 0 "
"5+6400 ",06500 0 5+6400 ",06500 "
,06500 5-6,50 0 ",06500 11,6500 "
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4 D4 hi4
0 0 0
0 0 0
"1410000 0 0
0 ",0500 5-,500 0 0
0 0 0
0 0 0
0 0 0
0 "5+400 ,0500
1410000 0 0
0 5+400 ",0500
0 ",0500 11,500
ek:
16-+36,3- "1065,5 "4536,3- "1065,5 " "
"1065,5 1,+6250 1065,5 263,5 " "
"4536,3- 1065,5 16-+36,3- 1065,5 " "
"1065,5 263,5 1065,5 1,+6250 " "
tanda%8'6%"' ditentkan arah gaCa th smb
tanda%8'6%"' ditentkan arah gaCa th smb
tanda%8'6%"' ditentkan arah gaCa th smb
tanda%8'6%"' ditentkan arah gaCa th smb
tanda%8'6%"' ditentkan arah gaCa th smb
tanda%8'6%"' ditentkan arah gaCa th smb
tanda%8'6%"' ditentkan arah gaCa th smb .!&s a = -0
tanda%8'6%"' ditentkan arah gaCa th smb .sin a = +0
tanda%8'6%"' ditentkan arah gaCa th smb
tanda%8'6%"' ditentkan arah gaCa th smb a = 3.125
tanda%8'6%"' ditentkan arah gaCa th smb b = 1.-,5
tanda%8'6%"' ditentkan arah gaCa th smb L = 5
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0
"30
"12.5
"2.-125
"+3.,5
Ae = "22.+5+25
2.-125
"12+.25
33.53,5
0
"+0
25
ata lang A!
$
2 "2.-125 d1
A! = hi2 "22.+5+25 d2
3 2.-125 d3
hi3 33.53,5 d4
1 0D1 "30
hi1 12 5
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D2 "+3.,5
D3 "12+.25
4 0
D4 "+0
hi4 25
D1 hi1 D2 D3 4 D4 hi4
0 0 33501+ "33501+ 0 0 0
,0500 5-,50 "5+400 "14100 0 0 0
0 0 "33501+ 33501+ "1410000 0 0
0 0 14100 5+400 0 ",0500 5-,50
0 0 0 0 0 0 0
5+400 ,0500 "5+400 0 0 0 0
,0500 11,500 ",0500 0 0 0 0
"5+400 ",0500 314,12 "25-312 0 0 0
0 0 "25-312 314,12 0 "5+400 ,0500
0 0 0 0 1410000 0 0
0 0 0 "5+400 0 5+400 ",0500
0 0 0 ,0500 0 ",0500 11,500
"
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0.30m
0.50m
1.50E"01 m2
2.35E80,kN/m2
3.13E"03 m4
2.5m
0 degrees
uan batang
4.EI / L 2.EI / L +.EI / L*2 12.EI / L*3
1.1-E805 5.--E804 ,.05E804 5.+4E804
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" " " " "
" " " " "" " " " "
" " " " "
" " " " "
" " " " "
" " " " "
" " " " "
" " " " "
" " " " "" " " " "
" " " " "
" " " " "
" " " " "
" " " " "
" " " " "
" " " " "
" " " " "
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" "
" "
" "
" "
kN 7aCa lateral = b/L =
kN 7aCa a?ial =
=
m = a/L =
m =
m =
A)L1
A)L2 b2%3a8b'/L3=
A)L3 ab2/L2=A)
L4
A)L5 a2%a83b'/L3=
A)L+ "a2b/L2=
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2
hi2
3
hi3
0
? 0
0
0
0
0
0
0
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22.5 kN a?ial %arah sb l&kal' ""# dibah arah sb gl&bal
25.3125 kN lateral %arah sb l&kal' ""# dibah arah sb gl&bal
35.15+25kN.m m&men %arah sb l&kal'
3,.5 kN a?ial %arah sb l&kal' ""# dibah arah sb gl&bal
54.+-,5 kN lateral %arah sb l&kal' ""# dibah arah sb gl&bal
"5-.53,5kN.m m&men %arah sb l&kal'
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? C
: 1- 13.5
: "15.1-,5 20.25
: 30 22.5
: "32.-125 43.,5
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+0
sin 0.-++0254 +0
cos 0.5 +0
1.,32050-1 +0
tan theta 0.,5 3+.-+-,+
0.,5
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