statika mekanika bahan 1
TRANSCRIPT
UNIVERSITAS BATANGHARI
FAKULTAS TEKNIK/SIPIL
TUGAS MATA KULIAH :
STATIKA & MEKANIKA BAHAN I
BOBDY ARDY. S
0500810203006
BOBDY ARDY. S 0500810203006
1.
=o
= x = T
= x = T
Perhitungan Reaksi Tumpuan
. 6 - q1 . 2 . 4 - q2 . 2 . 2 + P Sin a . 3 = 0
6 - 3 . 2 . 4 - 5 . 2 . 2 + . 3 = 0
6 = 0
P Sin a . 9 - . 6 + q2 . 2 . 4 + q1 . 2 . 2 = 0
. 9 - 6 + 5 . 2 . 4 + 3 . 2 . 2 = 0
- 6 + = 0
+ - q1 . 2 - q2 . 2 - P Sin a = 0
+ - 3 . 2 - 5 . 2 - = 0
- = 0
= 0
=
= T
HITUNG DAN GAMBARKAN BIDANG MOMEN, LINTANG DAN NORMAL
10 8,19
P Sin a 10 5,74
ΣMB = 0
P Cos a
a 35
35cos
cos 35
103,62
=103,62
= 17,27
T (↑)
ΣMA = 0
5,74
-26,79
26,79
6= = 4,47
T (↑)6
ΣV= 0
4,47 17,27 5,74
RBV
RAV RBV
RAV
RAV
RAV
RBV
RBV
RBV
RAV
RAH P Cos a
RAH 8,19 (→)
21,74 21,74
0,00
ΣH= 0
5,74
P = 10 T
3,00
A BC D E F
P Sin a
P Cos a
9,00
q2 = 5 T/m'q1 = 3 T/m'
1,00 2,00 2,00 1,00
35°
TEKNIK SIPIL
BOBDY ARDY. S 0500810203006
Perhitungan Gaya Dalam
*) Freebody AC, interval 0 ≤ x ≥ 1
= . x
= x
x = 0 , =
x = 1 , =
= / = (Tekan)
=
= = T = = T
*) Freebody CD, interval 1 ≤ x ≥ 3
= . x - . q1 . 2
= x - . 3 . ( x2
- 2 x + 1 )
= x - x2
+ x -
= - x 2 + x -
x = 1 , =
x = 3 , =
/ = 0
- x + = 0
x = m
Mmax =
= / = (Tekan)
= - x +
x = 1 , = = = T
x = 3 , =
Mx RAV
4,47
MA
MC
Dx Nx -8,19
NA NC -8,19DA
dMx dx
4,47
DC 4,47
0
4,47
T.m
T.m
4,47
-8,19
T.m
Dx dMx dx Nx
7,40
2,49
7,79
Mx RAV
4,47
MC
MD
1/2
1/2
4,47 1,50 1,50
1,50
dMx dx
3,00 7,47
(x-1)
3,00
1,50 7,47
T.m
DD -1,53 T
NC ND -8,19
T.m
3,00 7,47
DC 4,47 T
AC
xRAV
RAH
1,00
q1 = 3 T/m'
1,00 2,00
C DA
xRAV
RAH
TEKNIK SIPIL
BOBDY ARDY. S 0500810203006
*) Freebody DE, interval 3 ≤ x ≥ 5
= . x - q1 . 2 . - . q2 . 2
= x - 3 . 2 . ( x - 2 ) - . 5 . ( x2
- 6 x + 9 )
= x - x + - x2
+ x -
= - x2
+ x
x = 3 , =
x = 5 , =
/ = 0
- x + = 0
x = m < 3 m (Mmax tidak pada interval DE)
= / = (Tekan)
= - x +
x = 3 , = = = T
x = 5 , =
*) Freebody FB, interval 0 ≤ x ≥ 3
= - . x
= - x
x = 0 , =
x = 3 , =
= / = (Tekan)
=
= = T = = T
-17,21
-8,19
5,74
Nx -8,19
NF NB
Dx dMx dx
DF DB 5,74
Mx
5,74
MF 0 T.m
MB T.m
P Sin a
5,00 13,47
DD -1,53 T ND NE -8,19
DE T-11,53
5,00 13,47
2,69
Dx dMx dx Nx -8,19
22,50
MD 7,40 T.m
ME T.m-5,67
dMx dx
-10,502,50 13,47
1/2 (x-3)
1/2
2,50 15,00
Mx RAV (x-2)
4,47
4,47 6,00 12,00
q2 = 5 T/m'q1 = 3 T/m'
1,00 2,00 2,00
C D EA
RAV
RAH
P = 10 T
3,00
B F
P Sin a
P Cos a
x
TEKNIK SIPIL
BOBDY ARDY. S 0500810203006
*) Freebody BE, interval 3 ≤ x ≥ 4
= - . x + ( x - 3 )
= - x + ( x - 3 )
= - x + x -
= x -
x = 3 , =
x = 4 , =
= / = (Tekan)
=
= = T = = T-8,19
P Sin a
5,74
RBV
17,27
5,74 17,27 51,81
11,53 51,81
-17,21
-5,67
Dx dMx dx Nx -8,19
-11,53
DB DE -11,53 NB NE
Mx
MB T.m
ME T.m
1,00
P = 10 T
35°
3,00
BE F
P Sin a
P Cos a
RBVx
TEKNIK SIPIL
BOBDY ARDY. S 0500810203006
TEKNIK SIPIL
q2 = 5 T/m'q1 = 3 T/m'
1,00 2,00 2,00 1,00
P = 10 T35°
3,00A BC D E F
P Sin
P Cos
9,00
M
(+)
(-)
X = 2,49
(+)
(-) D
(-) N
+ 4,47 T.m
+ 7,79 T.m+ 7,40 T.m
- 5,67 T.m
- 17,21 T.m
+ 4,47 T
- 1,53 T
- 11,53 T
+ 5,74 T
- 8,19 T
(+)
BOBDY ARDY. S 0500810203006
2.
=o
= x = T
= x = T
Perhitungan Reaksi Tumpuan
. - q . 4 . - P Sin a . - P Cos a . = 0
- 5 . 4 . - . - . = 0
= 0
= T (↑)
- . - P Cos a . + P Sin a . + . q . 42
= 0
- - . + . + . 5 . 42
= 0
- + = 0
= T (↑)
+ - q . 4 - P Sin a = 0
+ - 5 . 4 - = 0
- = 0
= 0
=
= T
a 30
P Cos a 5
HITUNG DAN GAMBARKAN BIDANG MOMEN, LINTANG DAN NORMAL
ΣMB = 0
RAV 8,5 6,5 4,5
P Sin a 5
30 4,33
30 2,50
cos
cos
ΣMA = 0
RBV 8,5 10 4 1/2
10
8,5 RAV -184,55
RAV 21,71
RAV 4,338,5 6,5 2,50 4,5
RBV 0,79
ΣV= 0
RAV RBV
21,71 0,79 2,50
2,50
8,5 RBV 6,70
8,5 RBV 4,33 10
22,50 22,5
ΣH= 0
RAH P Cos a
RAH 4,33 (→)
0,00
10
4 1/2
P = 5 T
P Cos a
P Sin a
30°
BA
10,0
0
4,00 4,50
8,50
q = 5 T/m'
C D E
TEKNIK SIPIL
BOBDY ARDY. S 0500810203006
Perhitungan Gaya Dalam
*) Freebody AC, interval 0 ≤ y ≥ 10
= - . y
= - y
y = , =
y = , =
= / = = (Tekan)
=
= = T = = T
*) Freebody CD, interval 0 ≤ x ≥ 4
= - . + . x
- . q . x2
= - . + x
- . 5 . x 2
= - x 2 + x -
x = 0 , =
x = 4 , =
/ = 0
- x + = 0
x = m > 4
(Mmax tidak terdapat pada interval CD)
= / = = (Tekan)
= - x +
x = 0 , = = = T
x = 4 , =
Mx RAH
4,33
MA T.m0
DA DC 4,33 NA
MC T.m
Dy dMy dy
43,30
MC T.m
1/2
NC
21,71
0,00
-21,71
RAV
RAH
4,33
10
10
RAV
-21,71Ny
dMx dx
5,00 21,71
MD T.m
-43,30
3,55
10 -43,30
1/2
2,50 21,71
Mx
4,33
DD 1,71 T
4,34
Dx dMx dx
5,00 21,71
Nx RAH -4,33
NC ND -4,33DC 21,71 T
A
C
RAH
10,0
0
RAV
y
q = 5 T/m'
A
C
RAH
10,0
0
RAV
4,00x
D
TEKNIK SIPIL
BOBDY ARDY. S 0500810203006
*) Freebody BE, interval 0 ≤ y ≥ 10
= 0 = /
= 0
= = 0 = = 0
= = (Tekan)
= = T
*) Freebody ED, interval 0 ≤ x ≥ 4,5
= . x
= x
x = , =
x = , =
= /
=
= = T
= 0
= = 0
dMy dy
DB DE
Ny RBV -0,79
My
MB ME
Dy
T.m
4,5 ME 3,55 T.m
NB NE -0,79
Mx RBV
Nx
NE ND
-0,79
0,79
Dx dMx dx
-0,79
DB DE
0 MB 0,00
B
E
10,0
0
y
RBV
B
E
10,0
0
RBV
4,50
xD
TEKNIK SIPIL
BOBDY ARDY. S 0500810203006
TEKNIK SIPIL
P = 5 T
P Cos
P Sin
30°
BA
10,0
0
4,00 4,50
8,50
q = 5 T/m'
C D E
RAH
RAV RBV
(-)
- 43,30 T.m
- 43,
30 T
.m
+ 3,55T.m(+)
M
(-)
BOBDY ARDY. S 0500810203006
TEKNIK SIPIL
+ 4,
33 T
(+)
+ 21,71T
+ 1,71T
- 0,79T
(+)
- 21,
71 T
(-)
- 4,33T
- 0,79 T
(-)
D
N
BOBDY ARDY. S 0500810203006
3. PERHITUNGAN GARIS PENGARUH
Akibat beban P = 1 T bergerak dari A ke B
9
9
0
9
7
9
2
9
3
9
6
9
0
9
9
9
*) P = 1 T di A RA = x = 1,00 T
RB = x 1 = 0,00 T
1,
*) P = 1 T di C RA = x = 0,78 T
RB = x 1 = 0,22 T
1,
*) P = 1 T di D RA = x = 0,33 T
RB = x 1 = 0,67 T
1,
0,00 T
RB = x 1 = 1,00 T
1*) P = 1 T di B RA = x =,
A BC D2,00 4,00 3,00
P = 1 Tx
A BC D2,00 4,00 3,00
P = 1 Tx
1,00 0,78GP. RA
GP. RB
0,33
1,000,670,22
TEKNIK SIPIL
BOBDY ARDY. S 0500810203006
Akibat beban bergerak dan beban mati
a). RA Maksimum
Alternatif 1
y1 = T
RA = ( . y1 ) + ( . y2 ) + ( . y2 )
+ ( Luas Diagram GP . )
= ( 5 . ) + ( 5 . ) + ( 2 . )
+
= T
Alternatif 2
RA = ( . y2 ) + ( . y3 ) + ( . y2 )
+ ( Luas Diagram GP . )
= ( 5 . ) + ( 5 . ) + ( 2 . )
+
= T
PDL
+ (0,78 0,33
2x 4 x 5
3y4 = x 1,00 = 0,33 T9
19,33
+ (0,78 0,33
x 4 x
PDL
0,78 0,56 0,78
qDL
5 )2
1,00 = 0,56 T
0,78
)
= 7
9x 1,00 = 0,78 T
PLL PLL
1,00 0,78
21,56
PLL PLL
y3 =5
x9
qDL
1,00
y2
1,00 0,78GP. RA
PLL = 5 T
2,00
A B
3,004,00
qDL = 5 T/m
PLL = 5 T
PDL = 2 T
0,33
0,78GP. RA
PLL = 5 T
2,00
A B
3,002,00
qDL = 5 T/m
PLL = 5 T
PDL = 2 T
0,33
2,00
TEKNIK SIPIL
BOBDY ARDY. S 0500810203006
Alternatif 3
RA = ( . y3 ) + ( . y4 ) + ( . y2 ) +
+ ( Luas Diagram GP . )
= ( 5 . ) + ( 5 . ) + ( 2 . ) + ( . 4 . )
+
= T
Alternatif 4
RA = ( . y4 ) + ( . y5 ) + ( . y2 )
+ ( Luas Diagram GP . )
= ( 5 . ) + ( 5 . ) + ( 2 . )
+
= T
4 x 5 )2
qDL
PLL PLL
+ (0,78 0,33
x 4 x 5 )2
y5 =1
x 1,00 = 0,11 T9
PDL
0,780,33 0,11
0,00
PLL PLL PDL
0,56 0,33 0,78
0,78 0,33x
5
qDL
14,89
17,11
+ (
0,78GP. RA
PLL = 5 T
2,00
A B
3,002,00 2,00
0,56
qDL = 5 T/m
PLL = 5 T
PDL = 2 T
0,33
0,78GP. RA
PLL = 5 T
2,00
A B
4,00
qDL = 5 T/m
PLL = 5 T
PDL = 2 T
0,33
1,002,00
0,11
TEKNIK SIPIL
BOBDY ARDY. S 0500810203006
Alternatif 5
RA = ( . y6 ) + ( . y0 ) + ( . y2 )
+ ( Luas Diagram GP . )
= ( 5 . ) + ( 5 . ) + ( 2 . )
+
= T
Jadi, RA Maksimum = T
13,78
21,56
PDL
0,22 0,00 0,78
qDL
+ (0,78 0,33
x 4 x 5 )2
= 0,22 T9
PLL PLL
y6 =2
x 1,00
0,78GP. RA
PLL = 5 T
2,00
A B
qDL = 5 T/m
PLL = 5 T
PDL = 2 T
2,001,00
0,220,33
4,00
TEKNIK SIPIL
BOBDY ARDY. S 0500810203006
b). RB Maksimum
Alternatif 1
RB = ( . y0 ) + ( . y1 ) + ( . y1 )
+ ( Luas Diagram GP . )
= ( 5 . ) + ( 5 . ) + ( 2 . )
+
= T
Alternatif 2
RB = ( . y1 ) + ( . y2 ) + ( . y1 )
+ ( Luas Diagram GP . )
= ( 5 . ) + ( 5 . ) + ( 2 . )
+
= T
y2 =4
x 1,00 = 0,44 T9
qDL
+ (0,22 0,67
x 4 x 52
12,67
0,22 0,44 0,22
)
PDL
PLL PLL PDL
0,00 0,22 0,22
qDL
+ (0,22 0,67
x 4 x 5 )2
10,44
PLL PLL
= 0,22 T9
y3 =6
x 1,00 = 0,67 T9
y1 = 2 x 1,00
GP. RB
PLL = 5 T
2,00
A B
4,00
qDL = 5 T/m
PLL = 5 T
PDL = 2 T
3,00
1,000,22 0,67
GP. RB
PLL = 5 T
2,00
A B
2,00
qDL = 5 T/m
PLL = 5 T
PDL = 2 T
3,002,00
1,000,22 0,44 0,67
TEKNIK SIPIL
BOBDY ARDY. S 0500810203006
Alternatif 3
RB = ( . y2 ) + ( . y3 ) + ( . y1 )
+ ( Luas Diagram GP . )
= ( 5 . ) + ( 5 . ) + ( 2 . )
+
= T
Alternatif 4
RB = ( . y3 ) + ( . y4 ) + ( . y1 ) + ( . L . y2 )
+ ( Luas Diagram GP . )
= ( 5 . ) + ( 5 . ) + ( 2 . ) + ( . 4 . )
+
= T
+ (0,22 0,67
x 4 x 5 )2
17,11
0,00
+ (0,22 0,67
x 4 x 5 )2
qDL
PLL PLL PDL qDL
0,67 0,89 0,22 5
14,89
y4 =8
x 1,00 = 0,89 T9
PLL PLL PDL
0,44 0,67 0,22
qDL
GP. RB
PLL = 5 T
2,00
A B
2,00
qDL = 5 T/m
PLL = 5 T
PDL = 2 T
3,002,00
1,000,670,440,22
GP. RB
PLL = 5 T
2,00
A B
4,00
qDL = 5 T/m
PLL = 5 T
PDL = 2 T
1,002,00
1,000,890,670,22
TEKNIK SIPIL
BOBDY ARDY. S 0500810203006
Alternatif 5
RB = ( . y5 ) + ( . y6 ) + ( . y1 ) + ( . L . y2 )
+ ( Luas Diagram GP . )
= ( 5 . ) + ( 5 . ) + ( 2 . ) + ( . 4 . )
+
= T
Jadi, RB Maksimum = T
0,00
18,22
18,22
PLL PLL PDL qDL
0,78 1,00 0,22 5
qDL
+ (0,22 0,67
x 4 x 5 )2
y5 =7
x 1,00 = 0,78 T9
GP. RB
PLL = 5 T
2,00
A B
4,00
qDL = 5 T/m
PLL = 5 T
PDL = 2 T
2,001,00
0,220,78 1,00
TEKNIK SIPIL