Soal 1
12 kN 9,6 kN/m 32 kN
P1= q= P2=
E 2I 4I F 3I
A B C D
6 m 8 m 3 3
2m
L2 = L3=
Gambarkan Diagram M&D utk balok diatas !
Penyelesaian :
Membuat Diagram M akibat beban luar
1/8*q*L2 = 76,8 kNm
P2*L/4= 48 kNm
A1
A2
A1=1/12*q*L3 A2=1/2*L* 48
A1 = 409,6 m2 A2= 144 m2
Data :
P1 = 12 kN Jarak L1 = 2 m
P2 = 32 kN Jarak L2 = 6 m
q = 9,6 kNm Jarak L3 = 8 m a3 = 4m
Jarak L4 = 6 m a4 = 3m
Diperoleh :
A1 = 409,6 m2
A2 = 144 m2
MA = -24 kNm
MD = 0 kNm
L1 L1 L2 L2 6A1a1 6A2a 2 6EhA 6EhC
MA
I 2MB
I I
Mc
I
LI L I L
1 1 2 2 1
1
2 2 1
L2
akibat bbn luar akibat penurunan
Terapkan pers. Clapeyron pd bentang AB & BC:
MA [ ] 6
----
2I
+ 2MB [ 6
----
2I
+
8
----
4I
] + MC [ 8
----
4I
] =
6x 409,6 x
- -----------------
4 Ix 8
4
-24 [ ] 6
---- + 2MB [ 12
---- +
8
---- ] + MC [ 8
---- ] =
9830,4
- -----------------
2I 4I 4I 4I 32 I
-24 [ ] 6
----
2I
+ 2MB [ 20
----
4I
] + MC [ 8
----
4I
] =
307,2
- -----------------
I
--------------------------------------- dikali I
Sehingga Menjadi :
-24 ( 3 ) + 2MB ( 5 ) + MC ( 2 ) = -307,2
-72 + 10 MB + 2 MC = -307,2
10 MB + 2 MC = -307,2 + 72
10 MB + 2 MC = -235,2 ___________________ Pers. (1)
Bentang BC & CD
MB [ ] 8
----
4I
+ 2 MC [ 8
----
4I
+
6
----
3I
] =
6x 409,6 x
- -----------------------
4 Ix 8
4 6x 144,0 x
- ---------------------------
3I x 6
3
MB [ ] 8
----
4I
+ 2 MC [ 24
----
12 I
+
24
----
12 I
] =
9830,4
- -----------------
32 I
2592
- ------------------
18 I
MB [ ] 8
----
4I
+ 2 MC [ 48
----
12 I
] = - -----------------
307,2
I
144,0
- ------------
I
--------------------------------------- dikali I
Menjadi :
MB ( 2 ) + 2 MC ( 4 ) = -307,2 -144,0
2 MB + 8 MC = -307,2 -144,0
2 MB + 8 MC = -451,2
2 MB + 8 MC = -451,2 ___________________ Pers. (2)
Sederhanakan Per. (1) dan (2)
10 MB + 2 MC = -235,20 -------------- x 2
2 MB + 8 MC = -451,20 -------------- x 10
20 MB + 4 MC = -470,4
20 MB + 80 MC = -4512 -
-76 MC = 4041,6
MC = -53,18 kNm
Masukkan Nilai MC ke pers. (2)
2 MB + 8 MC = -451,20
2 MB + 8( -53,18 = -451,20
2 MB + ( -425,43 = -451,20
2,0 MB = -451,20 + 425,43
2 MB = -25,77
MB = -12,88 kNm
sehingga diperoleh :
MB = -12,88 kN.m
MC = -53,18 kN.m
Ceck
2 MB + 8 MC = -451,20
2 -12,88 + 8 ( -53,18 ) = -451,20
-25,77 + -425,4316 = -451,20
______ OK !!!
-451,20 = -451,20
Free Body
12 kN q= 9,6 kN/m 32 kN
MBc
P1= P2 =
MA= MA= 24 MBA=
12,88
MCB MDC
24 12,88 53,18 53,18
AA B C D
E B C
12/6 2 9,6*8/2 38,4 32*3/6 16
12/6 2 12,88/8 1,61 53,18/6 8,86
12,88/6 2,15 53,18/8 6,65
12 1,85 -1,85 43,44 24,86 7,14
33,36
6 8 6
MA= 24 kN.m MB= 12,88 kN.m MC= 53,18 kN.m
reaksi Ujung akibat 12,00 2,00 2,00 38,40 38,40 16,00 16,00
beban yg bekerja
reaksi ujung akibat 2,00 2,00 1,61 1,61 8,86 8,86
momen ujung 2,15 2,15 6,65 6,65
12,00 1,85 1,85 33,36 43,44 24,86 7,14
reaksi ujung Total
RA = 13,85 RB = 31,51 RC = 68,30 RD = 7,14
+
CHECK !!!
RA + RB + RC + RD = q.L + P1 + P2
13,85 + 31,51 + 68,30 + 7,14 = 76,8 + 32 + 12,0
120,80 = 120,80 ok !!!
Momen Maksimum
Bentang BC
Mx=RB(x)-q (x)2
Dx=RB-q.(x)=0
x= Rb
q
31,51
x= ------ = 3,28 m dari Titik B ke C
9,6
Mmak=Rb.(x)-(q)(x)2
Mmax = 31,51 x 3,28 - 0,5 x 9,6 3,28 ^2
= 103,43 - 51,714
= 51,714 kNm
Bentang CD
MC=RCD(3)-MCD
= 24,86 x 3 - 53,18
= 74,59 - 53,18
= 21,41 KnM
P1= 12 kN q= 9,6 kN/m P2= 32 kN
E 2Ic 4Ic F 3Ic
A B C D
2m L2= 6 L3= 8 m 3 3
53,18
24
Diagram
12,88
(-) Bidang Momen (-)
(+)
(+) 21,41
Mmax = 51,71
Diagram
31,51
Bidang Lintang 24,86
(+)
(+)
1,85 (+)
(-) (-)
(-) 7,14
12,00
43,44
CHECKING DENGAN APLIKASI STAAD_PRO
P2 =32 kN
q = 9,6 kN/m
P1 =12 kN
A B C D
DIAGRAM BID. MOMEN
DIAGRAM BID. LINTANG
REAKSI PERLETAKAN
A B C D
HASIL PERBANDINGAN CARA MANUAL DAN PROGRAM STAAD-PRO
GAYA MANUAL STAAD-PRO SELISIH
RA 13,853 13,846 0,007
RB 31,511 31,532 0,021
RC 68,300 68,273 0,027
RD 7,137 7,149 0,012
MA -24,000 -24 0,000
MB -12,884 -12,926 0,042
MC -53,179 -53,103 0,076
.OK !!!
