Konstruksi Balok Sederhana
Contoh soal dan penyelesaian MK Statika pelengkung tiga sendi
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S
c=4m
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L= 8m
Ditanyakan ; a. reaksi perletakan
b. bidang M,D,N
Penyelesaian:
Analisis
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S
c=4m
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RAV RBV
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L= 8m
a. Reaksi perletakan
Yang perlu selalu diingat dalam penyelasaian kasus-kasus pada statika tertentu adalah :
∑M=0
∑V=0
∑H=0
Dalam kasus ini yang diselesaikan adalah :
∑MA=0
∑MB=0
∑MS=0 (bagian A-S atau B-S, dan salah satunya digunakan untuk control atau mengecek apakah sudah ok apa belum)
Q=q.L = 1 t/m . 8m = 8t
∑MA=0
0 = -RBV.L + Q.a
0 = - RBV.8 + 8.4
RBV = 32/8 = 4t
∑MB=0
0 = -RAV.L + Q.b
0 = - RAV.8 + 8.4
RAV = 32/8 = 4t
∑MS=0 (bentang A-S)
0 = - q.a. ½ a + RAV.a – RAH.c
0 = - 1.4. ½ 4 + 4.4 – RAH.4
0 = - 8 + 16 – RAH.4
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∑H=0
0 = RAH – RBH
0 = 2 – RBH
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Control
∑MS=0 (bentang B-S)
0 = - q.b. ½ b + RBV.b – RBH.c
0 = - 1.4. ½ 4 + 4.4 – RBH.4
0 = - 8 + 16 – 2.4
0 = 0 ( ok )
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S
c=4m
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RAV= 4t RBV= 4t
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L= 8m
b. Bidang M,D,N
Untuk menyelesaikan bidang M,D,N selalu mengadakan Potongan…dan peninjauan
Adakan potongan sejauh x dari A (bentang A-S) dan tinjau kiri
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RAV= 4t
qx= q.x = x
Mx = +RAV.x – qx. ½ x – RAH.y
= + 4x – ½ x2 – 2y 0≤ x ≤ 4
X=0 Mx = 0
X=4 Mx = 0
Mx max atau titik puncak momen
dMx/dx = 0 = 4 – x – 2
puncak saat x = 2 dari A
Mx max = 4.2 – ½ 22 – 2.2 = 2 tm
Untuk bidang D
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RAV= 4t RAV cos α
Dx= RAVcosα – RAHsinα – qxcos2α
= 4. ½ √2 – 2. ½ √2 – x. ½
= 2√2 - √2 – ½ x 0≤x≤4
X=0 Dx= √2= 1,41
X=4 Dx = √2 – 2= - 0,59
Untuk bidang N
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+ RAVsinα X
RAV= 4t
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= 2. ½ √2 + 4. ½ √2 – ½ x
= √2 + 2√2 – ½ x
= 3√2 – ½ x 0 ≤x ≤4
x=0 Nx = 3√2=4,23
x=4 Nx = 2,23
karena kedua bentang sama (A-S dan B-S).. maka tidak perlu susah-susah menghitung lagi karena hasilnya akan sama. Hanya saja tandanya akan berbeda dalam proses penyelesaian.
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S
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C=4m
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RAV= 4t RBV=4t
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a=4m b=4m
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+ + Bidang Momen (M)
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- -
+ + Bidang Gaya Lintang (D)
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+ + Bidang Gaya Normal (N)
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