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8/18/2019 01 pendahuluan Statika Struktur.pdf
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PESAWAT ANGKAT
Tito Shantika, ST.,M Eng
Mechanical Engineeri ng Department I nstitut Teknologi Nasional
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Apa pentingnya mekanika (statik) /
eseim angan
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Keseimban an
=
eseim angan gaya
Keseimbangan Momen
= 0 M
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A a erbedaan Partikel dan Benda Te ar ?
Partikel: Benda Te ar:
Mempunyai suatu massanamun ukurannya dapat
Kombinasi sejumlahpartikel yang mana
diabaikan, sehingga
geometri benda tidak
semua partikel berada
pada suatu jarak tetap
analisis masalah
yang lain
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Apa Beda Partikel dengan Benda Tegar ?
In contrast to the forces on a particle, the
forces on a rigid-body are not usuallyconcurrent and may cause rotation of the
forces).
Forces on a particle
For a rigid body to be in equilibrium, the
net force as well as the net moment
equal to zero.
= =
Forces on a rigid body
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Benda Tegar Biasanya Memiliki Tumpuan
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Benda Tegar Biasanya Memiliki Tumpuan
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Contoh Partikel
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O erasi Vektor• Trapezoid rule for vector addition
• Triangle rule for vector addition
• Law of cosines,
B
C Q P R
B PQQ P Rrrr
+=
−+= cos2222
B
• Law of sines,
P
A
R
B
Q
C sinsinsin==
• Vector addition is commutative,
P QQ P rrrr
+=+
• Vector subtraction
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Penjumlahan gaya-gaya
S Q P Rrrrr
++=
concurrent forces,
S P iS P
jS iS jQiQ j P i P j Ri R y x y x y x y xrr
rrrrrrrr
+++++=
+++++=+
• eso ve eac orce into rectangu ar components
• The scalar components of the resultant are equal
to the sum of the corresponding scalar
components of the given forces.
∑=++=
x
x x x x
F
S Q P R
∑=++=
y
y y y y
F
S Q P R
x
y y x
R R R R R 122 tan−=+= θ
• To find the resultant magnitude and direction,
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Contoh Benda Te ar
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Macam-macam Tumpuan dan Reaksinya
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Contoh Menggambar FBD nya
Idealized modelFree body diagram
Lho kok ada beban yang segiempat, apa itu?
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Beban Terdistribusi
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Mencari Gaya Resultan pada Beban Terdistribusi
Mencari titik berat dari beban terdistribusi
terdistribusi Gaya resultan terletak pada titik berat beban terdisribusi
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Kalau beban terdistribusinya berbentuk segitiga ?
FR 100 N/m
=
m x
. R = ____________
A) 12 N B) 100 N
. __________.
A) 3 m B) 4 m
C) 600 N D) 1200 N m m
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Prosedur Menyelesaikan Soal
Gambar FBD dari soal
angan upa as per an an tan anya
Gambar gaya reaksi yang ada
,resultan, dan posisinya
Hitung besar gaya reaksi di tumpuan, menggunakan
∑ Fx = 0 ∑ Fy = 0 ∑ Mo = 0
titik O itu titik a a? Yan mana?
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Contoh Soal 1
Given: Weight of the boom =
125 lb, the center of
,
load = 600 lb.
Find: Su ort reactions at A
and B.
Plan:
1. Put the x and y axes in the horizontal and vertical directions,
respectively.
2. Draw a complete FBD of the boom.
. .
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Contoh Soal 1 (Jawaban)
A
AY
A
FBD of the boom:
1 ft1 ft 3 ft 5 ft
B G D
40°
600 lb125 lbFB
+ ∑MA = - 125 ∗ 4 - 600 ∗ 9 + FB sin 40° ∗ 1 + FB cos 40° ∗ 1 = 0
FB = 4188 lb or 4190 lb
→ + ∑FX = AX + 4188 cos 40° = 0; AX = – 3210 lb
+ ∑FY = AY + 4188 sin 40° – 125 – 600 = 0; AY = – 1970 lb
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Contoh Soal 2
SOLUTION:
• Create a free-body diagram for the crane.
• Determine B by solving the equation forthe sum of the moments of all forces
about . Note there will be no
contribution from the unknown
reactions at A.
A fixed crane has a mass of 1000 kg
and is used to lift a 2400 kg crate. It
is held in place by a pin at and a
solving the equations for the sum of
all horizontal force components and
rocker at B. The center of gravity ofthe crane is located at G.
.
• Check the values obtained for the
reactions by verifying that the sum of
reactions at A and B.the moments about B of all forces is
zero.
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Contoh Soal 2 (jawaban)
• Determine B by solving the equation for the
sum of the moments of all forces about A.
m2kN81.9m5.1:0 −+= B
( ) 0m6kN5.23 =−kN1.107+= B
• Create the free-bod dia ram.
• Determine the reactions at A by solving the
equations for the sum of all horizontal forcesan a ver ca orces.
0:0 =+=∑ B A F x x
kN1.107−=
0kN5.23kN81.9:0 =−−=∑ y y A F
kN3.33+= A
• Check the values obtained.
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Contoh Soal 3
−−==Σ )500(.225)275(.120)(400.0 y A m N m N Bm M
+
−
−−=
400.
.. y
m B
−+−==Σ
=
22575.3631200. y
N N N Ay Fy+
↓=
−=
N18.75
.
y
y
A
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Contoh Soal 4
Given: The loading on the beam as
shown.
Find: Support reactions at A and B.
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Contoh Soal 4 (jawaban)
SOLUTION:
• Taking entire beam as a free-body,
determine reactions at supports. :0∑ = y F
∑ = :0( ) ( )( ) ( )( )m.24kN45m.81kN90m.27
=−
−− D
=−+−− y
kN81= y A
.
kN117=
D
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Contoh Soal 5
Tentukan Reaksi di A dan B
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Soal Tantangan
Given: The loading on the beam as shown.
n : eact on at an
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Soal Tantangan (2)
Tentukan Reaksi di A dan C
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Analysis of Structures
Trusses
FramesMachines
Tito Shantika, ST.,M Eng
Mechanical Engineeri ng Department
I nstitut Teknologi Nasional
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•
joints. No member is continuous through a joint.
•
together to form a space framework. Each truss
carries those loads which act in its plane and may
be treated as a two-dimensional structure.
• Bolted or welded connections are assumed to be
pinned together. Forces acting at the member ends
reduce to a single force and no couple. Only two-force members are considered.
• When forces tend to pull the member apart, it is in
tension. When the forces tend to compress the
member, it is in compression.
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Definition of a Truss
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Sim le Trusses
• A ri id truss will not colla se under the
application of a load.
• A simple truss is constructed by
successively adding two members and
one connection to the basic triangular
truss.
• In a simple truss, m = 2n - 3 where
m is the total number of members
and n is the number of joints.
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Analysis of Trusses by the Method of
o n s
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Frames Which Cease To Be Rigid When
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