01 pendahuluan Statika Struktur.pdf

8/18/2019 01 pendahuluan Statika Struktur.pdf

1/35

PESAWAT ANGKAT

Tito Shantika, ST.,M Eng

Mechanical Engineeri ng Department I nstitut Teknologi Nasional


2/35

Apa pentingnya mekanika (statik) /

eseim angan


3/35


4/35

Keseimban an

=

eseim angan gaya

Keseimbangan Momen

= 0 M


5/35

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


6/35

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


7/35

Benda Tegar Biasanya Memiliki Tumpuan


8/35

Benda Tegar Biasanya Memiliki Tumpuan


9/35

Contoh Partikel


10/35

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


11/35

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,


12/35

Contoh Benda Te ar


13/35

Macam-macam Tumpuan dan Reaksinya


14/35

Contoh Menggambar FBD nya

Idealized modelFree body diagram

Lho kok ada beban yang segiempat, apa itu?


15/35

Beban Terdistribusi


16/35

Mencari Gaya Resultan pada Beban Terdistribusi

Mencari titik berat dari beban terdistribusi

terdistribusi Gaya resultan terletak pada titik berat beban terdisribusi


17/35

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


18/35

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?


19/35

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.

. .


20/35

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


21/35

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.


22/35

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.


23/35

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


24/35

Contoh Soal 4

Given: The loading on the beam as

shown.

Find: Support reactions at A and B.


25/35

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


26/35

Contoh Soal 5

Tentukan Reaksi di A dan B


27/35

Soal Tantangan

Given: The loading on the beam as shown.

n : eact on at an


28/35

Soal Tantangan (2)

Tentukan Reaksi di A dan C


29/35

Analysis of Structures

Trusses

FramesMachines

Tito Shantika, ST.,M Eng

Mechanical Engineeri ng Department

I nstitut Teknologi Nasional


30/35

•

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.


31/35

Definition of a Truss


32/35

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.


33/35

Analysis of Trusses by the Method of

o n s


34/35

Frames Which Cease To Be Rigid When


35/35

01 pendahuluan Statika Struktur.pdf

Documents

Transcript of 01 pendahuluan Statika Struktur.pdf