The document discusses portal structures, which are commonly used in warehouse, hangar, and bridge construction. It covers symmetric and asymmetric portal structures that carry various load combinations, including centered vertical loads, horizontal loads, and distributed loads. Methods for calculating the reactions, shear forces, bending moments, and normal stresses in the structural elements are presented. Free body diagrams are used to illustrate the distribution of internal forces.
Perkerasan Jalan Raya Lentur dan Kaku, metode Analisis dan Manual
ANGGOTA KELOMPOK :
DHANES PRABASWARA ( I 0112029)
AYU ISMOYO SOFIANA ( I 0113021)
MUHAMMAD BUDI SANTOSO( I 0113080)
RAKE ADIUTO ( I 0113105)
SITI DWI RAHAYU ( I 0113124)
Perkerasan Jalan Raya Lentur dan Kaku, metode Analisis dan Manual
ANGGOTA KELOMPOK :
DHANES PRABASWARA ( I 0112029)
AYU ISMOYO SOFIANA ( I 0113021)
MUHAMMAD BUDI SANTOSO( I 0113080)
RAKE ADIUTO ( I 0113105)
SITI DWI RAHAYU ( I 0113124)
From the Front LinesOur robotic equipment and its maintenanc.docxhanneloremccaffery
From the Front Lines
Our robotic equipment and its maintenance represent a fixed cost of $23,320 per month. The cost-effectiveness of robotic-assisted surgery is related to patient volume: With only 10 cases, the fixed cost per case is $2,332, and with 40 cases, the fixed cost per case is $583.
Source: Alemozaffar, Chang, Kacker, Sun, DeWolf, & Wagner (2013).
MATLAB sessions: Laboratory 5
MAT 275 Laboratory 5
The Mass-Spring System
In this laboratory we will examine harmonic oscillation. We will model the motion of a mass-spring
system with differential equations.
Our objectives are as follows:
1. Determine the effect of parameters on the solutions of differential equations.
2. Determine the behavior of the mass-spring system from the graph of the solution.
3. Determine the effect of the parameters on the behavior of the mass-spring.
The primary MATLAB command used is the ode45 function.
Mass-Spring System without Damping
The motion of a mass suspended to a vertical spring can be described as follows. When the spring is
not loaded it has length ℓ0 (situation (a)). When a mass m is attached to its lower end it has length ℓ
(situation (b)). From the first principle of mechanics we then obtain
mg︸︷︷︸
downward weight force
+ −k(ℓ − ℓ0)︸ ︷︷ ︸
upward tension force
= 0. (L5.1)
The term g measures the gravitational acceleration (g ≃ 9.8m/s2 ≃ 32ft/s2). The quantity k is a spring
constant measuring its stiffness. We now pull downwards on the mass by an amount y and let the mass
go (situation (c)). We expect the mass to oscillate around the position y = 0. The second principle of
mechanics yields
mg︸︷︷︸
weight
+ −k(ℓ + y − ℓ0)︸ ︷︷ ︸
upward tension force
= m
d2(ℓ + y)
dt2︸ ︷︷ ︸
acceleration of mass
, i.e., m
d2y
dt2
+ ky = 0 (L5.2)
using (L5.1). This ODE is second-order.
(a) (b) (c) (d)
y
ℓ
ℓ0
m
k
γ
Equation (L5.2) is rewritten
d2y
dt2
+ ω20y = 0 (L5.3)
c⃝2011 Stefania Tracogna, SoMSS, ASU
MATLAB sessions: Laboratory 5
where ω20 = k/m. Equation (L5.3) models simple harmonic motion. A numerical solution with ini-
tial conditions y(0) = 0.1 meter and y′(0) = 0 (i.e., the mass is initially stretched downward 10cms
and released, see setting (c) in figure) is obtained by first reducing the ODE to first-order ODEs (see
Laboratory 4).
Let v = y′. Then v′ = y′′ = −ω20y = −4y. Also v(0) = y′(0) = 0. The following MATLAB program
implements the problem (with ω0 = 2).
function LAB05ex1
m = 1; % mass [kg]
k = 4; % spring constant [N/m]
omega0 = sqrt(k/m);
y0 = 0.1; v0 = 0; % initial conditions
[t,Y] = ode45(@f,[0,10],[y0,v0],[],omega0); % solve for 0<t<10
y = Y(:,1); v = Y(:,2); % retrieve y, v from Y
figure(1); plot(t,y,’b+-’,t,v,’ro-’); % time series for y and v
grid on;
%-----------------------------------------
function dYdt = f(t,Y,omega0)
y = Y(1); v = Y(2);
dYdt = [ v ; -omega0^2*y ];
Note that the parameter ω0 was passed as an argument to ode45 rather than set to its value ω0 = 2
directly in the funct ...
enjoy the formulas and use it with convidence and make your PT3 AND SPM more easier..togrther we achieve the better:)
good luck guys and girls...simple and short ans also sweet formulas..
Multi-cluster Kubernetes Networking- Patterns, Projects and GuidelinesSanjeev Rampal
Talk presented at Kubernetes Community Day, New York, May 2024.
Technical summary of Multi-Cluster Kubernetes Networking architectures with focus on 4 key topics.
1) Key patterns for Multi-cluster architectures
2) Architectural comparison of several OSS/ CNCF projects to address these patterns
3) Evolution trends for the APIs of these projects
4) Some design recommendations & guidelines for adopting/ deploying these solutions.
1.Wireless Communication System_Wireless communication is a broad term that i...JeyaPerumal1
Wireless communication involves the transmission of information over a distance without the help of wires, cables or any other forms of electrical conductors.
Wireless communication is a broad term that incorporates all procedures and forms of connecting and communicating between two or more devices using a wireless signal through wireless communication technologies and devices.
Features of Wireless Communication
The evolution of wireless technology has brought many advancements with its effective features.
The transmitted distance can be anywhere between a few meters (for example, a television's remote control) and thousands of kilometers (for example, radio communication).
Wireless communication can be used for cellular telephony, wireless access to the internet, wireless home networking, and so on.
ER(Entity Relationship) Diagram for online shopping - TAEHimani415946
https://bit.ly/3KACoyV
The ER diagram for the project is the foundation for the building of the database of the project. The properties, datatypes, and attributes are defined by the ER diagram.
This 7-second Brain Wave Ritual Attracts Money To You.!nirahealhty
Discover the power of a simple 7-second brain wave ritual that can attract wealth and abundance into your life. By tapping into specific brain frequencies, this technique helps you manifest financial success effortlessly. Ready to transform your financial future? Try this powerful ritual and start attracting money today!
1. STATIKA I
MODUL 7
BANGUNAN PORTAL
Dosen Pengasuh :
Ir. Thamrin Nasution
Materi Pembelajaran :
1. Portal Simetris.
a) Memikul muatan terpusat tunggal.
b) Memikul muatan vertikal dan horisontal.
c) Memikul muatan campuran.
2. Portal Tidak Simetris.
a) Kolom miring sebelah, memikul muatan terpusat vertikal dan horisontal.
b) Kolom tinggi sebelah, memikul muatan terpusat vertikal dan horisontal.
c) Kolom tinggi sebelah, balok overhang, memikul muatan terbagi rata, terpusat
vertikal dan horisontal.
WORKSHOP/PELATIHAN
Tujuan Pembelajaran :
Mahasiswa memahami dan mengetahui tentang gaya-gaya dalam dari struktur portal
simetris,struktur portal tidak simetris, struktur portal dengan overhang, dengan berbagai
beban.
DAFTAR PUSTAKA
a) Soemono, Ir., “STATIKA 1”, Edisi kedua, Cetakan ke-4, Penerbit ITB, Bandung, 1985.
2. thamrinnst.wordpress.com
UCAPAN TERIMA KASIH
Penulis mengucapkan terima kasih yang sebesar-besarnya kepada
pemilik hak cipta photo-photo, buku-buku rujukan dan artikel, yang terlampir
dalam modul pembelajaran ini.
Semoga modul pembelajaran ini bermanfaat.
Wassalam
Penulis
Thamrin Nasution
thamrinnst.wordpress.com
thamrin_nst@hotmail.co.id
3. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
1
BANGUNAN PORTAL
Bangunan portal banyak dijumpai sebagai konstruksi bangunan gudang, hangar dan
jembatan. Prinsip perhitungan tetap sama dengan balok diatas dua perletakan selagi masih
dalam bentuk statis tertentu, yaitu V = 0, H = 0 dan M = 0.
1. PORTAL SIMETRIS.
a). Memikul muatan terpusat tunggal.
Gambar 1 : Bangunan portal simettris, memikul beban terpusat P
Penyelesaian :
a. Reaksi Perletakan.
MB = 0,
RAV . L - P . b = 0
RAV = P . b/L (ton).
MA = 0,
- RBV . L + P . a = 0
RBV = P . a/L (ton).
Kontrol :
V = 0,
RAV + RBV – P = 0
b. Gaya lintang.
DA-C = 0
DC-E = + RAV (ton).
DE-D = + RAV – P (ton).
DD-B = 0
P
ba
(A) (B)
h
RAV RBV
L
(C)
(E)
(D)
4. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
2
c. M o m e n .
MA = 0
MC = 0
ME = RAV . a = P.a.b/L
MD = RAV . L – P . b = + P.b/L . L – P . b = 0
MB = 0
d. Gaya Normal.
NA-C = – RAV (ton).
NC-D = 0
NB-D = – RBV (ton).
Gambar 2 : Bidang gaya lintang, momen dan gaya normal.
ba
(C)
(A) (B)
(E)
(D)
h
L
RAV
RBV
P
ba
(C)
(A) (B)
(E)
(D)
h
L
P.a.b/L
Bidang Gaya Lintang (D) Bidang Momen (M)
ba
(C)
(A) (B)
(E)
(D)
h
L
RBV
Bidang Gaya Normal (N)
+
–
+
– –
RAV
5. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
3
b). Memikul muatan terpusat vertikal dan horisontal.
Gambar 3 : Bangunan portal simettris, memikul beban terpusat vertikal P1 dan
beban horisontal P2.
Penyelesaian :
a. Reaksi Perletakan.
MB = 0,
RAV . L – RAH . 0 – P1 . b + P2 . d = 0
RAV = P1 . b/L – P2 . d/L (ton).
MA = 0,
– RBV . L + P1 . a + P2 . d = 0
RBV = P1 . a/L + P2 . d/L (ton).
H = 0,
– RAH + P2 = 0
RAH = P2 (ton, kekiri).
Kontrol :
V = 0,
RAV + RBV – P1 = 0
b. Gaya lintang.
DA-C = + RAH (ton).
DC-E = + RAV (ton).
DE-D = + RAV – P1 (ton).
DD-F = – RAH (ton).
DF-B = – RAH + P2 = 0 (ton).
c. M o m e n .
MA = 0
MC = + RAH . h (t.m’).
ME = RAV . a + RAH . h (t.m’).
MD = RAV . L + RAH . h – P1 . b (t.m’).
MF = RAV . L + RAH . d – P1 . b (t.m’) = 0 .
MB = 0
ba
(A) (B)
h
RAV RBV
L
(C)
(E)
(D)
(F)
P1
c
P2
RAH
d
6. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
4
d. Gaya Normal.
NA-C = – RAV ton (tekan, kalau reaksi RAV keatas).
NC-D = + RAH ton (tarik).
NB-D = – RBV ton (tekan).
Gambar 4 : Bidang gaya lintang (D), momen (M) dan gaya normal (N).
= mata, cara penglihatan, untuk menggambarkan gaya-gaya dalam dan menetapkan
tanda negatip/positip.
ba
(C)
(A) (B)
(E) (D)
h
L
Bidang Momen (M)
ba
(C)
(A) (B)
(E)
(D)
h
L
RBV
Bidang Gaya Normal (N)
+
– –
RAV
+
+ +
RAH
(F)
RAV
(D)(C)
(E)
(A)
RAH
+
(F)
ba
(B)
L
RBV
P
Bidang Gaya Lintang (D)
–
+
(F)
RAH
c
d
7. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
5
Gambar 5 : Distribusi gaya-gaya pada elemen struktur (balok/kolom) dengan cara freebody.
c). Memikul muatan campuran.
Gambar 6 : Bangunan portal simettris, memikul beban terpusat vertikal P1,
beban horisontal P2 dan q t/m’.
Diketahui : Konstruksi seperti tergambar.
L = 10 m, h = 5 m, a = 4 m, b = 6 m, c = 2m, d = 3 m.
P1 = 5 ton, P2 = 2 ton, q = 1 t/m’.
Diminta : Hitung dan gambarkan M, D dan N pada seluruh bentang.
Penyelesaian :
a. Reaksi Perletakan.
MB = 0,
RAV . L – RAH . 0 + q . h . ½ h – P1 . b + P2 . d = 0
RAV = P1 . b/L – P2 . d/L – ½ q . h2
/L = 5 . 6/10 – 2 . 3/10 – ½ . 1 . 52
/10
= 3 – 0,6 – 1,25
RAV = + 1,15 ton (keatas).
ba
(A)
(B)
h
RAV RBV
L
(C)
(E)
(D)
(F)
P1
c
P2
RAH
d
y
q t/m’
RAH
RAV
(A) RAH
RAV
(C)
(C)
RAV
RBV
RAHRAH
P2
P1
(D)
(D)
(B)
RBV
(E)
(F)
RAH
8. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
6
MA = 0,
– RBV . L + P1 . a + P2 . d + q . h . ½ h = 0
RBV = P1 . a/L + P2 . d/L + ½ q . h2
/L = 5 . 4/10 + 2 . 3/10 + ½ . 1 . 52
/10
= 2 + 0,6 + 1,25
RBV = + 3,85 ton (keatas).
H = 0,
– RAH + q . h + P2 = 0
RAH = P2 + q . h = 2 + 1 . 5
RAH = + 7 ton (kekiri).
Kontrol :
V = 0,
RAV + RBV – P1 = 0
1,15 + 3,85 – 5 = 0 (memenuhi).
b. Gaya lintang.
DAC = + RAH = + 7 (ton).
Dy = + RAH – q.y
Dy=5m = DCA = + RAH – q . h = + 7 – 1 . 5 = + 2 (ton).
DC-E = + RAV = + 1,15 (ton).
DE-D = + RAV – P1 = 1,15 – 5 = – 3,85 (ton).
DD-F = – RAH + q.h = – 7 + 1 . 5 = – 2 (ton), atau
DD-F = – P2 = – 2 (ton)
c. M o m e n .
MA = 0
My = + RAH . y – ½ . q . y2
MC = + RAH . h – ½ q . h2
= 7 . 5 – ½ . 1 . 52
= 35 – 12,5 = + 22,5 (t.m’).
ME = + RAH . h – ½ q . h2
+ RAV . a = + 7 . 5 – ½ . 1 . 52 + 1,15 . 4
= 35 – 12,5 + 4,6
ME = + 27,1 (t.m’).
MD = + RAH . h – ½ q . h2
+ RAV . L – P1 . b
= + 7 . 5 – ½ . 1 . 52 + 1,15 . 10 – 5 . 6 = 35 – 12,5 + 11,5 – 30
MD = + 4 (t.m’).
Atau,
MD = + P2 . c = + 2 . 2 = + 4 (t.m’).
MF = + RAH . d – q . h . (d – ½ h) + RAV . L – P1 . b
= + 7 . 3 – 1 . 5 . (3 – ½ . 5) + 1,15 . 10 – 5 . 6
= 21 – 2,5 + 11,5 – 30
MF = 0
MB = 0
d. Gaya Normal.
NA-C = – RAV = – 1,15 ton (tekan).
NC-D = + RAH – q . h = 7 – 1 . 5 = + 2 ton (tarik).
NB-D = – RBV = – 3,85 ton (tekan).
9. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
7
Gambar 7 : Bidang gaya lintang (D), momen (M) dan gaya normal (N).
= mata, cara penglihatan, untuk menggambarkan gaya-gaya dalam dan mene-
tapkan tanda negatip/positip.
ba
(C)
(A) (B)
(E) (D)
h
L
Bidang Momen (M)
ba
(C)
(A) (B)
(E)
(D)
h
L
Bidang Gaya Normal (N)
+
– –
+
+
(F)
(D)
(C) (E)
+
(F)
ba
(A) (B)
h
L
(F)
+
+ 7 t
+ 1,15 t
+ 2 t
+
- 3,85 t
–
–
- 2 t
c
d
Bidang Gaya Lintang (D)
+22,5 t.m’
+
+27,1 t.m’
+4 t.m’
- 1,15 t
+ 2 t
- 3,85 t
+ 2 t
10. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
8
Gambar 8 : Distribusi gaya-gaya pada elemen struktur (balok/kolom) dengan cara freebody.
2. PORTAL TIDAK SIMETRIS.
a). Kolom miring sebelah, memikul muatan terpusat vertikal dan horisontal.
Gambar 9 : Bangunan portal tidak simettris, memikul beban terpusat vertikal P1,
beban horisontal P2.
Diketahui : Konstruksi seperti tergambar.
L = 10 m, h = 5 m, a = 2 m, b = 2 m, c = 6 m, d = 2 m, e = 3 m.
P1 = 5 ton, P2 = 2 ton.
Diminta : Hitung dan gambarkan M, D dan N pada seluruh bentang.
Penyelesaian :
a. Reaksi Perletakan.
MB = 0,
ba
(A)
(B)
h
RAV RBV
L
(C)
(E) (D)
(F)
P1
d
P2
RAH
e
c
(A)
RAH = 7 t
RAV = 1,15 t
(C)
(C)
1,15 t
P2 = 2 t
P1 = 5 t
(D)
(D)
(B)
RBV = 3,85 t
(E)
(F)
q = 1 t/m
2 t 2 t
1,15 t 3,85 t
2 t 2 t
3,85 t
11. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
9
RAV . L – RAH . 0 – P1 . c + P2 . e = 0
RAV = P1 . c/L – P2 . e/L = 5 . 6/10 – 2 . 3/10 = 3 – 0,6
RAV = + 2,4 ton (keatas).
MA = 0,
– RBV . L + P1 . (a + b) + P2 . e = 0
RBV = P1 . (a+b)/L + P2 . e/L = 5 . (2 + 2)/10 + 2 . 3/10 = 2 + 0,6
RBV = + 2,6 ton (keatas).
H = 0,
– RAH + P2 = 0
RAH = P2 = + 2 ton (kekiri).
Kontrol :
V = 0,
RAV + RBV – P1 = 0
2,4 + 2,6 – 5 = 0 (memenuhi).
Gambar 10 : Komponen reaksi vertikal dan horisontal.
b. Gaya lintang.
= arc tan (h/a) = arc tan (5/2)
= 68o
11’ 55”
DA-C = + RAH sin + RAV cos = 2 . sin(68o
11’55”) + 2,4 . cos(68o
11’55”)
= + 1,857 + 0,891 = + 2,748 (ton).
DC-E = + RAV = + 2,4 (ton).
DE-D = + RAV – P1 = 2,4 – 5 = – 2,6 (ton) = – RBV.
DD-F = – RAH = – 2 (ton).
c. M o m e n .
MA = 0
MC = + RAH . h + RAV . a = 2 . 5 + 2,4 . 2 = + 14,8 (t.m’).
ME = RAH . h + RAV . (a + b) = 2 . 5 + 2,4 . (2 + 2) = + 19,6 (t.m’).
MD = RAH . h + RAV . L – P1 . c = 2 . 5 + 2,4 . 10 – 5. 6 = + 4 (t.m’).
Atau,
MD = + P2 . d = + 2 . 2 = + 4 (t.m’).
a
(A)
h
(C)
RAH
RAV
RAV Cos
RAV Sin
RAH Sin
RAH Cos
12. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
10
MF = RAH . e + RAV . L – P1 . c = 2 . 3 + 2,4 . 10 – 5. 6 = 0 + P’).
MB = 0
d. Gaya Normal.
= 68o
11’ 55”
NA-C = + RAH cos + RAV sin
= 2 . cos(68o
11’55”) – 2,4 . sin(68o
11’55”)
= + 0,743 – 2,228
= – 1,485 ton (tekan).
NC-D = + RAH = + 2 ton (tarik).
NB-D = – RBV = – 2,6 ton (tekan).
Gambar 11.a. : Bidang gaya lintang.
Gambar 11.b. : Bidang momen.
ba
(A) (B)
h
L
(C) (E) (D)
(F)
d
e
c
Bidang Momen (M)
+
+4 t.m’+ 14,8 t.m’
+
+
+ 19,6 t.m’
+ 14,8 t.m’
+4 t.m’
ba
(A) (B)
h
L
(C) (E)
(D)
(F)
d
e
c
Bidang Gaya Lintang (D)
+
–
- 2,60 t
2,40 t
- 2,0 t+
2,748 t
2,748 t
13. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
11
Gambar 11.c. : Bidang gaya normal.
Gambar 12 : Distribusi gaya-gaya pada elemen struktur (balok/kolom) dengan cara freebody.
(A)
(B)
RAV = 2,4 t RBV = 2,6 t
(C)
(E)
(D)
(F)
P1 = 5 t
P2 = 2 t
RAH = 2 t
2 t
1,485 t
2,748 t
2,4 t
2 t
2,748 t
2,4 t
1,485 t
(C) (D) 2 t 2 t
2,6 t
2,6 t
ba
(A) (B)
h
L
(C)
(E)
(D)
(F)
d
e
c
Bidang Gaya Normal (N)
+
- 2,60 t
2,0 t
- 1,485 t
––
- 1,485 t
2,0 t
14. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
12
b). Kolom tinggi sebelah, memikul muatan terpusat vertikal dan horisontal.
Gambar 13 : Portal dengan kaki tinggi sebelah.
Diketahui : Konstruksi seperti tergambar.
L = 10 m, h1 = 5 m, h2 = 6 m, a = 4 m, b = 6 m, c = 2m, d = 4 m.
P1 = 5 ton, P2 = 2 ton.
Diminta : Hitung dan gambarkan M, D dan N pada seluruh bentang.
Penyelesaian :
a. Reaksi Perletakan.
H = 0,
– RAH + P2 = 0
RAH = + P2 = + 2 ton (kekiri).
MB = 0,
RAV . L – RAH . (h2 – h1) – P1 . b + P2 . d = 0
RAV = P1 . b/L + RAH . (h2 – h1)/L – P2 . d/L (ton).
= 5 . 6/10 + 2 . (6 – 5)/10 – 2 . 4/10 = 3 + 0,2 – 0,8
RAV = + 2,4 ton (keatas).
MA = 0,
– RBV . L + P1 . a + P2 . (h1 – c) = 0
RBV = P1 . a/L + P2 . (h1 – c)/L
= 5 . 4/10 + 2 . (5 – 2)/10 = 2 + 0,6
RBV = + 2,6 ton (keatas).
Kontrol :
V = 0,
RAV + RBV – P1 = 0
2,4 + 2,6 – 5 = 0 …..(memenuhi).
b. Gaya lintang.
ba
(A)
h1
RAV
RBV
L
(C)
(E)
(D)
(F)
c
P2
RAH
d
h2
(B)
P1
15. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
13
DA-C = + RAH = + 2 ton.
DC-E = + RAV = + 2,4 ton.
DE-D = + RAV – P1 = + 2,4 – 5 = – 2,6 ton.
DD-F = – RAH = – 2 ton, atau
DD-F = – P2 = – 2 ton.
c. M o m e n .
MA = 0
MC = + RAH . h = + 2 . 5 = + 10 t.m’.
ME = + RAV . a + RAH . h = + 2,4 . 4 + 2 . 5 = 19,6 t.m’.
MD = + RAV . L + RAH . h – P1 . b = 2,4 . 10 + 2 . 5 – 5 . 6
= 24 + 10 – 30
MD = + 4 t.m’.
Atau,
MD = + P2 . c = + 2 . 2 = + 4 t.m’.
MF = + RAV . L + RAH . (h1 – c) – P1 . b (t.m’) (dari kiri)
= + 2,4 . 10 + 2 . (5 – 2) – 5 . 6 = 24 + 6 – 30
MF = 0
MB = 0
d. Gaya Normal.
NA-C = – RAV = – 2,4 ton (tekan).
NC-D = + RAH = + 2 ton (tarik).
NB-D = – RBV = – 2,6 ton (tekan).
Gambar 14.a : Bidang gaya lintang dan momen.
ba
(C)
(A) (B)
(E) (D)
L
Bidang Momen (M)
++
+
(D)
(C) (E)
ba
(A)
(B)
h2
L
(F)
+ 2,4 t
+ 2 t
+
- 2,6 t
–
–
- 2 t
c
d
Bidang Gaya Lintang (D)
+10 t.m’
+
+19,6 t.m’
+4 t.m’
+ 2 t
+ h1
(F)
16. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
14
Gambar 14.b : Bidang gaya normal.
Gambar 15 : Distribusi gaya-gaya pada elemen struktur (balok/kolom) dengan cara freebody.
(D)(C)
(E)
(A)
(B)
(F)
+ 2 t + 2 t
2,4 t 2,6 t
P1 = 5 t
+ 2 t + 2 t
2 t
2,4 t
2,4 t
2,6 t
2,6 t
P2 = 2 t
(C) (D)
(D)(C)
(E)
ba
(A)
(B)
h2
L
(F)
+ 2 t
+
- 2,6 t
–
c
d
Bidang Gaya Normal (N)
+ 2 t
h1
- 2,4 t
–
17. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
15
c). Kolom tinggi sebelah, balok overhang, memikul muatan terbagi rata, terpusat
vertikal dan horisontal.
Gambar 16 : Portal dengan overhang pada kiri kanan.
Diketahui : Konstruksi seperti tergambar.
L = 10 m, h1 = 5 m, h2 = 6 m, a = 1 m, b = 4 m, c = 6 m, d = 2 m,
e = 2m, f = 4 m, q = 1 t/m’, P1 = 5 ton, P2 = 2 ton.
Diminta : Hitung dan gambarkan M, D dan N pada seluruh bentang.
Penyelesaian :
a. Reaksi Perletakan.
R1 = q . a = 1 . 1 = 1 ton.
R2 = q . b = 1 . 4 = 4 ton.
H = 0,
– RAH + P2 = 0
RAH = + P2 = 2 ton (kekiri).
MB = 0,
RAV . L – RAH . (h2 – h1) – R1 . (L + ½a) – R2 . (L – ½ b) + P1 . d + P2 . f = 0
RAV = RAH . (h2 – h1)/L + R1 . (L + ½a)/L + R2 . (L – ½ b)/L – P1 . b/L
– P2 . d/L
= 2 . (6 – 5)/10 + 1 . (10 + ½ . 1)/10 + 4 . (10 – ½ . 4)/10 – 5 . 2/10
– 2 . 4/10
= 0,2 + 1,05 + 3,2 – 1 – 0,8
RAV = + 2,65 ton (keatas).
MA = 0,
– RBV . L + P1 . (L + d) + P2 . (h1 – e) – R1 . ½a + R2 . ½ b = 0
RBV = P1 . (L + d)/L + P2 . (h1 – e)/L – R1 . ½ a/L + R2 . ½ b/L
= 5 . (10 + 2)/10 + 2 . (5 – 2)/10 – 1 . ½ . 1/10 + 4 . ½ . 4/10
= 6 + 0,6 – 0,05 + 0,8
ba
(A)
h1
RAV
RBV
L
(C) (E)(D) (F)
c
P2
RAH
d
h2
(B)
P1
(G)
(H)
e
f
q t/m’
R1 R2
18. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
16
RBV = + 7,35 ton (keatas).
Kontrol :
V = 0,
RAV + RBV – R1 – R2 – P1 = 0
2,65 + 7,35 – 1 – 4 – 5 = 0 …..(memenuhi).
b. Gaya lintang.
DA-D = + RAH = + 2 ton.
DD-A = DA-D = + 2 ton.
DDC = – q . a = – 1 . 1 = – 1 ton.
DDG = + RAV + DDC = + RAV – q . a = + 2,65 – 1 = + 1,65 ton.
DG-E = DDG – q . b = + RAV – q . (a + b) = + 2,65 – 1 . (1 + 4) = – 2,35 ton.
DE-G = DG-E = – 2,35 ton.
DE-F = DE-G + RBV = + RAV – q . (a + b) + RBV = – 2,35 + 7,35 = + 5 ton.
Atau,
DE-F = + P1 = + 5 ton.
DE-G = – RAH = – 2 ton.
Atau,
DE-G = – 2 ton.
c. M o m e n .
MA = 0
MDA = + RAH . h = + 2 . 5 = + 10 t.m’.
MDC = – ½ q . a2
= – ½ . 1 . 12
= – 0,50 t.m’.
MDG = + RAH . h – ½ q . a2
= + 2 . 5 – ½ . 1 . 1 = + 9,50 t.m’.
MG = + RAV . b + RAH . h – ½ q . (a + b)2
= + 2,65 . 4 + 2 . 5 – ½ . 1 . (1+4)2
= + 10,60 + 10 – 12,50
MG = + 8,10 t.m’.
MEG = + RAV . L + RAH . h1 – R1 . (L + ½ a) – R2 . (L – ½ b)
= + 2,65 . 10 + 2 . 5 – 1 . (10 + ½ . 1) – 4 . (10 – ½ . 4)
= + 26,5 + 10 – 10,5 – 32
MEG = – 6 t.m’.
MEF = – P1 . d = – 5 . 2 = – 10 t.m’.
MEB = + RAV . L + RAH . h1 – R1 . (L + ½ a) – R2 . (L – ½ b) + P1 . d
= + 2,65 . 10 + 2 . 5 – 1 . (10 + ½ . 1) – 4 . (10 – ½ . 4) + 5 . 2
= + 26,5 + 10 – 10,5 – 32 + 10
MEB = + 4 t.m’.
Atau,
MEB = + P2 . e = + 2 . 2 = + 4 t.m’.
MH = + RAV . L + RAH . (h1 – e) – R1 . (L + ½ a) – R2 . (L – ½ b) + P1 . d
= + 2,65 . 10 + 2 . (5 – 2) – 1 . (10 + ½ . 1) – 4 . (10 – ½ . 4) + 5 . 2
= + 26,5 + 6 – 10,5 – 32 + 10
MH = 0 t.m’.
MF = 0 t.m’ ;
MB = 0 t.m’.
Momen maksimum positip pada daerah D-E,
MX1 = RAV . x1 + RAH . h1 – R1 . (a + x1) – ½ . q . x1
2
DX1 = dMX1/dX1 = 0
RAV – R1 – q . x1 = 0
x1 = (RAV – R1)/q
19. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
17
= (2,65 – 1)/1
x1 = 1,65 m (dari D)
Mmaks = 2,65 . 1,65 + 2 . 5 – 1 . (1+ 1,65) – ½ . 1 . (1,65)2
= + 10,36125 t.m’.
Letak titik dimana momen sama dengan nol (dari kiri),
MX2 = RAV . (b + x2) + RAH . h1 – R1 . (½ . a + b + x2) – R2 . (½ . b + x2) = 0
2,65 . (4 + x2) + 2 . 5 – 1 . (½ . 1 + 4 + x2) – 4 . (½ . 4 + x2) = 0
2,65 x2 – x2 – 4 x2 + 2,65 . 4 + 2 . 5 – 1 . 4,5 – 4 . 2 = 0
– 2,35 x2 + 10,6 + 10 – 4,5 – 8 = 0
x2 = (20,6 – 12,5)/2,35
= 3,45 m (dari G).
Maka letak momen sama dengan nol dari titik D,
x3 = b + x2 = 4 + 3,45 = 7,45 m.
Apabila letak momen sama dengan nol ini dihitung dari kanan kekiri,
MX4 = RBV . x4 + P2 . e – P1 . (d + x4) = 0
7,35 . x4 + 2 . 2 – 5 . (2 + x4) = 0
7,35 x4 – 5 x4 + 4 – 10 = 0
2,35 x4 – 6 = 0
x4 = 6/2,35 = 2,55 m (dari E)
(L – x4) = 10 – 2,55 = 7,45 m (dari D) ….(memenuhi)
d. Gaya Normal.
NA-D = – RAV = – 2,65 ton (tekan).
NC-D = 0
ND-E = + RAH = + 2 ton (tarik).
NE-B = – RBV = – 7,35 ton (tekan).
NE-F = 0
Gambar 17 : Putaran momen pada titik D dan titik E.
ba
(A)
h1
RAV
RBV
L
(C)
(E)
(D) (F)
c
P2
RAH
d
h2
(B)
P1
(G)
(H)
e
f
q t/m’
MDA
MDG
MDC
MEG
MEB
MEF
20. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
18
Gambar 18.a : Bidang gaya lintang.
Gambar 18.b : Bidang momen.
(A)
h1
RBV
(C) (E)(D) (F)
h2
(B)
(G)
(H)
e
f
ba c d
+
x1
L
Bidang Momen (M)
+ 10 t.m’
- 0,5 t.m’
- 10 t.m’+ 8,1 t.m’
+ 9,5 t.m’
+ 10,36 t.m’
- 6 t.m’
+ 4 t.m’
–+
(A)
h1
RBV
(C)
(E)(D)
(F)
h2
(B)
(G)
(H)
e
f
ba c d
+
+ 2 t
- 2,35 t
+ 1,65 t
+ 5 t
- 1 t
- 2 t
–
+
–
x1
L
Bidang Gaya Lintang (D)
21. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
19
Gambar 18.c : Bidang gaya normal.
Gambar 19 : Distribusi gaya-gaya pada elemen struktur (balok/kolom) dengan cara freebody.
(A)
(C)
(E)(D)
(F)
P2 = 2 t
(B)
P1 = 5 t
(G)
(H)
q = 1 t/m
1 t
1 t
1,65 t
2,65 t
2 t
(D)
1,65 t
2 t 2 t
q = 1 t/m
2,35 t
(E)
2,35 t
7,35 t
5 t
5 t
2 t 2 t
(D) (E)
(A)
h1
RBV
(C)
(E)(D)
(F)
h2
(B)
(G)
(H)
e
f
ba c d
- 2,65 t
+ 2 t
–
–
L
Bidang Gaya Normal (N)
+
- 7,65 t
+ 2 t
22. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
20
WORKSHOP/PELATIHAN
Diketahui : Struktur portal seperti tergambar.
Data-data,
PANJANG DAN TINGGI
L a b c d e h1 h2
m m m m m m m m
12 10 2 2 3 4 8 7
BEBAN KERJA
No. q Q P1 P2 Q + P1
Stb. t/m' ton ton ton ton
-1 1.0 10.0 2.0 2.0 12.000
0 1.2 12.0 2.2 2.1 14.200
1 1.4 14.0 2.4 2.2 16.400
2 1.6 16.0 2.6 2.3 18.600
3 1.8 18.0 2.8 2.4 20.800
4 2.0 20.0 3.0 2.5 23.000
5 2.2 22.0 3.2 2.6 25.200
6 2.4 24.0 3.4 2.7 27.400
7 2.6 26.0 3.6 2.8 29.600
8 2.8 28.0 3.8 2.9 31.800
9 3.0 30.0 4.0 3.0 34.000
Diminta : Gambarkan bidang-bidang gaya lintang, momen dan gaya normal pada seluruh
bentang.
Penyelesaian :
a. Reaksi Perletakan.
Q = q . a = (1 t/m’) . (10 m) = 10 ton.
H = 0,
RAH – P2 = 0
RAH = P2 = 2 ton (ke kanan).
ba
(A)
RAV
RBV
L
(C)
(E)
(D) (F)
c
P2
RAH
h2
(B)
P1
(G)
e
d
q t/m’
Q
h1
23. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
21
MB = 0,
RAV.L – RAH.(h1 – h2) – Q.(1/2a + b) + P1.c – P2.e = 0
RAV = RAH.(h1 – h2)/L + Q.(1/2a + b)/L – P1.c/L + P2.e/L
RAV = 2 . (8 – 7)/12 + 10 . (1/2.10 + 2)/12 – 2 . 2/12 + 2 . 4/12
RAV = 0,167 + 5,833 – 0,333 + 0,667
RAV = 6,333 ton (ke atas).
MA = 0,
– RBV . L – P2 . (h1 – d) + P1 . (L + c) + Q . (1/2a) = 0
RBV = – P2 . (h1 – d)/L + P1 . (L + c)/L + Q . (1/2a)/L
RBV = – 2 . (8 – 3)/12 + 2 . (12 + 2)/12 + 10 . (1/2 . 10)/12
RBV = – 0,833 + 2,333 + 4,167
RBV = 5,667 ton (ke atas).
Kontrol :
V = 0,
RAV + RBV = Q + P1
6,333 t + 5,667 t = 10 t + 2 t
12 ton = 12 ton (memenuhi).
b. Gaya lintang.
DA-C = – RAH = – 2 ton.
DCD = RAV = + 6,333 ton
DD-E = RAV – Q = 6,333 – 10 = – 3,667 ton.
DE-F = RAV – Q + RBV = + 6,333 – 10 + 5,667 = + 2 ton = P1.
DE-G = + RAH = + 2 ton
c. M o m e n .
MA = 0
MCA = – RAH . h1 = – 2 . 8 = – 16 t.m’.
MC = – RAH . h1 = – 2 . 8 = – 16 t.m’.
MD = – RAH . h1 + RAV . a – Q . 1/2a = – 2 . 8 + 6,333 . 10 – 10 . ½ . 10
= – 16 + 63,33 – 50
MD = – 2,67 t.m’.
MED = – RAH . h1 + RAV . L – Q . (1/2a + b)
= – 2 . 8 + 6,333 . 12 – 10 . (½.10 + 2)
= – 16 + 76 – 70
MED = – 10,0 t.m’.
MEG = – P2 . d = – 2 . 3 = – 6 t.m’.
MEF = – P1 . c = – 2 . 2 = – 4 t.m’.
Atau,
MED = MEG + MEF = – 6 + (– 4) = – 10 t.m’.
Letak titik dimana terdapat momen maksimum positip pada C-D,
Mx = RAV . x – RAH . h1 – ½ q . x2
d(Mx)/dx = Dx = 0,
RAV – q . x = 0
x = RAV / q = 6,333/1 = 6,333 m (dari titik C).
Mmaks= 6,333 . (6,333 m) – 2 . (8 m) – ½ . (1 t/m’) . (6,333 m)2
= + 4,053 t.m’.
Letak titik dimana momen sama dengan nol pada bentang C-E,
Mx = RAV . x – RAH . h1 – ½ q . x2
= 0
x2
– (RAV /1/2q) . x + RAH . h1/1/2q = 0
x2
– (6,333/0,5) . x + 2 . 8/0,5 = 0
x2
– 12,667 . x + 32 = 0
24. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
22
)1(.2
)32(.)1(.4)667,12(667,12
2
.4 22
2,1
a
cabb
x
x1,2 = 6,333 2,848
x1 = 6,333 + 2,848 = 9,181 m (dari C).
x2 = 6,333 – 2,848 = 3,485 m (dari C).
d. Gaya normal.
NA-C = – RAV = – 6,333 ton (tekan).
NC-E = – RAH = – 2 ton (tekan).
NE-B = – RBV = – 5,667 ton (tekan).
e. Gambar Bidang Gaya Lintang.
f. Gambar Bidang Momen.
ba
(A)
L
(C) (E)(D)
(F)
c
h2
(B)
(G)
e
d
h1
- 16 t.m’
- 10 t.m’
+
–
- 16 t.m’
- 6 t.m’
- 4 t.m’
ba
(A)
RAV
RBV
L
(C) (E)(D)
(F)
c
h2
(B)
(G)
e
d
h1
- 2 t
- 2 t
6,333 t
- 3,667 t
–
+
2 t
2 t
x
25. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
23
g. Gambar Bidang Gaya Normal.
KUNCI JAWABAN
REAKSI PERLETAKAN
No. RAH RAV RBV RAV + RBV
Stb. ton ton ton ton
-1 2.0 6.333 5.667 12.000
0 2.1 7.508 6.692 14.200
1 2.2 8.683 7.717 16.400
2 2.3 9.858 8.742 18.600
3 2.4 11.033 9.767 20.800
4 2.5 12.208 10.792 23.000
5 2.6 13.383 11.817 25.200
6 2.7 14.558 12.842 27.400
7 2.8 15.733 13.867 29.600
8 2.9 16.908 14.892 31.800
9 3.0 18.083 15.917 34.000
GAYA LINTANG
No. DA-C DCD DD-E DE-F DE-G
Stb. ton ton ton ton ton
-1 -2.000 6.333 -3.667 2.000 2.000
0 -2.100 7.508 -4.492 2.200 2.100
1 -2.200 8.683 -5.317 2.400 2.200
2 -2.300 9.858 -6.142 2.600 2.300
3 -2.400 11.033 -6.967 2.800 2.400
4 -2.500 12.208 -7.792 3.000 2.500
5 -2.600 13.383 -8.617 3.200 2.600
6 -2.700 14.558 -9.442 3.400 2.700
7 -2.800 15.733 -10.267 3.600 2.800
8 -2.900 16.908 -11.092 3.800 2.900
9 -3.000 18.083 -11.917 4.000 3.000
ba
(A)
L
(C) (E)(D)
(F)
c
h2
(B)
(G)
e
d
h1
- 2 t- 2 t
- 6,333 t
–
- 6,333 t
- 5,667 t
- 5,667 t
26. Modul kuliah “STATIKA 1” , Modul 7, 2012 Ir. Thamrin Nasution
Departemen Teknik Sipil, FTSP. ITMI.
24
M O M E N
No. MCA MD MED MEG MEF MEG+MEF x Mmaks
Stb. t.m' t.m' t.m' t.m' t.m' t.m' m t.m'
-1 -16.000 -2.667 -10.000 -6.000 -4.000 -10.000 6.333 4.056
0 -16.800 -1.717 -10.700 -6.300 -4.400 -10.700 6.257 6.690
1 -17.600 -0.767 -11.400 -6.600 -4.800 -11.400 6.202 9.329
2 -18.400 0.183 -12.100 -6.900 -5.200 -12.100 6.161 11.971
3 -19.200 1.133 -12.800 -7.200 -5.600 -12.800 6.130 14.615
4 -20.000 2.083 -13.500 -7.500 -6.000 -13.500 6.104 17.261
5 -20.800 3.033 -14.200 -7.800 -6.400 -14.200 6.083 19.908
6 -21.600 3.983 -14.900 -8.100 -6.800 -14.900 6.066 22.555
7 -22.400 4.933 -15.600 -8.400 -7.200 -15.600 6.051 25.203
8 -23.200 5.883 -16.300 -8.700 -7.600 -16.300 6.039 27.852
9 -24.000 6.833 -17.000 -9.000 -8.000 -17.000 6.028 30.501
GAYA NORMAL
No. NA-D NC-E NE-B
Stb. ton ton ton
-1 -6.333 -2.000 -5.667
0 -7.508 -2.100 -6.692
1 -8.683 -2.200 -7.717
2 -9.858 -2.300 -8.742
3 -11.033 -2.400 -9.767
4 -12.208 -2.500 -10.792
5 -13.383 -2.600 -11.817
6 -14.558 -2.700 -12.842
7 -15.733 -2.800 -13.867
8 -16.908 -2.900 -14.892
9 -18.083 -3.000 -15.917