Mechanical English Static
Nur Salam
i
ii
MECHANICAL ENGLISH STATIC
Penulis : Nur Salam
Penerbit : Polinema Press
iii
Mechanical English Static
Hak Cipta © Nur Salam
Hak Terbit pada POLINEMA PRESS
Penerbit POLINEMA PRESS, Politeknik Negeri Malang Jl. Soekarno-Hatta no.09 PO BOX 04 Malang 65141 Telp. (0341) 404424, 404425
Fax. (0341) 404420
UPT. Percetakan dan Penerbitan Gedung AU ground floor [email protected] www.polinemapress.org press.polinema.ac.id
Anggota APPTI (Asosiasi Penerbit Perguruan Tinggi Indonesia) no.
207/KTA/2016
Anggota IKAPI (Ikatan Penerbit Indonesia) no. 177/JTI/2017
Cetakan Pertama, Juni 2021
ISBN : 978-623-6562-94-9
xiv;63 hlm.; 15,5 x 23 cm
Setting & Layout : Putra Fanda Hita Cover Design : Putra Fanda Hita Penyunting : Abd. Muqit
Hak cipta dilindungi undang-undang. Dilarang memperbanyak karya tulis ini dalam bentuk dan dengan cara apapun, termasuk fotokopi, tanpa izin tertulis dari penerbit. Pengutipan harap menyebutkan sumber.
iv
Sanksi Pelanggaran Pasal 113 Undang-Undang Nomor 28 Tahun 2014
Tentang Hak Cipta
1) Setiap Orang yang dengan tanpa hak melakukan pelanggaran hak ekonomi sebagaimana dimaksud dalam Pasal 9 ayat (1) huruf i untuk Penggunaan Secara Komersial dipidana dengan pidana penjara paling lama 1 (satu) tahun dan/atau pidana denda paling banyak Rp100.000.000 (seratus juta rupiah).
2) Setiap Orang yang dengan tanpa hak dan/atau tanpa izin Pencipta atau pemegang Hak Cipta melakukan pelanggaran hak ekonomi Pencipta sebagaimana dimaksud dalam Pasal 9 ayat (1) huruf c, huruf d, huruf f, dan/atau huruf h untuk Penggunaan Secara Komersial dipidana dengan pidana penjara paling lama 3 (tiga) tahun dan/atau pidana denda paling banyak Rp500.000.000,00 (lima ratus juta rupiah).
3) Setiap Orang yang dengan tanpa hak dan/atau tanpa izin Pencipta atau pemegang Hak Cipta melakukan pelanggaran hak ekonomi Pencipta sebagaimana dimaksud dalam Pasal 9 ayat (1) huruf a, huruf b, huruf e, dan/atau huruf g untuk Penggunaan Secara Komersial dipidana dengan pidana penjara paling lama 4 (empat) tahun dan/atau pidana denda paling banyak Rp1.000.000.000,00 (satu miliar rupiah).
4) Setiap Orang yang memenuhi unsur sebagaimana dimaksud pada ayat (3) yang dilakukan dalam bentuk pembajakan, dipidana dengan pidana penjara paling lama 10 (sepuluh) tahun dan/atau pidana denda paling banyak Rp4.000.000.000,00 (empat miliar rupiah).
v
FOREWORD
This Engineering Mechanics I (Statics) teaching material module is prepared to be used as a learning medium in Engineering Mechanics I courses, especially statics material at D-IV Construction Engineering Management (MRK) - Study Program – Civil Engineering Department - State Polytechnic of Malang. The basis for the development of this teaching material module is the existence of a teaching module improvement program which is one of the efforts of D-IV MRK to standardize international class management.
The compilers would like to express their gratitude to the State Polytechnic of Malang, especially the Director and Head of the Civil Engineering Department. The authors hope that this teaching material module can be used for teaching materials for Engineering Mechanics I course and as a support for other courses.
Finally we, the writers, express our big thanks to Dr. Nur Salam, who have translated this module perfectly. We hope this translated module can help our international students understand the teaching materials completely.
Malang, July 2020
Authors,
vi
TABLE CONTENT
COVER ... i
FOREWORD ... v
TABLE OF CONTENT ... vi
LIST OF FIGURES ... vii
LIST OF TABLES ... x
CHAPTER I FORCE IN 2D ... 1
1.1 Unit System ... 1
1.2 Definition of Force... 4
1.3 Resultant Force ... 4
Exercise ... 12
CHAPTER II MOMENT IN 2D ... 13
2.1 Definition of Moment ... 13
2.2 Magnitude of Moment ... 14
2.3 Kopel Moment... 16
2.4 Position of the resultant Parallel Force ... 17
Exercise ... 19
CHAPTER III SUPPORT AND INTERNAL FORCE... 21
3.1 Support ... 21
3.2 Types of Loads ... 24
3.3 Internal Force ... 26
Excercise ... 30
CHAPTER IV SIMPLE BEAM ... 31
4.1 Properties of Stucture ... 31
vii
4.2 Definition of Simple Beam ... 31
4.3 Simple Simple Beam Analysis with Concentrated Load ... 32
4.4 Simple Beam Analysis with Uniformly Distributed Load ... 34
4.5 Simple Beam Analysis with Combined Loads... 36
Exercise ... 41
CHAPTER V CANTILEVER BEAM ... 43
5.1 Definition of Cantilever Beam... 43
5.2 Analysis of Cantilever Beam with Concentrated Load ... 43
5.3 Analysis of Cantilever Beam with Uniformly Distributed Load ... 45
5.4 Analysis of Cantilever Beam with Combined Load... 46
Exercise ... 50
CHAPTER VI OVERHANGING BEAM ... 52
6.1 Definition of Overhanging Beam ... 52
6.2 Overhanging Beam Analysis ... 52
Exercise ... 62
REFERENCES ... 63
viii
LIST OF FIGURE
Figure 1.1 Force Vector ... 4
Figure 1.2 Concurrent Forces... 4
Figure 1.3 Addition of the vector direction and line becomes the resultant force ... 5
Figure 1.4 Addition of Force using the Parallelogram Method .... 5
Figure 1.5 Addition of Force using the Triangle Method ... 6
Figure 1.6 Addition of Forces using the Polygon Method ... 6
Figure 1.7 Two Forces Flanking the Angles ... 6
Figure 1.8 Vector Force on Cartesian Coordinates ... 7
Figure 1.9 Force Decomposing ... 8
Figure 2.1 Illustration of a Moment ... 13
Figure 2.2 Moments with Varignon Theory ... 15
Figure 2.3 Kopel Moments ... 17
Figure 3.1 Application of Roller Support... 22
Figure 3.2 Application of pinned support ... 23
Figure 3.3 Application of Fixed Support ... 24
Figure 3.4 Idealized concentrated load in engineering mechanics 25 Figure 3.5 Idealization of uniformly distributed load in engineering mechanics ... 25
Figure 3.6 Idealization of uniformly varying load in engineering mechanics ... 26
Figure 3.7 Idealization of moment loads in engineering mechanics ... 26
Figure 3.8 Concept of Moments as an internal force ... 27
Figure 3.9 Sign Convention for Bending Moment ... 27
Figure 3.10 Concept of Shear Force ... 28
ix
Figure 3.11 Concept of Axial Force ... 28
Figure 4.1 Simple Beam ... 31
Figure 5.1 Cantilever Beam ... 43
Figure 6.1 One sided Overhanging Beam ... 52
Figure 6.2 Two sided Overhanging Beam ... 52
x
LIST OF TABLES
Table 1.1 Principal Quantities in SI ... 1
Table 1.2 Quantities of Derivatives in SI ... 2
Table 1.3 List of British and SI Units Conversions ... 2
1
CHAPTER I FORCE IN 2D Learning Objectives:After studying this chapter, students are expected to be able to:
1.
Answer the questions on unit conversion2.
Explain the meaning of force3.
Determine the resultant force using graphic and analytical methods1.1
Unit SystemStandard units can be expressed in two ways in the unit system, namely mks (meters, kilograms and seconds) or known as the metric system and cgs (centimeters, grams, and seconds) or known as the gaussian system. The mks system uses meters for length, kilograms for mass, and seconds for time. Meanwhile, the cgs system uses units of centimeters for length, grams for mass, and seconds for time. It is not necessary to choose which system to use, but the mks system is a system that is widely used.
As science and technology advance, scientists are constantly trying to create unit systems. The International System of Units (SI) is a modern version of the metric system by international conventions.
The SI system has seven principal quantities and many other quantities which can be derived from a base quantity called a derived quantity. The seven principal quantities are shown in the following table.
Table 1.1 Principal Quantities in SI Principal Quantity Quantity
Symbol
Unit Simbol Satuan
Length l meter m
Mass m kilogram kg
Time t second s
Temperature T kelvin K
electric current i ampere A
light intensity I candela cd
the amount of substance n mole mol
2
The derivative quantities that are often used in mechanics are shown in the following
Table 1.2 Quantities of Derivatives in SI
However, there are still many books, especially on mechanics using the British system of units, so it is necessary to convert them. The following table lists the conversion of British to SI units.
Table 1.3 List of British and SI Units Conversions
Magnitude Conversion Factor
Length 1 cm = 0,3937 in
1 m = 3,2808 ft
1 in = 2,54 cm
1 ft = 0,3048 m
Volume 1 cm3 = 0,061024 in3
1 m3 = 35,315 ft3
1 in3 = 16,387 cm3
1 ft3 = 0,028317 m3
Derivative Quantity Derivative Units
Name Symbol Name Symbol
area A meter squared m2
volume V cubic meter m3
velocity v meter per second m/s
acceleration a meter per second squared
m/s2
force F Newton kg.m/s2
Moment of force
M Newton meter N.m
stress P / σ Pascal N/mm2
3
Mass and Density 1 kg = 2,2046 lb
1 g/cm3 = 62,428 lb/ft3
1 lb = 0,4536 kg
1 lb/ft3 = 16,018 kg/m3
Force 1 N = 0,22481 lbf
1 lbf = 4,4482 N
Table 1.3 (Continued)
Quantity Conversion Factor
Pressure / Stress 1 MPa = 145 psi
1 kPa = 0,145 psi
1 psi = 6,895 kPa
1 ksi = 6,895 MPa
Moment 1 N.m = 0,7376 lb.ft
1 N.m = 8,8496 lb.in
1 lb.ft = 1,356 N.m
1 lb.in = 0,113 N.m
Example 1.1
Convert the unit moment of force expressed in M = 25 lbf.in into SI!
Solution:
M = 25 lbf.in = 25 x (4,4482 N) x (2,54 cm)
= 282,4607 N.cm = 2,825 N.m
Example 1.2
Convert the unit spring constant k = 12
4
N/cm into British units! Solution:k = 12 N/cm = 12 x (0,22481 lb) x (0,3973 ⅈ𝑛1 )
= 6,852 lb/in
1.2
Definition of ForceForce is a cause or action that causes things to change. Force has two effects, namely:
1)
causing an object to move if it is at rest or changes in motion if it has moved and 2) causing deformation. Force is a vector quantity that has both magnitude and direction. A force is completely expressed in terms of magnitude, direction, and point of application.1.
Magnitude, refers to the size or magnitude of the force. The 1000 N force is larger than the 500 N force.2.
Direction, refers to an infinite line that shows the work force, called a line of work. The force can be vertical, horizontal or angular against vertical or horizontal.Meanwhile, the direction of the force is indicated by an arrow.
3.
Point of application, refers to the point of the object where the force acts.Figure 1.1 Force Vector
1.3
Resultan ForceSeveral forces acting on a structure can be reduced to a single force which is called the resultant force. Calculating the resultant force depending on the number and direction of these forces. Several forces whose work lines pass through the same point are said to be concurrent.
5
Figure 1.2 Concurrent Forces
The results of two or more forces that are on the same line of force can be added directly (if the direction is the same) or subtracted (if the direction is opposite).
Figure 1.3 Addition of the vector direction and line becomes the resultant force
Several methods to calculate or find the resultant force with different lines of work, namely:
1.
Graphical Methoda.
The Parallelogram MethodThis method is a method of adding two forces that are placed at the same starting point, so that the resultant force result is a diagonal line.
Figure 1.4 Addition of Force using the Parallelogram Method
b.
Triangle MethodThe triangle method is a method of adding forces by placing the base of the second force at the end of the first force. The resultant result of the force is the force which has the origin at the point of the first force and at the end of the second force.
6
Figure 1.5 Addition of Force using the Triangle Method
c.
Polygon MethodThe polygon method is a graphical way of adding three or more forces by connecting the base of the force to the other end of the force until the last force. This method is in principle the same as the triangle method.
Figure 1.6 Addition of Forces using the Polygon Method
2.
Analytical Methoda.
Law of Sines and CosinesIf there are two force vectors flanking the angles, then using the parallelogram method, the resultant force can be obtained as shown in Figure 1.7.
Figure 1.7 Two Forces Flanking the Angles
Analytically, the magnitude and direction of the resultant force can be calculated according to the law of sines and cosines. The magnitude of the resultant force can be used in
7
the following equation.Whereas the direction of the resultant force (ϕ), the angle between the force vector and the x-axis, can be determined using the law of sines such as.
𝑅
𝑠ⅈ𝑛 𝜃 = 𝐴
𝑠ⅈ𝑛 𝑎 = 𝐵 𝑠ⅈ𝑛 𝛽
with :𝜃
= angle between force vectors A and B𝑎
= angle between force vector A and resultant β = angle between force vector B and resultant b. Force Decomposing MethodThe resultant force can be done by decomposing each force vector into its components to the x-axis and the y-axis at the Cartesian coordinates. For example, the three force vectors A, B, and C which are located at the Cartesian coordinates are shown in Figure 1.8.
Figure 1.8 Vector Force on Cartesian Coordinates
The steps for finding the resultant force using the decomposition method are as follows:
1.
Decomposing each force vector into its vector components on the x-axis and the y-axis.R
=
8
Figure 1.9 Force Decomposing
2.
Adding up all the components on the x-axis to Rx and all the components on the y-axis to be Ry.𝑅𝑥 = 𝐴𝑥 + 𝐵𝑥 − 𝐶𝑥 𝑅𝑥 = 𝐴𝑦 + 𝐵𝑦 − 𝐶𝑦
3.
The resultant vector of the sum is obtained by adding up the vector components Rx and Ry with the Pythagorean theory.𝑅 = √𝑅𝑥2+ 𝑅𝑦2 With the direction of the resultant force:\
𝜙 = 𝑎𝑟𝑐 𝑡𝑎𝑛𝑅𝑦
𝑅𝑥 Example 1.3
Calculate the magnitude and direction of the resultant forces in the following figure using an analytical method!
9
Solution:Resultant Force:
𝑅 = √𝐹12+ 𝐹22+ 2𝐹1𝐹2𝑐𝑜𝑠 𝜃 = √102+ 10𝑧+ 2.10.10 𝑐𝑜𝑠 1200 = 10 N
The direction of the resultant force:
ϕ = 𝛼 = 𝑠ⅈ𝑛−1 . (𝐹2⋅𝑠𝑖𝑛 𝜃
𝑅 )
= 𝑠ⅈ𝑛−1(10⋅𝑠𝑖𝑛 1200
10 )
= 60°
Problem 1.4
Calculate the magnitude and direction of the resulting forces in the following figure using the method of breaking down the forces!
10
Solution:Component of Force on x axis:
F1x = F1 . cos 37° = 7,986 N F2x = 0
F3x = − F3 . cos 53° = − 9,027 N 𝐹𝑥 = 7,986 − 9,027 = − 1,041 N
Component of Force on y axis:
F1y = F1 . sin 37° = 6,018 N F2y = −25 N F3y = F3 . sin 53° = 11,980 N
𝐹𝑦 = 6,018 − 25 + 11,980 = − 7,002 N Resultant Force:
𝑅 = √𝐹𝑥2+ 𝐹𝑦2
= √(−1,041)2+ (7,002)2
= 7,079 N
The direction of the resultant force:
𝜙 = 𝑎𝑟𝑐 𝑡𝑎𝑛𝐹𝑦 𝐹𝑥
11
= 𝑎𝑟𝑐 𝑡𝑎𝑛−7,002−1,041
= 81,540
12
Exercise1.
Convert the following units into SI units!a.
450 lbb.
225 lb/in2c.
75 ft3d.
160 lb fte.
90 lb/ft32.
Convert the following units into British units!a.
105 kN/m3b.
80 kg/cm2c.
315 Nmd.
0,5 cm2/detike.
45 kN/m23.
Two people are moving the big rock by pulling it up and down as shown. Determine the magnitude and direction of the resultant force (in a graphic and analytical way)!4.
Determine the magnitude and direction of the resultant forces!13
CHAPTER II MOMENT IN 2D Learning Objectives:
After studying this chapter, students are expected to be able to:
1.
Explain the meaning and concept of moments2.
Calculating the magnitude of the moment3.
Explain the Kopel moment4.
Determine the location of the resultant parallel forces2.1
Definition of MomentMoment is the magnitude of the tendency or tendency of a force to rotate an object / rigid object to a certain point or axis.
Figure 2.1 Illustration of a Moment Moment arm; Force work line
The amount of moment depends on two factors, namely the moment arm (distance) and the working force. If the working force is constant, the moment magnitude is directly proportional to the moment arm. Big moment arm, then the resulting moment is also big and vice versa. The moment is obtained from the result of the force with the perpendicular distance (moment arm), it can be written in the following equation:
M = F . d with:
M = Moment F = Force
D = distance / moment arm
Moment arm
Force work line
14
The sign convention for the direction of the moment is positive for moments that are clockwise and negative for moments that are anti- clockwise.
2.2
Magnitude of Moment Calculation2.2.1
Magnitude of Moment based on Existing Force (Actual Force)Based on the magnitude, direction of the arrow force, and depending on the relative place of the force and axis.
Example:
Three forces act on a triangular plate as shown in the picture.
Determine:
a.
Moment in C due to FBb.
Moment in B due to FCc.
Total moment in BSolution:
𝜙 = 𝑡
𝑎𝑛−1(20/15) = 53,13
2So, L
AC= √(20
2+ 15
2) =
25 ⅈ𝑛dC/B = LAC – 10 = 15 in dB/C = 15 sin 53,130 = 12 in dB/D = 15 – 15 sin 53,130 = 3 in
15
a.
MC = FB . dC/B b. MB = - FC . dB/c= 400 . 15. = - 300 . 12.
= 6000 lb.in = - 3600 lb.in
c.
MB = - FC . dB/C - FD . dB/D= -3600 - 750
= - 4350 lb.in
2.2.2
Magnitude of Moment with Varignon Theory According to the Varignon theory, the moment of a force on an axis is equal to the number of moments of its components about that axis.Figure 2.2 Moments with Varignon Theory
𝑀 = (𝐹𝑥 . 𝑏) − (𝐹𝑦 . 𝑎) Example:
Calculate the moment at point O due to the force of 600 N as shown in the Figure using the Varignon Theory.
16
Solution:MO = (460 . 4) + (386 . 2) = 2610 N.m
2.3
Kopel MomentThe Kopel Moment is two parallel forces that are equal in and in opposite directions which lie on separate lines of work. The properties of a Kopel include the following:
1.
The magnitude of the resultant force of a kopel is zero.2.
The kopel moment is the product of one of the forces and the distance between the two working lines of the force.3.
The kopel moments for all points on the kopel area are equal.F1 = 600 cos 400 = 460 N F2 = 600 sin 400 = 386 N
17
Figure 2.3 Kopel Moments MA = F . d (clockwise)
MB = F . d (clockwise)
2.4
Position of the Resultant Parallel ForcesThe position of the resultant parallel force can be
determined using the moment theory. As in the case of determining the resultant parallel forces on the following beam:
R = 12 + 20 + 15
= 47 T
MA = 12 . 2 + 20 . 6 + 15 . 10 47 . x = 294
x = 6,255 m from A
18
19
Exercise1.
Calculate the moment in A due to the 250 lb force as in the following picture, and convert the resulting moment into SI units!2.
Two force act on the structure below. Find:a.
Moment in A is due to F1b.
Moment in B is due to F23.
The beam AE is 12 m long and is divided into 4 equal sections. Find the moments at points A, B, C, D, and E as a result of the forces!4.
Determine the location of the following parallel resultant forces:20
21
CHAPTER IIISUPPORT AND INTERNAL FORCES Learning Objectives:
After studying this chapter, students are expected to be able to:
1.
Describe the nature of each type of support2.
Describe the types of loads3.
Explain the internal forces3.1
SupportA building construction must be located on a support. The support function is to transmit external forces acting on the construction and the construction's own weight to the lower part. So that the support will respond to reaction forces to maintain balance and maintain the structure so that its conditions remain stable There are 3 (three) types of support, including:
1. Roller Support
The symbol of the roller support is as follows:
Roller support properties:
-
Can only withstand forces which are perpendicular to the support area-
Cannot withstand forces which are parallel to the support area and moment-
Has one support reaction, namely RV22
Figure 3.1 Application of Roller Support 2. Pinned Support
The symbol of the joint support is as follows:
Pinned support properties:
-
Can withstand forces in the same direction and forces that are perpendicular to the support area-
Can't hold the moment-
Has two support reactions, namely RV and RH23
Figure 3.2 Application of pinned support 3. Fixed Support
The symbol of the fixed support is as follows:
Fixed support properties:
-
Can withstand forces in the same direction and forces that are perpendicular to the support area-
Can hold the moment-
Has three support reactions, namely RV, RH, and MPhysically, the fixed support is obtained by building a beam into a brick wall, casting it into the concrete, or welding it into the main building.
24
Figure 3.3 Application of Fixed Support
3.2
Types of LoadsIn general, the loads that work on the building structure are dead loads, live loads, earthquake loads, wind loads, temperature loads and so on.
1. Dead Load (DL)
Is a load that remains in a construction and does not change, for example selfweight of the structure and other attached parts.
2. Live Load (LL)
Live loads consist of moving or variable loads like people, furniture, temporary stores, etc. It is also called super-imposed load.
3. Wind Load (W)
The wind acts horizontally on the surface of the walls, roofs, and inclined roof of the structure. That means it exerts uniform pressure on the structural components on which it acts and tend to disturb the stability of the structure. The value of wind loads varies depending on several factors such as geographical location of the structure, height of the structure, duration of wind flow, etc.
4. Earthquake Load (E)
These types of loads are internal forces which act on the structure due to earthquake developed ground movements.
5. Temperature Load (T)
Is a special load caused by changes in temperature.
25
According to its shape, loads can be classified as follows:
a.
Concentrated Load (Point Load) is a load that is concentrated somewhere. Example: a human standing on a bridge, a vehicle that stops on a bridge.Figure 3.4 Idealized concentrated load in engineering mechanics
b.
Uniformly Distributed Load is a load that is evenly distributed both in the longitudinal direction and in the broad direction.Example: self-weight of a bridge.
Figure 3.5 Idealization of uniformly distributed load in engineering mechanics
c.
Uniformly Varying Load (Non-Uniformly Distributed Load) is a load that whose magnitude varies along the loading length with a constant rate. This load can be a triangular load or trapezoidal load.Example: the pressure load of water on the wall of the water tank or the water gate.
Concrete bridge
Unformly distributed load Concentrated load
26
Figure 3.6 Idealization of uniformly varying load in engineering mechanics
d.
Load in the form of moments.Figure 3.7 Idealization of moment loads in engineering mechanics
3.3
Internal ForcesInternal forces are the forces that work in a construction due to external forces (loads on the construction and support reactions). If there is no external force acting, then in the construction there is also no internal force. In making constructions made of
concrete, wood, steel, and others, the size or dimensions of each structural element (beams, columns, plates, etc.) are definitely required.
This internal force is used to determine the dimensions of the structural elements.
The internal forces that exist in a structure are as follows:
1. Moment / Bending Moment (M)
Due to the external force P, the beam will bend. Internal force is in the form of moments that will oppose this condition so that there will be balance. Moment is the product of force and perpendicular distance between line of action of the force and the point about which moment is required to be calculated.
water water
27
Figure 3.8 Concept of Moments as an internal force A bending moment is considered positive if it tends to cause concavity upward (sagging). If it bending moment tends to cause concavity downward (sagging), it will be considered a negative bending moment.
The diagram which shows the variation of bending moment along the length of the beam is called Bending Moment Diagram (BMD). As a convention, the positive bending moments are drawn below the x-centroidal axis of the structure, while the negative bending moments are drawn above the axis.
Figure 3.9 Sign Convention for Bending Moment 2. Shear Force (D)
Shear Force is defined as the algebraic sum of all the vertical forces acting on the beam either to the left or right of the section is known as the shear force at a section. The shear force is perpendicular to the normal force.
Bending line
sagging
M +
sagging
M ˗
hagging hagging
28
Figure 3.10 Concept of Shear Force
A shear force that tends to move the left of the section upward or the right side of the section downward will be regarded as positive. Similarly, a shear force that has the tendency to move the left side of the section downward or the right side upward will be considered a negative shear force.
The diagram which shows the variation of shear force along the length of the beam is called Shear Force Diagram (SFD). As a convention, the positive shear force are drawn above the x- centroidal axis of the structure, while the negative bending moments are drawn below the axis.
3. Axial/ Normal Force (N)
The normal force at any section of a structure is defined as the algebraic sum of the axial forces acting on either side of the section. The normal force runs in the direction of the beam axis.
The force causes a change in the beam length through an internal reaction to tensile or compressive external forces acting on the beam.
Figure 3.11 Concept of Axial Force
29
Positive normal forces act normal to the face and are tensile forces (pointing away from the face) – stretching the material.
Then negative normal forces are compressive forces (toward the face).
The diagram which shows the variation of normal force along the length of the beam is called Normal Force Diagram (NFD).
As a convention, the positive normal force are drawn below the x-centroidal axis of the structure, while the negative normal force are drawn above the axis.
30
ExerciseSummarize:
1.
Types and properties of support2.
Types of loads acting on a structure3.
Internal forces that exist in the structure31
CHAPTER IV SIMPLE BEAM Learning Objectives:
After studying this chapter, students are expected to be able to:
1.
Explain the notion of a simple beam2.
Calculating the support reaction on a simple beam3.
Calculating and drawing the Bending Moment Diagram (BMD), Shear Force Diagram (SFD), and Normal Force Diagram (NFD) on a simple beam4.1
Properties of StructureAn object is in equilibrium if the system of forces acting on it does not cause translation or rotation of the object. A structure must always be in a balanced state, so the structure must remain at rest when holding external loads. There are three equation of equilibrium, namely:
1.
∑M = 0 (The number of moments is equal to zero)2.
∑V = 0 (The sum of the vertical forces is equal to zero)3.
∑H = 0 (The sum of the horizontal forces is equal to zero)There are two structural characteristics, namely Statically Determinate and Statically Indeterminate. A statically determinate is a structure which can be solved by three equilibrium equations.
Meanwhile, a statically indeterminate cannot only use equilibrium equations, but requires additional equations.
4.2
Definition of Simple BeamA simple beam is one of the simplest structural elements that both ends rest on support but is free to rotate. It contains pinned support at one end and a roller support at the other end. Simple beam include statically determinate which can be solved by three equilibrium equations.
Figure 4.1 Simple Beam
32
4.3
Simple Beam Analysis with Concentrated LoadA simple beam with a concentrated / point load in the middle of the span is known as in the image below. Determine the magnitude of the support reaction and calculate and draw Bending Moment Diagram (BMD), Shear Force Diagram (SFD), and Normal Force Diagram (NFD)!
Solution:
1.
Calculating Support Reaction∑𝑀
𝐵= 0 ∑𝑀
𝐴= 0
𝑅
𝑣𝐴⋅ 𝑙 = 𝑃 ⋅
12𝑙 𝑅
𝑣𝐵⋅ 𝑙 + 𝑃 ⋅
12𝑙 = 0 𝑅
𝑣𝐴=
12
𝑝(𝑘𝑒𝑎𝑡𝑎𝑠) 𝑅
𝑣𝐵= 𝑃.
12
𝑙 𝑅
𝑣𝐵=
12
𝑃(𝑘𝑒𝑎𝑡𝑎𝑠)
Control: 𝛴𝑣 = 0 RVA + RVB = 0
1
2𝑃 +12𝑃 − 𝐷 = 0
0 = 0 (OK!)
2.
Calculating the Bending Moment (M)33
MA = 0MC = RVA . 1
2 l = 1
4 P l MB = RVA . l – P . 1
2 l = 0
3.
Calculating the Shear ForceDA ki = 0 DA ka = RVA = 1
2 P DC ki = 1
2 P DC ka = 1
2 P - = −12 P DB ki = -1
2 P DB ki = -1
2 P + RVB = 0
4.
Calculating Axial Force (N)N = 0 (because there is no horizontal force acting)
5.
BMD, SFD, and NFD34
4.4
Simple Beam Analysis with Uniformly Distributed Load There is a simple beam with a Uniformly Distributed Load across the span as in the figure below. Determine the magnitude of the support reaction and calculate and draw the Bending Moment Diagram (BMD), Shear Force Diagram (SFD), and Normal Force Diagram (NFD)!Solution:
1.
Calculating Support Reaction∑𝑀
𝐵= 0 ∑𝑀
𝐴= 0
𝑅
𝑣𝐴⋅ 𝐿 + 𝑞. 𝐿 ⋅
12𝐿 =
0𝑅
𝑣𝐵⋅ 𝑙 + 𝑃 ⋅
12𝑙 = 0
35
RVA . l = 12
𝑞𝐿
2 RVB . l = 12
𝑞𝐿
2𝑅
𝑣𝐴=
12
𝑞L 𝑅
𝑣𝐵=
12
𝑞L
Control: 𝛴𝑉 = 0 RVA + RVB - Q = 0
1 2𝑞𝐿 +1
2𝑞𝐿 − 𝑞𝐿 = 0 0 = 0 (OK!) 𝛴𝐻 = 0
RHB = 0
2.
Calculating the Bending Moment (M)\Interval A-B ((0 ≤ 𝑥 ≤ 1) MX = RVA . x – 1
2𝑞𝑥2
Terms of location Mmax: (first derivative Mx = 0) 𝑑𝑀𝑥
𝑑𝑥 = 0 RVA – q.x = 0 x = 𝑅𝑉𝐴
𝑞 =1
2 𝐿 from A x = 0 → MA = 0 𝑥 =1
2𝐿 → 𝑀𝑚𝑎𝑥= (1
2𝑞𝐿 ⋅1
2𝐿) - (1
2𝑞𝐿 ⋅ (1
2⋅ 𝐿)2) = 1
8𝑞𝐿2 x = L2 → MB = (1
2𝑞𝐿 . 𝐿) - (1
2𝑞𝐿 . 𝐿2) = 0
3.
Calculating the Shear ForceDA ki = 0 DA ka = RVA = 1
2 qL DB ki = 1
2 P – qL = −1
2𝑞𝐿 DB ka = -1
2 qL + RVB = 0
4.
Calculating Axial Force (N)N = 0 (because there is no horizontal force akting)
5.
BMD, SFD, and NFD36
4.5
Simple Beam Analysis with Combined LoadsA simple beam with a combined load (concentrated and uniformly distributed load) is known as shown below. Determine the magnitude of the support reaction and calculate and draw the Bending Moment Diagram (BMD), Shear Force Diagram (SFD), and Normal
Force Diagram (NFD)!
Solution:
37 1.
Calculating Support Reaction∑𝑀𝐵 = 0
𝑅𝑉𝐴 . 8 − 2 . 5 . 5,5 − 5 . 𝑠ⅈ𝑛 30° . 3 = 0 8 𝑅𝑉𝐴 − 55 − 7,5 = 0
8 𝑅𝑉𝐴 = 62,5 𝑅𝑉𝐴 = 7,8125 𝑡
∑𝑀𝐴 = 0
−𝑅𝑉𝐵 . 8 + 2 . 5 . 2,5 + 5 . 𝑠ⅈ𝑛 30° . 5 = 0
− 8 𝑅𝑉𝐵 + 25 + 12,5 = 0 8 𝑅𝑉𝐵 = 37,5
𝑅𝑉𝐵 = 4,6875 𝑡
Control:
∑𝑉 = 0 𝑅𝑉𝐴 + 𝑅
𝑉𝐵− 𝑄 − 𝑃 . 𝑠ⅈ𝑛 30° = 0 7,8125 + 4,6875 − 2 .5 − 5 . 𝑠ⅈ𝑛 30° = 0 12,5 − 10 − 2,5 = 0 0 = 0
∑𝐻 = 0
−𝑅𝐻𝐴 + 𝑃 . 𝑐𝑜𝑠 30° = 0
−𝑅𝐻𝐴 + 4,330 = 0
𝑅𝐻𝐴 = 4,330 𝑡
38
2.
Calculating the Bending Moment (M) Interval A-C (0 ≤ x ≤ 5)MX = RVA. x − 1 q x2
= 7,8125 . x − 1 . 2 . 𝑥2 = 7,8125 x − 𝑥2 Terms of location Mmax: (first derivative Mx = 0)
𝑑𝑀
𝑥𝑑
𝑥= 0
7,8125 − 2 . x = 0 x =
7,8125
2 = 3,906 m from A x = 0
→
MA = 0x = 3,906
→
Mmax = (7,8125 . 3,906) − 3,9062 = 15,259 𝑡𝑚 x = 5→
MC = (7,8125 . 5) − 52 = 14,0625 tm39
Interval C-B (5 ≤ x ≤ 8)MX = RVA . x − 2 . 5 . (x − 2,5) − 𝑃. 𝑠ⅈ𝑛 30° . (𝑥 − 5)
= 7,8125 . x − 10 (𝑥 − 2,5) − 2,5 (𝑥 − 5)
x = 5
→
MC = (7,8125.5)−10.(5−2,5)−2,5. (5 − 5) = 14,0625 tm x = 8→
MC = (7,8125 . 8) − 10 . (8 − 2,5) − 2,5 . (8 − 5) = 03.
Calculating the Shear ForceDAki = 0
DAka = RVA = 7,8125 𝑡
DCki = 7,8125 − 2 . 5 = −2,1875 t DCka = − 2,1875 − P . sin 30° = − 4,6875 𝑡 DBki = − 4,6875 𝑡
DBka = −4,6875 + RVB = 0
Another way to determine the Shear Force is the First Derivative of the Moment
Interval A-C (0 ≤ x ≤ 5) 𝐷𝑥 = 7,8125 − 2 . x x = 0
→
DA = 7,8125 tx = 5
→
DCki = 7,8125 − 2 . 5 = −2,1875 t Interval C-B (5 ≤ x ≤ 8)𝐷𝑥 = 7,8125 − 10 − 2,5 = − 4,6875 𝑡
4.
Calculating Axial Force (N)NAki = 0
NAka = −RHA = − 4,330 𝑡 NCki = − 4,330 t
NCka = − 4,330 + P . cos 30° = 0 NBki = NB ka = 0
5.
BMD, SFD, and NFD40
41
ExerciseCalculate and Draw the Bending Moment Diagram (BMD), Shear Force Diagram (SFD), and Normal Force Diagram (NFD)!
42
43
CHAPTER V CANTILEVER BEAM Learning Objectives:After studying this chapter, students are expected to be able to:
1.
Explain the meaning of a cantilever beam2.
Calculate the support reaction on a cantilever beam3.
Calculating and drawing the Bending Moment Diagram (BMD), Shear Force Diagram (SFD), and Normal Force Diagram (NFD) on a cantilever beam5.1
Definition of Cantilever BeamA cantilever beam is a beam with a laterally and rotationally fixed support at one end and with no support at the other end. It is one of the statically determinate beam.
Figure 5.1 Cantilever Beam
5.2
Analysis of Cantilever Beam with Concentrated LoadIt is known that a beam is fixeded with a concentrated load as in the image below. Determine the magnitude of the support reaction and calculate and draw the Bending Moment Diagram (BMD), Shear Force Diagram (SFD), and Normal Force Diagram (NFD)!
Solution:
1.
Calculating Support Reaction44
∑𝑀𝐴 = 0
−𝑀𝐴 + 𝑃 . 𝐿 = 0
𝑀𝐴 = 𝑃 𝐿 (anticlockwise)
∑𝑉 = 0
𝑅𝑉𝐴 − 𝑃 = 0 𝑅𝑉𝐴 = 𝑃
∑𝐻 = 0 𝑅𝐻𝐴 = 0
2.
Calculating the Bending Moment (M) MA = - P LMB = − 𝑀𝐴 + RVA . L = − P L + 𝑃 L = 0
3.
Calculating the Shear ForceDAki = 0
DA ka = RVA = P DBki = P
DBka = P − P = 0
4.
Calculating Axial Force (N)N = 0 (because there is no horizontal force acting)
5.
BMD, SFD, and NFD45
5.3
Analysis of Cantilever Beam with Uniformly Distributed Load It is known that a beam is fixeded with an uniformly distributes load across the span as in the image below. Determine the magnitude of the support reaction and calculate and draw the Bending Moment Diagram (BMD), Shear Force Diagram (SFD), and Normal Force Diagram (NFD)!Solution:
1.
Calculating Support Reaction∑𝑀𝐴 = 0 -MA + q . L. 1
2 L = 0 MA = 1
2 q L2 (anticlockwise)
∑V = 0 RVA – qL = 0 RVA = q L
2.
Calculating the Bending Moment (M) MA = 12 q L2
MB = -MA + RVA . L - 1
2 q L2 = -1
2 q L2 + q L2 - 1
2 q L2 = 0
3.
Calculating the Shear ForceDA ki = 0
DA ka = RVA = q L DA ki = q L – q L = 0 DB ka = 0
4.
Calculating Axial Force (N)N = 0 (because there is no horizontal force acting)
5.
BMD, SFD, and NFD46
5.4
Analysis of Cantilever Beam with Combined LoadIt is known that a beam is fixeded with a combined load as shown below. Determine the magnitude of the support reaction and calculate and draw the Bending Moment Diagram (BMD), Shear Force Diagram (SFD), and Normal Force Diagram (NFD)!
Solution:
1.
Calculating Support Reaction47
∑𝑀𝐵 = 0
− 𝑃 . 10 − 𝑞 .4 . 2 + 𝑀𝐵 = 0
− 10 . 10 − 3 .4 . 2 + 𝑀𝐵 = 0
− 124 + 𝑀𝐵 = 0
𝑀𝐵 = 124 𝑁𝑚 (clockwise)
∑𝑉 = 0
𝑅𝑉𝐵 − 𝑃 − 𝑞 . 4 = 0 𝑅𝑉𝐵 − 10 − 12 = 0
𝑅𝑉𝐵 = 22 𝑁
∑𝐻 = 0 𝑅𝐻𝐵 = 0
2.
Calculating the Bending Moment (M) Interval A-C (0 ≤ x ≤ 6)MX = − 𝑃 . x = −10 x x = 0
→
MA = 0x = 6
→
MC = −10 . 6 = −60 Nm Interval C-B (6 ≤ x ≤10)48
MX = − 𝑃 . x − 12 𝑞 (𝑥 − 6)2
= − 10 x − 3
2 (𝑥 − 6)2
Terms of location Mmax: (first derivative Mx = 0)
𝑑𝑀
𝑥𝑑
𝑥= 0
−10 − 3 . (x − 6) = 0
−28 − 3 . x = 0
x = − 9,333 m (Not OK) x
→ 6 M
C= -10. 6 -
32 (6 − 6)2 = - 60 Nm x
→ 10 M
C= -10. 10 -
32 (10 − 6)2 = - 124 Nm
3.
Calculating the Shear ForceDA ki = 0
DA ka = - P = -10 N DC ki = DC ka = -10 N DB ki = - 1- 3 .4 = -22 N DB ka = -22 + RVB = 0
4.
Calculating Axial Force (N)N = 0 (because there is no horizontal force acting)
5.
BMD, SFD, and NFD49
50
ExerciseCalculate and Draw the Bending Moment Diagram (BMD), Shear Force Diagram (SFD), and Normal Force Diagram (NFD)!
51
52
CHAPTER VI OVERHANGING BEAM Learning Objectives:After studying this chapter, students are expected to be able to:
1.
Explain the meaning of the overhanging beam2.
Calculate the support reaction on the overhanging beam3.
Calculating and drawing the Bending Moment Diagram (BMD), Shear Force Diagram (SFD), and Normal Force Diagram (NFD) on the overhanging beam6.1
Definition of Overhanging BeamOverhanging Beam is a beam simply supported at two points and having one end or both ends extended beyond the support. The types of overhanging beam include one sided overhangs (right or left) and two sided overhangs.
Figure 6.1 One sided Overhanging Beam
Figure 6.2 Two sided Overhanging Beam
6.2
Overhanging Beam Analysis Example 6.1It is known that overhanging beam with a load as shown below.
53
Determine the magnitude of the support reaction and calculate and draw Bending Moment Diagram (BMD), Shear Force Diagram (SFD), and Normal Force Diagram (NFD)! (unit of length in meters)
Solution:
1.
Calculating Support Reaction𝛴𝑀𝐵 = 0
RVA . 8 – 5 .sin 60°.6 – 1 . 6. 1 = 0 8 RVA – 25,981 – 6 = 0
8 RVA = 31,981 RVA = 3,998 kN 𝛴𝑀𝐴= 0
–RVB . 8 + 5 . sin 60°. 2 + 1 . 6 . 7 = 0 –8 RVB + 8,660 + 42 = 0
8 RVB = 50,660 RVB = 6,332 kN
54
Control: ∑V = 0
RVA + RVB – Q – P .sin 60° = 0 3.998 + 6,332 – 1.6 – 5 . sin 60° = 0
10,330 – 6 – 4,330 = 0
0 = 0 (OK!)
∑H = 0
RHA – P . cos 60° = 0 RHA – 2,5 = 0
RHA = 2,5 kN
2.
Calculating the Bending Moment (M) Interval A-C (0 ≤ x ≤ 2)Mx = RVA . x = 3,998 x x = 0 MA = 0
x = 2 MC = 3,998 . 2 = 7,996 kNm Interval C-D (2 ≤ x ≤ 4)
MX = RVA . x P . sin 60° . ( x – 2) = 3,998 x – 4,330 (x – 2)
x = 2 MC = 3,998 . 2 – 4,330 (2 – 2) = 7,996 kNm x = 4 MC = 3,998 . 4 – 4,330 (4 – 2) = 7,332 kNm
55
Interval D-B (4 ≤ x ≤ 8)MX = RVA . x – P . sin 60° . (x – 2) – 1
2𝑞 (𝑥 − 4)2 = 3,998 . x – 4,330 ( x – 2 ) – 1
2 (𝑥 − 4)2 Terms of location Mmax: (first derivative Mx = 0) 𝑑𝑀𝑥
𝑑𝑥 = 0
3,998 – 4,330 – ( x – 4 ) = 0 –x + 3,668 = 0
x = 3,668 m dari titik A (TM/ Not OK) x = 4 → MD = 3,998 . 4 – 4,330 ( 4 – 2 ) – 1
2 (4 − 4)2 x = 8 → MB = 3,998 . 8 – 4,330 ( 8 – 2 ) – 1
2 (8 − 4)2 = –1,996 kNm ≈ – 2 kNm
Location M = 0 MX = 0
3,998 . x – 4,330 ( x – 2 ) 1
2 (𝑥 − 4)2= 0
−1
2 𝑥2+ 3,668 − 0,660 = 0 𝑥2− 7,336𝑥 + 1,320 = 0 x1 = 7,151 m from point A (OK)
56
x2 = 0,185 m from point A (Not OK)Interval E-B (0 ≤ x ≤ 2)→counted from right to left
𝑀𝑋 = − [1 2𝑞 𝑥2] = −1
2 𝑥2
Terms of location Mmax: (first derivative Mx = 0) 𝑑𝑀𝑥
𝑑𝑥 = 0 −x = 0 (TM) x = 0→ ME = 0 x = 2→ MB = −1
222= −2 𝑘𝑁𝑚
3.
Calculating the Shear Force DA ki = 0DA ka = RVA = 3,998 kN DC ki = 3,998 kN
DC ka = 3,998 – P . sin 60° = −0,332 𝑘𝑁 DD ki = DDka = – 0,332 kN
DB ki = –0,332 – 1 . 4 = –4,332 kN DB ka = –4,332 + RVB = 2 kN DE ki = DE ka = 2 – 1 . 2 = 0
4.
Calculating Axial Force (N) NA ki = 0NA ka = RHA = 2,5 kN NC ki = 2,5 kN
57
NC ka = 2,5 – P. cos 60° = 0 ND – NE = 05.
BMD, SFD, and NFDExample 6.2
It is known that a beam is hitched with a load as shown below. Determine the magnitude of the support reaction and calculate and draw the
Bendling Moment Diagram (BMD), Shear Force Diagram (SFD), and Normal Force Diagram (NFD)! (unit of length in meters)
58
Solution :1.
Calculating Support Reaction∑MB = 0
RVA . 4 – 4 . 5 – 5. 4 . 2 + 6 .1 = 0 4 RVA – 20 – 40 + 6 = 0 4 RVA = 54 RVA = 13,5 kg
∑MA = 0
–RVB . 4 + 6 . 5 + 5 . 4 . 2 – 4 . 1 = 0 –4 RVB + 30 + 40 – 4 = 0 4 RVB = 66 RVB = 16,5 kg
Control : ∑V = 0
RVA + RVB + P1 – Q – P2 = 0 13,5 + 16,5 – 4 – 5 .4 – 6 = 0
0 = 0 (OK!)
∑H = 0 RHB = 0
2.
Calculating the Bending Moment (M) Interval C-A (0 ≤ x ≤ 1)59
MX = – P1 . x = – 4 xx = 0 →MC = 0
x = 1 → MA = – 4 . 1 = – 4 kgm Interval A-B (0 ≤ x ≤ 4)
MX = – P1 . (x+1) + RVA . x – 1
2𝑞 𝑥2 = – 4 (x + 1) + 13,5x – 5
2𝑥2
Terms of location Mmax: (first derivative Mx = 0) 𝑑𝑀𝑥
𝑑𝑥 = 0
–4 + 13,5 – 5 x = 0 –5x + 9,5 = 0
x = 1,9 m from point A (OK) x = 0 → MA = – 4 (0 + 1) + 13,5 . 0 – 5
2 . 02= – 4 kgm x = 1,75→ Mmax = – 4 (1,9 + 1) + 13,5 . 1,9 – 5
2 . 1, 92= 5,025 kgm
x = 4→MB = – 4 (4 + 1) + 13,5 . 4 – 5
2 . 42= – 6 kgm Location M = 0
MX = 0
–4 (x + 1) + 13,5 x – 5
2𝑥2= 0 –5
2𝑥2+ 9,5 𝑥 – 4 = 0 5𝑥2− 19 𝑥 + 8 = 0
x1 = 0,482 m from point A (OK) x2 = 3,318 m from point A (OK)
Interval D-B (0 ≤ x ≤ 1) → counted from right to left
60
MX = – [ P2 . x]= – 6 x x = 0→ME = 0
x = 1→MB = –6 . 1 = –6 kgm 3. Calculating the Shear Force
DC ki = 0
DC ka = – P1 = – 4 kg DC ki = – 4 kg
DC ka = – 4 + RVA = 9,5 kg DB ki = 9,5 – 5 . 4 = –10,5 kg DBka = –10,5 + RVB = 6kg DE ki = 6 kg
DE ka = 6 – P2 = 0
4. Calculating Axial Force (N)
N = 0 (because there is no horizontal force acting)
61
5. BMD, SFD, and NFD62
Exercise
Calculate and Draw the Bending Moment Diagram (BMD), Shear Force Diagram (SFD) and Normal Force Diagram (NFD)
63
REFERENCESFrick H. 2006. Mekanika Teknik I. Yogyakarta: Kanisius, Gunawan T., Margaret, 1985. Teori, Soal dan Solution Mekanika
Teknik I. Jakarta: Delta Teknik Group.
Soemono. 1997. Statika 1. Bandung: Penerbit ITB.
Wesli. 2010. Mekanika Rekayasa. Yogyakarta: Graha Ilmu.
POLINEMA PRESS POLITEKNIK NEGERI MALANG Jl. Soekarno-Hatta no.09 PO BOX 04 Malang 65141 Telp. (0341) 404424, 404425 Fax. (0341) 404420 UPT. Percetakan dan Penerbitan Gedung AU ground floor [email protected] www.polinemapress.org
9 786236 562741
ISBN : 978-602-66956-9-7
