Civil Engineering Dept.
CE 201 Engineering Statics
Dr. Rashid Allayla
1.1 , 1.2, 1.3, 1.4) Fundamentals:
Scalar versus Vector:
Scalar quantity is a quantity that has magnitude only and is independent of direction. Examples include: Time, Speed, Volume and Temperature. On the other hand, vector quantity has both magnitude and direction. Examples include: Force, Velocity and Acceleration.
Graphical representation of a vector:
The symbol → above the letter q indicates that q is vector. The magnitude of q is designated as by the symbol ׀q׀.
Basic definitions:
Length: Designated by the letter L (cm, mm, m, km, inch, ft, mile)
Mass: Designated by the letter M (kg, lb)
Force: Designated by the letter F (N “Newton”, lbf “pound force”)
Particle: A particle is a mass of negligible size with no particular geometry.
Rigid Body: It is a combination of large number of particles that occupy more than one point in space and located a fixed distance from each other both before and after applying a load.
Concentrated Force: All loads are acting on a point on a very small body.
Newton
Newton’s Laws of Motion:
First Law:
“A particle in a state of uniform motion or at rest tends to remain in that state unless subjected to an external force".
Example:
A 10 N object is moving at constant speed of 10 km / hr on a friction free surface. Which one of the horizontal forces is necessary to maintain this state of motion?
a) 0 N b) 1 N c) 2 N ?
Answer:
It does not take any force to maintain the motion as long as the surface is friction free. Any additional force will accelerate or decelerate the motion depending on the force applied.
Second Law:
“The acceleration of a particle is proportional to the resultant force acting on it and moves in the same direction of this force”
f = ma
Where “f” is the force, “m” is the mass and “a” is the acceleration. In this notes, instead of placing arrows above forces, they will be written in bold letters instead.
Third Law:
“For every action there is reaction. The mutual forces of action and reaction are equal in magnitude and opposite in direction and collinear in orientation".
F (Action) F (Reaction)
Online Conversion Unit: Go to http://www.onlineconversion.com/
SI Units:
SI is known as the International System of Units where Length is in meters (m), time is in seconds (s), and mass is in kilograms (kg) and force is in Newton (N) (1 Newton is the force required to give 1 kilogram of mass an acceleration of 1 m/s2).
US Customary System of Units (FPS); is the system of units where length is in feet (ft), time is in seconds (s), and force is in pounds (lb). The unit mass is called a slug (1 pound is the force required to give one slug of mass an acceleration of 1 ft/s2).
Conversion of Units:
Force; 1 lb (FPS Unit) = 4.4482 N (SI Unit)
Mass; slug (FPS Unit) = 14.5938 kg (SI Unit)
Length; ft (FPS Unit) = 0.304 m (SI Unit)
Prefixes:
Giga = G = 109 = 1 000 000 000 Milli = m = 10-3 = 0.001
Mega = M = 106 = 1 000 000 Micro = μ = 10-6 = 0.000 001
Kilo = k = 103 = 1 000 Nano = η = 10-9 = 0.000 000 001
Example:
If one lb of an object has a mass of 0.4536 kg, find the weight in Newton's.
Solution: Mass Acceleration Force
Weight in Newton's: (0.4536 kg) (9.807 m / s2) = 4.448 N
Civil Engineering Dept.
CE 201 Engineering Statics
Dr. Rashid Allayla
Force Vectors
2.1, 2.2) 2.3) Vectors, Vector Operations and Vector Addition of Forces:
A force represents an action of one body on another. A force is defined by the following components:
a) Point of application b) Magnitude c) Direction
Forces F1 and F2 acting on a particle may be replaced by a single (resultant) force R which will have the same effect on the particle. The resultant force R can be found by constructing a parallelogram. So it is evident that vector addition does not obey ordinary arithmetic addition, that is, two forces of 9 and 3 lb magnitudes do not add up to 12 lb. On the other hand, if the two vectors are collinear (i.e. acting on the same line), arithmetic addition (or scalar addition) will apply.
Vector Addition Using Triangular Construction:
Required: Add the two vectors A and B
Method: We can add the two vectors by connecting the tail of B to the head of A or connecting the head of B to the tail
Vector Subtraction Using Triangular Construction:
Vector subtraction is a special case of vector addition. It is carried out by reversing the sign of the vector to be subtracted and performing the same rule of vector addition
Required: Subtract vector B from A
Resolution of a Vector:
Resolution of a vector into two vectors acting along any two given lines is carried out by constructing parallelogram as shown in the illustration below:
Vector Addition of Number of Forces:
Vector addition of n forces is accomplished by successive application of parallelogram
law as described above and as shown in the following illustration:
Law of Sine and Cosine:
The magnitude of the resultant force can be obtained using the law of cosines and the direction can be obtained using the law of sines.
Given: force A and Force B as shown below
Required: The resultant force and its direction using Sine & Cosine laws.
Cosine Law: R = SQRT (A2 + B2 – 2 AB Cos β)
Sine Law: A/Sin γ = B / Sin α = R/ Sin β
Resolving Resultant to Components Using Law of Sine:
’
Ax = - A Cos α = A Cos (180 - α)
Ay = A Sin α = A Sin (180 – α) Note that: Ax’ ≠ A Cos α
EXAMPLE:
Determine the magnitude and direction of force P such that the resultant of the two forces on the pulling tug boat ( P & T ) is equal to 4.00 kN.
Solution:
Using Cosine Law: P = SQRT[ 42 + (2.6)2 - 2 x 4 x 2.6 cos 20o] Gives: P = 1.8 kN
Using Sine Law: 2.6 / Sin θ = 1.8 / Sin 20o Gives: θ = 30o
The resultant is found using triangular law (see figure) R = 4.0 KN
EXAMPLE: (Beer & Johnston)
Two forces A = 40N and B = 60N acting on bolt C. Determine the magnitude and the direction of the resultant R using law of Cosine & Sine.
B = 60 N
25o
A = 40 N
20o
Solution:
Drawing the system using triangular rule and applying the law of cosine: A = 40 N
25O
R2 = A2 + B2 – 2 AB Cos [β)] But: β = 180-25=155
B=60 N R
= (402) + (602) – 2 (Cos 155) β
θ α = 97.7 N
Applying the law of Sines:
A / Sin α = R / Sin 155 where α is the angle opposite to vector A.
40 / Sin α = 97.7 / sin 155o then α = Sin-1 (40) Sin 155 / 97.7 = 0.173 = 10o
Then θ = (25+20) – 10 = 35o
EXAMPLE: (Beer & Johnston)
Two forces are applied as shown to a hook support. Using trigonometry and knowing that the magnitude of P is 14lb, determine (a) the required angle α if the resultant R of the two forces applied to the support to be horizontal, (b) the corresponding magnitude of R.
Solution: 20 lb 30o
Force Triangle: R α α
20 lb P =14 lb
P = 14 β
R α 30o
P = 14 lb
Using law of sines:
20 / Sin α = P / Sin 30 = R / Sin β
Since P = 14 lb, then: Sin α = (20 / 14) Sin 30 = 0.71428 → α = 45.6o
The value of β: β + α + 30 = 180 → β = 104.4 then 14 / Sin 30 = R / Sin 104.4 Gives
R = 27.1 lb
2.3) Vector Addition of Forces
The successive application of parallelogram method to find the resultant of set of forces is often tedious. Instead, it would be easier to find the components of the forces along specified axis algebraically and then find the resultant.
It is often desirable to resolve a force into two components which are perpendicular to each other as shown below.
Unit Vector:
A unit vector is a vector directed along the positive x and y axis having dimensionless magnitude of unity. Any vector can be expressed in terms of the unit vector as, F = Fx i + Fy j
Where i and j are the unit vectors in x and y direction and Fx and Fy are the “scalar” magnitudes of F in x and y direction. The two magnitudes can be either positive or negative depending on the sense of Fx and Fy.
If θ is measured counterclockwise from the positive x axis, the magnitude of the force is measured as
Fx = F Cos θ and Fy = F Sin θ
2.4) Resultant of Coplanar Forces:
In order to obtain the resultant of a set of coplanar forces, each force is resolved into x and y components and then added algebraically to obtain the resultant. In the figure below, F1, F2 and F3 are a set of coplanar forces. In Cartesian vector notation, the forces are written as
F1 F2
F3y F3x
F3
F1 = - F1x + F1y , F2 = F2x + F2y , F3 = F3x - F3y
The resultant is: FR = F1 + F2 + F3 Angle resultant makes with + x axis
FR = (-F1x + F2x + F3x) i + (F1y + F2y – F3y) j & ІFR І = SQRT ( FRx2 + FRy2 ) θ = Tan-1 (ІFRy І / ІFRx І)
EXAMPLE: (Beer & Johnston)
Four forces act on bolt A, determine the resultant of the forces on the bolt.
F2 F1
F4
F3 F1 Cos 30 i
F4 Cos 15 i
-F2 Cos 20 i
θ = Tan-1 ( 14.3/199.9) = 4.1o
EXAMPLE:
Determine magnitude and direction cosine of resultant (R) of the following force vectors:
F1 = 5i + 15 j + 30 k (N)
F2 = 25i + 30 j - 40 k (N)
F3 = - 25j - 50 k (N)
Solution:
R = ∑ Fi = F1 + F2 + F3
R = 30 i + 20 j - 60 k R = SQRT [(30)2 + (20)2 + (60)2] = 70 N
Cos α = Rx / │R │ = 30 / 70 = 0.42857 α = 64.6o
Cos β = Ry / │R │ = 20 / 70 = 0. 28571 β = 73.4o
Cos γ = Rz / │R │ = -60 / 70 = -0.8571 γ = 149.0o
Check the result Cos2 α + Cos2 β + Cos2 α = 1 (0.42857)2 + (0.28571)2 + (0.8571)2 = 1 OK
28
2.5, 2.6) Cartesian Vectors & Position Vectors:
Cartesian vector is a set of unit vectors i, j and k that defines the direction of a given vector. It locates a point in space relative to a second point. Unit vector in the direction of a given vector (such as the one shown in the figure) is obtained by dividing the position vector rAB by the magnitude of rAB:
C
Unit vector is useful to express a force in a vector form. When a unit vector acting in the same direction of the force is multiplied by the magnitude of the force, a vector representation of the force is accomplished.
F = │F│uAB and, therefore, ux = Fx / │F│ uy = Fx / │F│ ux = Fz / │F│
UF = (Fx / │F│) i + (Fy / │F│) j + (Fz / │F│) k Then: U = Cos α i + Cos β j + Cos γ k
Note that the sum of squares of direction cosines is unity because │uF│ = 1
Cos α2 i + Cos2 β j + Cos2 γ k = 1
EXAMPLE: (From umr)
Write a unit vector in the direction from B to A
Solution:
The unit vector from B toward A UBA = rBA / │ rBA │
rBA = (XA – XB) i + (YA – YB) j + (ZA – ZB) k
= (-6 – 3) i + (8 – (-4)) j + (5- (-2)) k
= -9 i + 12 j + 7 k m
The magnitude of uBA:
rBA = SQRT [(9)2 + (12)2 + (7)2] = 16.553
uBA = (-9 i + 12 j + 7 k) / 16.553 = -0.5437 i + 0.7249 j + 0.4229 k
EXAMPLE: (From umr)
Determine the distance between point A and B located as shown using a position vector.