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
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 000Milli = 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 F2acting 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 Cosα
EXAMPLE:
Determine the magnitude and direction of force P such that theresultant 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 = Fxi + 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
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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 locatesa 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 + Cosγk
Note that the sum of squares of direction cosines is unity because │uF│ = 1
Cos α2i + Cos2βj + Cos2γk = 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.
4 m
Solution:
The position vector in the direction AB is
rAB = (XB – XA ) i + ((YB – YA ) j + (ZB – ZA ) k
= (4 – (-2)) i +(12 – (-6)) j + (-2 – 3 ) k = 6 i - 18 j – 5 k m
The distance from A to B is │r │ = ( 62 +182 + 52 ) =19.62 m
EXAMPLE:(From Hibbeler)
Determine the magnitude and the coordinate direction angle of the resultant force acting on the ring.
The resultant force FR = 50 i – 40 j + 180 k
The magnitude = SQRT [ (50)2 + (-40)2 + (180)2 ] = 191.0 lb
UFR = (50 / 191.0 i – (40 / 191.0) i + ( 180 / ( 191.0) k = 0.2617 i – 0.2094 j + 0.9422 k
Then Cos α = 0.2617 α =74.8o and: Cos β = - 0.2094 β =102o
and: Cos γ = 0.9422 γ=19.6o
Example: (Hibbeler)
A roof is supported by cables as shown. If the cables exert forces FAB = 100 N and FAC 120 N on the wall hook at A as shown. Determine the resultant force at A.
The position vector AB rAB = (4 m – 0) i + (0 – 0) j + (0 – 4 m) k = 4 i – 4 k
׀rAB׀= SQRT [ (4)2 + (-4)2] = 5.66 m
Then: FAB = (100 N) [ rAB / ׀rAB׀] = (100 N) [ (4 / 5.66) i – (4 / 5.66) k ]
FAB = [70.7 i – 70.7 k] N
The position vector AC rAB = (4 m – 0) i + (2 m – 0) j + (0 – 4 m) k = 4 i + 2j – 4 k
׀rAC׀= SQRT [ (4)2 + (2)2 + (-4)2] = 6 m
Then: FAB = (120 N) [ rAC / ׀rAC׀] = (120 N) [ (4 / 6) i + (2 / 6) – (4 / 6) k ]
FAB = [80 i + 40 j – 80 k] N
The resultant force is:
FR = FAB + FAC = [70.7 i – 70.7 k] + [80 i + 40 j – 80 k]
= [150.7 i + 40 j – 150.7 k] N
׀FR׀= SQRT [ (150.7)2 + (40)2 + (-150.7)2 ] = 217 N
2.7, 2.8)Position Vector Along a Line:
We have shown that the position vector along a line AB is:
u = (XB – XA ) i + ((YB – YA ) j + (ZB – ZA ) k / SQRT{(XB – XA )2+ ((YB – YA )2 + (ZB – ZA )2} or: u = r / │r│
If we have a force F with magnitude of │F│ acting along the line AB, then the vector F is defined as:
F = u │F│ Where: u is the unit vector acting along the line AB as defined above.
EXAMPLE 1:(Beer & Johnston)
A towe guy wire is anchored by means of bolt at A. The reaction in the wire is 2500 N. Determine a) the components Fx, Fy and Fz, b) The angles α, β and γ
The distance from A to B = SQRT [ (40)2 + (80)2 + (30)2] = 94.3 m Then:
Position vector: AB = - 40 i+ 80 j+ 30 k , The unit vector uAB = - (0.4242) i + (0.8484) j + (0.3181) k
The vector Along AB = (2500) uAB =- (1060.5 N) i + (2121 N) j + (795.33 N) k
Direction of force: α = Cos-1 [-1060 / 2500] =115.1o, β = Cos-1 [2120 / 2500] = 32.0o, γ = Cos-1 [795 / 2500] = 71.5o
2.9)Dot Product:
Dot product of two vectors P and Q (otherwise known as scalar product) is defined as the product of their two magnitudes and the cosine of the angle formed by P & Q. Dot product of two vectors is useful for:
a) determining the angle between two vectors, and,
b) determining the projection of a vector along a specified line.
Let: P = Pxi + Pyj + Pzk and: Q = Qx i + Qyj + Qzk
Then: P.Q = │P││Q│ cos θ Ξ Px Qx + PyQy + Pz Qz
Rules:
1) Dot product follows commutative law: Q . P = P.Q
2) Dot Product follows distributive law:P. (Q1 + Q2) = P. Q1 + P. Q2
3) Multiplication by a scalar: a (P.Q) = (a P) . (Q) = (P) . (a Q) = (P.Q)a
uP = P/│P│
P.Q = │P││Q│ cos θ or: cos θ = [P.Q / │P││Q│] or: θ Ξ cos-1 (uP . uQ)
Since │Pa-a │=│P│cos θ Then: │Pa-a│= P . ua-a and: Pa-a = [ P . ua-a ]uaa
Usefulness of Dot Product:
- Angle between two intersecting vectors can be determined:
θ = cos-1 [P.Q / │P││Q│]
- The component of a vector parallel and perpendicular to a line can be determined if the unit vector along this line is known:
F ║ = F cos θ = F.u
Since F = F ║+ F ┴ Then: F ┴ = F - F ║
Projecting a Force Along a Line:
Given: A force FAB = A i + B j + C k along line AB
Required: The projection of this force along line AC
Method of solution:
1) Find the unit vector along the line AC
UAC = [(xC – xA) i + (yC – yA) j + (zC – zA)] / SQRT [[(xC–xA)2+[(yC – yA)2+ [(zC – zA)2]
2) Use the dot product to find the projection of the force along AC:
׀FAC׀= UAC . FAB
׀FAC׀= { [(xC – xA) i + (yC – yA) j + (zC – zA)] / SQRT [[(xC–xA)2+[(yC – yA)2+ [(zC – zA)2]} . { A i + B j + C k }
This is scalar value which is the projection of force F into line AC
The Cartesian vector of the projection is:
FAC = UAC׀FAC׀
EXAMPLE:(from umr)
The force F = 50i + 75 j + 100 k acts on a pole as shown. Determine the projected component of F along and perpendicular to the pole.
The unit vector along AB = rAB /│ r │ = {(4-3) i + [4-(-2)] j + (6-0) k} / [12 + 62 + 62 ]
Then: uAB = (0.117 i + 0.702 j + 0.702 k)(This is the unit vector along AB)
FAB (the projection of F on AB) = F . uAB = ( 50i + 75 j + 100 k) . (0.117 i + 0.702 j + 0.702 k)
=23.41 lb
The Cartesian vector from the parallel component is FAB . uAB = 23.41 (0.117 i + 0.702 j + 0.702 k)
F║ =2.74 i + 16.44 j + 16.44 k lb
The component of the force perpendicular to AB is = F – FAB = (50i + 75 j +100 k) – (2.74 i + 16.44 j + 16.44 k) lb
F┴ =47.3 I – 91.4 j + 85.6 k lb
EXAMPLE 2:
Find the a) angle between cable BD and the boom AB and b) the projection on AB of cable BD at point B.
Unit vector in AB direction = ( 6 i + 4.5 j ) / SQRT ( 62 + 4.52 ) = 0.8 I + 0.6 j
The angle between BD and AB is = cos-1 [ uAB . uBD ] = cos-1 [(0.8i +0.6j) . (-0.67i +0.33j -0.67 k) ]
= cos-1[- 0.536 + 0.19] =110.24o
Force BD = (180) (UBD) = (180) (-0.67i +0.33j -0.67 k) = - 120.6 i + 59.4 j + 120.6 k
The projection of BD on AB = UAB . FBD = (-0.8i +0.6j ).(- 120.6 I + 59.4 j + 120.6 k) = - 96.48 + 35.6
=60.88‘
1) Equilibrium of A Particle:
A particle is said to be at equilibrium if the resultant of all forces acting on it is zero. Another case of equilibrium is illustrated in the figure below. If the four forces acting on a particle at point O are at equilibrium, then starting from point O with F1 and arranging the forces in tip to tail fashion, the tip of F4 will coincide with the tail of force F1 and the resultant of the four forces will be zero. The graphical representation is expressed mathematically as:
∑ F = 0
Free-Body Diagram:
What ? - It is a drawing that shows all external forces acting on the particle.
Why ? - It helps you write the equations of equilibrium used to solve for the unknowns (usually forces or angles) (Hibbeler)
Therefore: - Free body diagram is a method of isolating the forces acting on a body from its surroundings and drawing the forces acting on the body.
Procedure for Drawing Free Body Diagram (FBD):
1) Isolate the particle from its surroundings.
2) Sketch all forces that act on the particle while observing Newton’s third law which notes the existence of equal and opposite reaction to every action.
3)Known forces are labeled with their magnitudes and directions. Assign letters to the unknown forces with assumed directions. The body’s weight must be included if applicable.
EXAMPLE:
Draw the free body diagram of the two structures shown
Springs:
The magnitude of the force exerted on the linear elastic spring is:
F = Ks
where K is the stiffness of the spring (measured in N/m), s is the deformation (which is a measure of the difference between the deformed length L and the undeformed length L0). Note that if s is negative, F must push on the spring and if s is positive, F must pull on the spring to bring it to the desired length. K is also defined as the force required to deform the spring a unit distance.
EXAMPLE:
A spring has undeformed length of 0.4 meters and stiffness k = 500 N/m. What is the force needed to stretch the spring to a length of 0.6 m? and what force is required to compress the spring to a length = 0.2 m?
SOLUTION:
F = K s
F = (500 N/m) (0.6 m – 0.4 m) = 100 N (s is positive, force is pulling spring)
F = (500 N/m) (0.2 m – 0.4 m) = -100 N (s is negative, force is pushing spring)
Cables and Pulleys:
When a cable is passing over a frictionless pulley, the force along the cable is always in tension and constant in magnitude. This is necessary condition to keep the cable in equilibrium.
EXAMPLE:
The spring has stiffness of 500 N/m and outstretched length of 6 m. Determine the horizontal force F applied to the cord so that the displacement of the pulley from the wall is d = 1.5 m.
SOLUTION:
AC = 3.3541 m
Σ Fx = 0 If the tension at each spring is T Then: Tx = (1.5/3.3541) T
Then, 2(1.5 / 3.3541) (T) – F = 0 = 3.3541-3=0.3541 m
T = K S = (500)( 0.3541) = 177.05 N Then: F = 158 N
2) Coplanar Force System:
Procedure for Solution of Problems in Equilibrium:
- Establish x-y Coordinate system. & Draw free body diagram.
- Label all known forces and unknown forces and assume the direction of unknown forces.
- Apply the equations of equilibrium ∑ Fx = 0 and ∑ Fx = 0.
- Compare the number of unknowns on the free body diagram with the number of independent equations of equilibrium available.
- If there are more unknowns to be evaluated than the number of equations, draw a free body diagram of another body and repeat the steps described above.
EXAMPLE : (From Higdon & Stiles)
A 500 N shaft A and 300 N shaft B are supported as shown. Neglecting friction at all contact points find the reactions at points R and S at shaft A.
The first FBD (Shaft A ) has three unknowns Q, R and S and only two independent equations of equilibrium. The next step is to draw FBD of shaft B. The force on shaft B exerted by shaft A is Q directed to the upper right. Writing the equation:
∑ Fy = 0 then: Q sin 40o – 300 = 0 then Q = 467 N on B
From the FBD of shaft A:
∑Fy = 0 then: S – 500 – Q sin 40o = 0 then S = 800 N directed upward.
∑Fx = 0 then: R – Q cos 40o = 0 then: R = 467 cos 40o = 358 N directed to the right
EXAMPLE : (From Hibbeler)
Determine the required length of cord AC so that the 8 kg lamp is suspended in the position shown. The unreformed length of the spring is L’AB = 0.4 m and its stiffness is 300 N/m
Weight of lamp W = 8 (9.81) = 78.5 N
∑Fx = 0 TAB – TAC cos 30o = 0
∑Fy = 0 TAC sin 30o – 78.5 = 0
Then: TAC = 157.0 N and TAB = 136.0 N which is the stretch of spring AB
TAB = K s or: 136.0 = 300 s Then s = 0.453 m
The stretch length is LAB = LAB’ + SAB = 0.4 + 0.453 = 0.853 m
The horizontal distance CB requires that 2 = LAC cos 30o + 0.853 Then LAC = 1.32.
EXAMPLE: (from umr)
The pulley system is used to rise a 50 lb weight. Determine the tension T necessary to hold the weight in equilibrium.
∑ Fy = 0 Then: from FBD (3): T + T – TB = 0 Or: TB = 2 T
∑ Fy = 0 Then from the FBD (4): TB + T – 50 = 0 or: T = 50 / 3 = 16.67 lb
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Example:
Calculate the tension T in the cable which supports the 1000 lb load with the system of pulley shown. Ignore the weight of the pulley. Find F at Pulley C.
Solution:
Start with the free body diagram (pulley A) because it has the only known force:
∑ Fy = 0 Then: T1 + T2 = 1000
Since the cord around pulley A is continuous, T1 = T2 Then: T1 = T2 = 500 lb↑
Since the cord around pulley B is continuous, T3 = T4 = T2 / 2 =250 lb↑
Again, since the cord around pulley C is continues T3 = T Then: T = T3 =250 lb ↑
Evaluation of reaction forces at pulley C:
∑ Fx = 0 250 cos 30o – Fx = 0 Then: Fx =217 lb ←
∑ Fy = 0 - Fy + 250 sin 30o -250 = 0 Then Fy = - 125 lb ↓ or: +125 lb ↑
F = SQRT [ (217)2 + (125)2 ] =250 l
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