Civil Engineering Dept.

CE 201 Engineering Statics

Dr. Rashid Allayla

Lecture 1

General Principles

(1.1-1.6)

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: http://scienceworld.wolfram.com/biography/photo-credits.html - Newton

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

Lecture 2

Force Vectors

Vectors, Vector Operations and Vector Addition of Forces (2.1, 2.2 & 2.3)

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

Resultant

Resultant

B

A

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

B

Resultant

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

B

B

Resolving Resultant to Components Using Law of Sine:

A_y^’

β

α

Ax^’

Y^’

A

α

^’

A_x = - A Cos α = A Cos (180 - α)

A_y = A Sin α = A Sin (180 – α) Note that: A_x^’ ≠ 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 20^o] Gives: P = 1.8 kN

Using Sine Law: 2.6 / Sin θ = 1.8 / Sin 20^o Gives: θ = 30^o

P

θ P 2.6 N

20^o

2.6 N θ 4.0 KN

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

25^o

A = 40 N

20^o

Solution:

Drawing the system using triangular rule and applying the law of cosine: A = 40 N

25^O

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 155^o then α = Sin^-1 (40) Sin 155 / 97.7 = 0.173 = 10^o

Then θ = (25+20) – 10 = 35^o

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 30^o

Force Triangle: R α α

20 lb P =14 lb

P = 14 β

R α 30^o

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 = 45.6^o

The value of β: β + α + 30 = 180 → β = 104.4 then 14 / Sin 30 = R / Sin 104.4 Gives

R = 27.1 lb

Lecture 3

Force Vectors

Addition of System of Coplanar Forces (2.4)

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

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, F_1, F₂ and F₃ are a set of coplanar forces. In Cartesian vector notation, the forces are written as

F₁ F₂

F_3y F_3x

F₃

F₁ = - F_1x + F_1y , F₂ = F_2x + F_2y , F₃ = F_3x - F_3y

The resultant is: F_R = F₁ + F₂ + F_{3 Angle resultant makes with + x axis}

F_R = (-F_1x + F_2x + F_3x) i + (F_1y + F_2y – F_3y) j & ІF_R І = SQRT ( F_Rx² + F_Ry² ) θ = Tan^-1 (ІF_Ry І / ІF_Rx І)

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 = F_x i + F_y j

Where i and j are the unit vectors in x and y direction and F_x and F_y 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 F_x and F_y.

If θ is measured counterclockwise from the positive x axis, the magnitude of the force is measured as

F_x = F Cos θ and F_y = F Sin θ

EXAMPLE: (Hibbeler)

Determine the x and y components of F₁ and F₂ acting on the boom and express each force as a Cartesian vector.

1) Scalar Notation:

F_1x = -200 Sin 30 = -100 N

F_1y = 200 Cos 30 = 173 N

F_2x / 260 = 12/13 Gives: F_2x = 240 N

F_2y / 260 = 5/13 Gives: F_2y = 100 N

2) Cartesian Notation:

Having determined the magnitudes of forces and their directions, then:

F₁ = --100 i + 173 j

F₂ = 240 i – 100 j

EXAMPLE: (Beer & Johnston)

Four forces act on bolt A, determine the resultant of the forces on the bolt.

Force Magnitude N X-Component N Y-Component

F₁ 150 +129.9 + 75.0

F₂ 80 -27.4 + 75.2

F₃ 110 0.0 - 110.0

F₄ 100 +96.6 - 25.9

R_x = +199.9 R_y = +14.3

F₂ F₁

F₄

F₃ F₁ Cos 30 i

F₄ Cos 15 i

-F₂ Cos 20 i

θ = Tan^-1 ( 14.3/199.9) = 4.1^o

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 = ∑ F_i = F₁ + F₂ + F₃

R = 30 i + 20 j - 60 k R = SQRT [(30)² + (20)² + (60)²] = 70 N

Cos α = R_x / │R │ = 30 / 70 = 0.42857 α = 64.6^o

Cos β = R_y / │R │ = 20 / 70 = 0. 28571 β = 73.4^o

Cos γ = Rz / │R │ = -60 / 70 = -0.8571 γ = 149.0^o

Check the result Cos² α + Cos² β + Cos² α = 1 (0.42857)² + (0.28571)² + (0.8571)² = 1 OK

Lecture 4

Force Vectors

Cartesian Vectors (2.5, 2.6)

Cartesian 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 r_AB by the magnitude of r_AB:

z

B (x_B, y_B, z_B)

A

Y r_AB = (X_B – X_A) i + (Y_B – Y_A) j + (Z_B – Z_A) k

│r_AB│ = SQRT [ (X_B – X_A)² + (Y_B – Y_A)² + (Z_B – Z_A)² ]

Unit Vector u_AB = r_AB / │r_AB│ = (X_B – X_A) i + (Y_B – Y_A) j + (Z_B – Z_A) k /

SQRT [ (X_B – X_A)² + (Y_B – Y_A)² + (Z_A – Z_B)² ]

z_A x_A

y_A

x

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│u_AB and, therefore, u_x = F_x / │F│ u_y = F_x / │F│ u_x = F_z / │F│

U_F = (F_x / │F│) i + (F_y / │F│) j + (F_z / │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 │u_F│ = 1

Cos α² i + Cos² β j + Cos² γ k = 1

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

r_AB = (X_B – X_A ) i + ((Y_B – Y_A ) j + (Z_B – Z_A ) 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 │ = SQRT[( 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 F_R = (50+0) i + (-100+40) j + (100+80) k = 50 i – 40 j + 180 k

The magnitude = SQRT [ (50)² + (-40)² + (180)² ] = 191.0 lb

U_FR = (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 = 74.8^o and: Cos β = - 0.2094 β = 102^o

and: Cos γ = 0.9422 γ= 19.6^o

Lecture 5

Force Vectors

Position Vector, Force alona Line (2.7, 2.8)

We have shown that the unit vector along a line AB is:

Called position vector r

u ₌ (X_B – X_A ) i + ((Y_B – Y_A ) j + (Z_B – Z_A ) k / SQRT{(X_B – X_A )² + ((Y_B – Y_A )² + (Z_B – Z_A )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: (From umr)

Write a unit vector in the direction from B to A

Solution:

The unit vector from B toward A U_BA = r_BA / │ r_BA │ and the position vector is:

r_BA = (X_A – X_B) i + (Y_A – Y_B) j + (Z_A – Z_B) k

= (-6 – 3) i + (8 – (-4)) j + (5- (-2)) k

= -9 i + 12 j + 7 k m

The magnitude of the unit vector u_BA is: U_BA = r_BA / │ r_BA │

Where, r_BA = SQRT [(9)² + (12)² + (7)²] = 16.553 Then:

U_BA = (-9 i + 12 j + 7 k) / 16.553 = -0.5437 i + 0.7249 j + 0.4229 k

EXAMPLE : (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 F_x, F_y and F_z, 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: r_AB = - 40 i + 80 j + 30 k , The unit vector u_AB = - (0.4242) i + (0.8484) j + (0.3181) k

The vector Along AB = [ (2500) u_AB ] = - (1060.5 N) i + (2121 N) j + (795.33 N) k

Direction of force: α = Cos^-1 [-1060 / 2500] = 115.1^o, β = Cos^-1 [2120 / 2500] = 32.0^o, γ = Cos^-1 [795 / 2500] = 71.5^o

Example: (Hibbeler)

A roof is supported by cables as shown. If the cables exert forces F_AB = 100 N and F_AC 120 N on the wall hook at A as shown. Determine the resultant force at A and its magnitude.

The position vector AB: r_AB = (4 m – 0) i + (0 – 0) j + (0 – 4 m) k = 4 i – 4 k

׀r_AB׀ = SQRT [ (4)² + (-4)²] = 5.66 m

Then: F_AB = (100 N) [ r_AB / ׀r_AB׀ ] = (100 N) [ (4 / 5.66) i – (4 / 5.66) k ]

F_AB = [70.7 i – 70.7 k] N

The position vector AC r_AB = (4 m – 0) i + (2 m – 0) j + (0 – 4 m) k = 4 i + 2j – 4 k

׀r_AC׀ = SQRT [ (4)² + (2)² + (-4)²] = 6 m

Then: F_AC = (120 N) [ r_AC / ׀r_AC׀ ] = (120 N) [ (4 / 6) i + (2 / 6) – (4 / 6) k ]

F_AC = [80 i + 40 j – 80 k] N

The resultant force is:

F_R = F_AB + F_AC = [70.7 i – 70.7 k] + [80 i + 40 j – 80 k]

= [150.7 i + 40 j – 150.7 k] N

׀F_R׀ = SQRT [ (150.7)² + (40)² + (-150.7)² ] = 217 N

Lecture 6

Force Vectors

Dot Product (2.9)

Dot product of two vectors P and Q (otherwise known as scalar product) is defined as the product of the scalar magnitudes of the two vectors and the cosine of the angle formed by the vectors. Dot product is useful for:

a) Determining the angle between two vectors, and,

b) Determining the projection of a vector along a specified line.

Let: P = Px i + P_y j + P_z k and: Q = Q_x i + Q_y j + Q_z k

Then: P.Q = │P││Q│ cos θ Ξ Px Q_x + P_yQ_y + P_z Q_z

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

u_{P = P /│P│}

P.Q = │P││Q│ cos θ or: cos θ = [P.Q / │P││Q│] or: θ Ξ cos-1 (u_P . u_Q)

Since │P_a-a│ = │P│cos θ Then: │P_a-a│ = P. u_a-a and P_a-a = [ P . u_a-a ] u_aa

Usefulness of Dot Product: Vector form of projection of F into x axis

- 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 F_AB = 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

U_AC = [(x_C – x_A) i + (y_C – y_A) j + (z_C – z_A)] / SQRT [[(x_C–x_A)²+[(y_C – y_A)²+ [(z_C – z_A)²_]

2) Use the dot product to find the projection of the force along AC:

׀F_AC׀ = U_AC . F_AB

׀F_AC׀ = { [(x_C – x_A) i + (y_C – y_A) j + (z_C – z_A)] / SQRT [[(x_C–x_A)²+[(y_C – y_A)²+ [(z_C – z_A)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 of F into AC is:

F_AC = U_AC ׀F_AC׀

EXAMPLE 1: (from umr)

The force F = 50i + 75 j + 100 k acts on a pole AB as shown. Determine the projected component of F along AB and component of the force perpendicular to AB.

The unit vector along AB = r_AB /│ r │ = {(4-3) i + [4-(-2)] j + (6-0) k} / SQRT[1² + 6² + 6² ]

Then: u_AB = (0.117 i + 0.702 j + 0.702 k) (This is the unit vector along AB)

F_AB (the projection of F on AB) = F . u_AB = ( 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 F_AB . u_AB = 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 – F_AB = (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 [ u_AB . u_BD ] = cos^-1 [(0.8i +0.6j) . (-0.67i +0.33j -0.67 k) ]

= cos^-1 [- 0.536 + 0.19] = 110.24^o

Force BD = (180) (u_BD) = (180) (-0.67i +0.33j -0.67 k) = - 120.6 i + 59.4 j + 120.6 k

The projection of BD on AB = u_AB . F_BD = (-0.8i +0.6j ).(- 120.6 I + 59.4 j + 120.6 k) = - 96.48 + 35.6

= 60.88^‘

Lecture 7

Equilibrium of Particle

Free Body Diagram (3.1, 3.2)

Conditions for 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 all external forces acting on a body from its surroundings.

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.

T T

EXAMPLE:

The springs have stiffness of 500 N/m each and length of 3 m each. 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:

T

AC = 3.3541 m F 2T Cos θ

Σ F_x = 0 If the tension at each spring is T Then: T_x = (1.5/3.3541) T T_x

Then, 2(1.5 / 3.3541) (T) – F = 0 since: s = 3.3541-3=0.3541 m T

T = K S = (500)( 3.3541 - 3) = 177.05 N Then: F = 158 N

Lecture 8

Equilibrium of Particle

Coplanar Force System (3.3)

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 ∑ F_x = 0 and ∑ F_y = 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:

∑ F_y = 0 then: Q sin 40^o – 300 = 0 then Q = 467 N on B From the FBD of shaft A:

∑F_y = 0 then: S – 500 – Q sin 40^o = 0 then S = 800 N directed upward.

∑F_x = 0 then: R – Q cos 40^o = 0 then: R = 467 cos 40^o = 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 undeformed length of the spring is L^’_AB = 0.4 m and its stiffness is 300 N/m

Y

T_AC

₃₀⁰ T_AB

78.5

78.5 N

Weight of lamp W = 8 (9.81) = 78.5 N

∑Fx = 0 T_AB – T_AC cos 30^o = 0

∑F_y = 0 T_AC sin 30^o – 78.5 = 0

Then: T_AC = 157.0 N and T_AB = 136.0 N which is the stretch of spring AB

T_AB = K s or: 136.0 = 300 s Then s = 0.453 m (this is the amount of stretch

on the spring)

The stretch length is L_AB = L_AB^’ + S_AB = 0.4 + 0.453 = 0.853 m

The horizontal distance CB requires that 2 = L_AC cos 30^o + 0.853 Then L_AC = 1.32.m

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.

∑ F_y = 0 Then: from FBD (3): T + T – T_B = 0 Or: T_B = 2 T Since: ∑ F_y = 0 Then from the FBD (4) T_B + T – 50 = 0 or: T = 50 / 3 = 16.67 lb

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:

∑ F_y = 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 30^o – F_x = 0 Then: F_x = 217 lb ←

∑ F_y = 0 - F_y + 250 sin 30^o -250 = 0 Then F_y = - 125 lb ↓ or: +125 lb ↑

F = SQRT [ (217)² + (125)² ] = 250 l

Example: (Hebbeler)

The sack at A weighs 20 lb, determine the weight of the sack at B and the force in each cord needed to hold the system in the equilibrium position shown.

There are three forces acting on E: T_EG, T_EC and 20 lb and on C: T_CE, T_CD and W_B

From FBD No. 1

∑ F_X = 0 Then: T_EG sin 30^O – T_EC cos 45^O = 0

∑ F_y = 0 Then: T_EG cos 30^O – T_EC sin 45^O – 20 lb = 0 Then: T_EC = 38.6 lb and T_EG = 54.6 lb

From FBD No. 3: T_EC = T_CE = 38.6 lb

From FBD No. 2

∑ F_X = 0 Then: 38.6 cos 45^O – (4/5) T_CD = 0

∑ Fy = 0 Then: (3/5) T_CD + 38.6 sin 45^O – W_B = 0 Then: T_CD = 34.2 lb and: W_B = 47.8 lb

Lecture 9

Equilibrium of Particle

Three Dimensional Force Systems (3.4)

The requirement for equilibrium of a particle is:

∑F = 0 or: ∑F_x + ∑F_y + ∑F_z = 0 The method of solution is summarized as follows:

Draw Free body diagram and:

- Establish appropriate x,y and z axis.

- Label all known and unknown forces and assume the direction of unknown forces.

- Apply the equations of equilibrium ∑F_x = 0, ∑F_y = 0, ∑F_z = 0.

- Reverse the directions of unknown forces if solution yields negative result.

Example: (From Hibbeler)

Determine the magnitudes of F₁, F₂ and F₃ for equilibrium of the particle.

∑ F_z = 0 F₁ sin 30^O – 2.8 = 0 Then: F₁ = 5.6 KN

∑ Fy = 0 - F₂ (24/25) + (8.5) cos 15 = 0 Gives: F₂ = 8.55 k

F₃ – 5.6 cos 30^O – 8.55 (7/25) – 8.5 sin 15^O = 0 Then: F₃ = 9.44 k

EXAMPLE: (From Hibbeler)

The 100 kg crate is supported by three cords, one of which is connected to a spring. Determine the tensions in AC and AD and the stretch of the spring

Note: F_C is directed in - x, - y, +z zone, F_D is directed in –x, +y, +z zone and F_B is directed in +x zone, then:

F_{B =} F_B i

F_C = F_C cos 120 i + F_C cos 135 j + F_C cos 60 k

= -0.5 F_C i - 0.707 F_C j + 0.5 F_C k

F_D = F_D ( -1 i + 2j + 2k) / SQRT (-1² +2² + 2²) = -0.333 F_D i + 0.667 F_D j + 0.667 F_D k

W = - 981 k

Σ F = 0 Then:

Σ F = 0, then: F_B i - 0.5 F_C i – 0.707 F_C j + 0.5 F_D i + 0.667 F_D j + 0.667 F_D k – 981 k = 0

Setting Σ F_x = 0 , Σ F_y = 0 , Σ F_z = 0 We have three unknowns and three equations, then:

F_C = 813 N

F_D = 862 N

F_B = 693.7 N

The stretch of the spring is F_B = k s

693.7 = 1500 s Then s = 0.462 m

EXAMPLE: (From Hibbeler)

A 200 lb box is supported by cables DA, DB and DC find the forces in these cables

Σ F_x = 0 Then:

(4.5–1.5) i / SQRT[( 4.5-1.5)²+(-1.5)²+(3)² ] F_DA + (-1.5) i / SQRT[ (-1.5)² + (2.5-1.5)² +(3)² ] F_DC

= 3 / 4.5 F_DA – 1.5 / 3.5 F_DC = 0.667 F_DA - 0.429 F_DC = 0

Σ F_y=0 Then:

(–1.5) j / SQRT [( 4.5-1.5)²+(-1.5)²+(3)² ] F_DA + (2.5-1.5) j / SQRT [(1.5)²+(3)²+(2.5-1.5)²] F_DC

₊ (-1.5) j / SQRT[( 1.5 )² + 0 + 0) ] F_DB

= -1.5/ 4.5 F_DA + 1 / 3.5 F_DC - F_DB = - 0.33 F_DA + 0.286 F_DC - F_DB = 0

Σ F_z = 0 then:

(3) k /SQRT[(4.5-1.5)²+(1.5)²+(3)²] F_DA+(3) k /SQRT[(1.5)²+(3)²+(2.5-1.5)²] F_DC – W k

= 3 / 4.5 F_DA + 3 / 3.5 F_DC - 200 = 0.667 F_DA + 0.857 F_DC – 200 = 0

The above are three equations and three unknowns F_DA, F_DB and F_DC. Solving yields:

F_DA = 100 lb F_DB = 11.1 lb , and: F_DC = 155.6 lb

Lecture 10

Force System Resultant

Moment of a Force-Scalar Formulation (4.1)

A moment of a force around a point (say O) is a measure of the tendency of the force to rotate the body around the point. The magnitude of the moment is found by multiplying the magnitude of the force with the shortest distance from the force to that point. The sense the moment vector is determined by the right-hand rule defined earlier.

If F is a force along A-A line, the magnitude of the moment of the force around point O is:

M_o = │F│ d = │F││r│sin θ Where θ is the angle between the two vectors

Here: d is the “moment arm” which is the perpendicular distance from O to the line A-A.

F is parallel to x-y axis F is parallel to z-y axis F passes through 0

EXAMPLE (From Hibbeler)

Determine the magnitude and the directional sense of the resultant moment of the force about point P.

Solution:

+

M_P = (260) (5/13) [3] +(260) (12/13) [2] – (400) (sin 30^o) [4-2] + (400) (cos 30^o) [3+5]

= 300 + 480 – 400 + 2771.28= 3151 N

EXAMPLE:

A 100 lb vertical force applied to the end of a lever attached to a shaft. Determine 1) The moment about O; b) The magnitude of horizontal force applied at A which creates the same moment about O; c) The smallest force applied at A which creates the same moment about O; d) How far from the shaft a 240 lb vertical force must act to create the same M_O

a) d = 24 cos 60 = 12 in

M_O = (100 lb) (12 in) = 1200 lb-in Perpendicular to the paper and pointed into the paper.

b) d = (24 in) sin 60 = 20.8 in Then: M_O = 1200 = (20.8) F or: F = 57.8 lb

c) The smallest value of F occurs when d is maximum which is 24 inches, then

1200 = F (24) Then: F = 50 lb

d) 1200 = (240) d Then: d = 5 in

EXAMPLE: (From Hibbeler)

Determine the resultant moment of the four forces acting on the rod shown below.

Assuming that positive moment acts on the +k direction (counterclockwise), then:

+M _R0 = Σ F d

M _R0 = - 50 N (2 m) + 60 N (0) + 20 N (3 Sin 30^o m) – 40 N (4 m + 3 Cos 30^o m)

= - 334 N.m

= 334 N.m

Example: (From Hibbeler)

The 70 N force acts on the end of the pipe at B. Determine the angle θ (0^o < θ > 180) of the force that will produce maximum moment about point A. What is the magnitude of the moment? (a = 0.9 m, b = 0.3 m and c = 0.7 m)

M_A = F sin (θ) c + F cos (θ) a

dM_A/d θ = 0 = F cos (θ) c - F sin (θ) a

θ_max = atan (c/a) Gives: θ = 37.9^o

M_max = (70 N) sin (37.9^o) (0.7 m) + (70 N) cos (37.9^o) (0.9 m)

M_max = 79.812 N.m

Lecture 11

Force System Resultant

Cross Product (4.2)

Direction of Moment:

The direction of moment is determined using the right–hand rule. Starting from the position r and curling fingers into the force F, the direction of the moment is parallel to the direction of the thumb (i.e.: perpendicular to the plane formed by the vectors r and F). We can also use the principle of screw where the direction of the movement of the screw as it is driven, is the same as the direction of the moment when the turning action of the screw is directed from r to F.

Direction of moment

z

Direction of fingers from r to F

x y

r

Position Vector r F

Cross Product:

Cross product of vectors Q and P is a vector V = Q X P directed perpendicular to the plane containing the two vectors. The magnitude of the resulting vector is the product of the magnitudes of the two vectors multiplied by the sine of the angle formed by Q and P. The direction of the resultant vector is determined using the right-hand rule. This suggests that the commutative rule does not apply in cross products as:

V = P X Q

Q P V = Q X P

Q

P X Q ≠ Q X P

However, distributive law can be applied as:

A X (B + C) = (A X B) + (A X C)

The cross product of two vectors P X Q in Cartesian coordinate system is:

P X Q = (P_x i + P_y j + P_z k ) X ( Q_x i + Q_y j + Q_z k)

= P_x Q_x (i X i) + P_x Q_y (i X j) + P_x Q_z (i X k) + P_y Q_x (j X i) + P_y Q_y (j X j) + P_y Q_z (j X k)

+ P_z Q_x (k X i) + P_z Q_y (k X j) + P_z Q_z (k X k)

Noting that: (i X i) = (j X j) = (k X k) = 0 and: (i X j) = (j X k) = (k X i) = 1

(j X i) = (k X j) = (i X k) = -1

i j k

P X Q = (P_y Q_z- P_z Q_y) i – (P_x Q_z – P_z Q_x) j + (P_x Q_y – P_y Q_x) ki ₌ │ P_x P_y P_z │

Where the symbol │ … │ is known as the determinant Q_x Q_y Q_z

EXAMPLE:

A 300 lb force acts on bracket as shown. Find the moment M_A about point A

The moment is the vector M_A = r X F where r is the vector pointing from A to C

M_x = 300 cos 30^Oi(6) =1558.8 in-lb M_y =300 sin 30^O (7) = 1050

Then M_A = 1558-1050 = 508.8 in-lb Pointing into paper

i j k

M_A = r X F = | 7 6 0 | = (7x150 – 6x259.8) k = – 508.8 k (-) Pointing into the paper

259.8 150 0

Example:

Determine the direction θ for 0^O ≤ θ ≥ 180^O of the force F so that F produces a) the maximum moment about point A, b) the minimum moment about point A.

The maximum moment produced when F ┴ BA

M_A = 400 [√ ( 3² + 2²) ] = 1442 N-m Since: Ф = tan^-1 (2/3) = 33.69^O Then θ = 90^O – 33.69= 56.3^O

b) M_A = 0 Then: θ = 180^O – 33.69^O = 146^O

Lecture 12,13

Force System Resultant

Moment of a Force-Vector Formulation (4.3,4.4)

Principle of Transmissibility:

The force applied at point A (shown above) produces a moment M about point O at the direction shown. The moment produced is:

M = F x r₁ = F x r₂ = ׀ F ׀ ׀ r ׀ sin θ = F d

It is evident that the moment produced is the same as long as the position vector r is taken from point O to any point along the line of action of force F (This is called principle of transmissibility)

Resultant Moment of a Set of Forces:

The total moment above is:

(M_R)_O = ∑(r X F)

The moment M of F about an axis through O is a vector normal to a plane formed by F and r and is calculated as:

i j k

M = │ r_x r_y r_z │= [r_y F_z – r_zF_y] i - [r_x F_z – r_zF_x] j + [r_x F_y – r_yF_x] k Ξ r X F

F_x F_y F_z

│M │ = │r│ │F│ sin θ = │F│d

The moment about point O in any given space can be interpreted as the moment about an axis passing through point O and perpendicular to a plane containing point O and force F. The Cartesian representation of moment vector is:

i j k

M_O = r X F = │ r_x r_y r_z │= [r_y F_z – r_zF_y] i - [r_x F_z – r_z(-Fx)] j + [r_x F_y – r_y(-F_x)] k

-F_x F_y F_z

Where: r is the position vector connecting point O to any point on the line of action of F.

F

z

(M_O)_z

r

r_z y

r_x (M_O)_y

r_y

x (M_O)_x

The i component of the moment is the result of three moments: one due to F_x about the x axis (but producing no moment since F_x is parallel to x-axis), the second is due to F_z about the x-axis (Ξ +r_y F_z acts in the positive x-axis), the third is due to F_y about the x-axis (Ξ - r_zF_y acts in the negative x-axis).

Similarly:

The j component of the moment is the result of three moments: one due to F_y about the y axis (but producing no moment since F_y is parallel to y-axis), the second is due to F_z about the y-axis (Ξ - r_x F_z acts in the negative y-axis), the third is due to F_x about the y-axis (Ξ - r_zF_x acts in the negative y-axis). Also: The k component of the moment is the result of three moments: one due to F_z about the z axis (but producing no moment since F_z is parallel to z-axis), the second is due to F_y about the z-axis (Ξ + r_x F_y acts in the positive z-axis), the third is due to F_x about the z-axis (Ξ + r_yF_x acts in the positive y-axis).

Example:

A force P and Q of magnitude 50 N and 100 N act in the directions shown. Determine the moment of P about point F.

2 i - (9-5) j + 6 k

P = 50 N [ ---------------------------] = 50 (0.267 i – 0.535 j + 0.802 k) = 13.35 i – 26.75 j + 40.10 k

SQRT( 4+16+36)

Q = - 100 N j

r_FB = (0-10) i +(9-0) j + (0-4) k = - 10 i + 9 j – 4 k

r_FD = (0-10) i +(0-0) j +(0-4) k = - 10 i – 4 k

i j k i j k

M_F = r_FB x P + r_FD x Q = | -10 +9 - 4 | + | -10 0 - 4 |

13.35 -26.75 40.10 0 -100 0

= [9x40.1 –(-4x(-26.75)]i–[-10x40.1–(-4x13.5) ] j+[-10x(-26.75)-9x13.35]

+[-(-4)x(-100)] i +0 j +[-10x(-100)] k

= (360.9 - 107) i +(400.1 - 54) j + (267.5 – 120.15) k - 400 i + 1000 k

= 253.9 i + 346.1 j + 147.35 k - 400 i + 1000 k

= - 146.1 i+346.1 j+1147.4 k

Alternative way: Using r_FB as an arm for both forces:

i j k i j k

M_F = r_FB x P + r_FD x Q = | -10 +9 -4 | + | -10 +9 -4 |

13.35 -26.75 40.10 0 -100 0

= (360.9 - 107) i +(400.1 - 54) j + (267.5 – 120.15) k - 400 i + 1000 k

= - 146.1 i+346.1 j+1147.4 k

Lecture 14

Force System Resultant

Moment of a Force about Specified Axes (4.5)

Approaches Used to Obtain Moments about Specified Axis:

Consider a force F acting on a rigid body containing point O as shown. The moment of this force about a given axis, say a-a, passing through point O, can be found by applying the cross product between the position vector OA with the force F producing moment M_O perpendicular to the plane formed by F and OA and projecting the resulting moment into axis a-a.

1) Scalar Approach:

Alternatively, the moment about the axis can be found by resolving F into its three components, finding the individual moments about O by multiplying the force components with shortest distance between the force and the point in question. Finally, the moment about the axis is computed by projecting the moments into the axis in question and adding them algebraically.

2) Triple Scalar Product:

It was pointed out earlier that the projection of any vector into a specified axis is found by performing the dot product of the vector with a unit vector in the direction of the axis. In the above, if M_O is the resulting moment of F about point O, the moment of F about the axis a-a is obtained by performing the dot product of the moment of F about point O with u _a-a where u _a-_a is the unit vector in the direction of the line a-a. The magnitude of M_a-a is:

M_a-a = M_O cos θ, which, in general, can be written as:

M_aa = u_a-a . M_O = u_a-a . (r x F). Writing the equation in matrix form:

(u_a-a )_x (u_a-a )_y (u_a-a )_z

M_a-a = u _a-a . (r x F) Ξ │ r_x r_y r_z │ This quantity known as the “triple scalar product”

F_x F_y F_z

The vector form of the moment about a-a is: M_{aa =} [u_a-a . (r x F)] u_aa

If there is more than one force, the moment about a-a:

M_a-a = ∑ [u_a-a . (r x F)] = u_a-a . ∑(r x F)]

Example:

What is the moment of the 100 N force about the x axis

1) Vector Method:

Taking r from any point in x - axis to any point in vector force F:

r = 4i + 3k

F = (100) [ j ] = 100 j

i j k

M_O = ׀ 4 0 3 ׀ = - 300 i + 400 k

0 100 0

׀ M_x ׀ = u_x . M_O = (i) . (- 300 i + 400 k) = - 300

M_x = ׀ M_x ׀u_x = - 300 i

2) Triple Scalar Product Method:

The unit vector along x- axis is i’ then:

׀ M_x ׀ = [ i . (r x F)]

1 0 0

׀ M_x ׀ = ׀ 4 0 3 ׀ = - 300

0 100 0

M_x = ׀ M_x ׀u_x = - 300 i

Example:

A cube of sides 3 meters is acted upon a force F = 1000 N. Determine the moment about the diagonal AG.

F

x

z

Length of CF = √ [(3)² + (3)²] = 4.24

The force F in rectangular components: (1000 / 4.24) [ 3j – 3k] = 707.1 j – 707.1 k

And: r_AF = 3i -3j

i j k

The moment about A is: M_A = r_AF x F = │ 3 -3 0 │ = 2121 i + 2121 j + 2121 k

0 +707.1 -707.1

Moment about AG = u_AG . M_A

Unit vector along AG u_AG = (1/SQRT ( 3² + 3² + 3² )[ 3i - 3j – 3k] = 0.577 i - 0.577 j – 0.577k

M_AG = u_AG . M_A = [0.577 i – 0.577 j – 0.577 k ] . [ 2121 i + 2121 j + 2121 k] = 1223.8 - 1223.8 – 1223.8 m-N

= – 1223.8 m-N

Alternative method:

The moment about AG is:

u_x u_y u_z 0.577 -0.577 -0.577

M_AG = │ r_x r_y r_z │ = │ 3 -3 -3 │ = [ ( 0.577)(-3)(-707.1) – (-0.577)(3)(-707.1)

F_x F_y F_z 0 707.1 - 707.1 + (-0.577)(3)(-707.1)]

= – 1223.8 m-N

Example:

The jib crane is oriented so that the boom DA is parallel to the x axis. At the instant shown the tension in the cable AB is 15 KN. Determine the moment about each of the coordinate axis of the force exerted by the cable AB.

The rectangular components of T_AB = (15) {1/ [5.2² + 3²]^1/2 ( - 5.2 j + 3 k ] = (15)(- 0.87 j + 0.5 k)

= - 13.1 j + 7.5 k

Position vector OA = + 3.8 i + 5.2 j

1 0 0

a) Moment about x – axis = u_x . (r_OA x T) = │ 3.8 5.2 0 │ = [(5.2)(7.5) ] = +39 KN-m

0 -13.1 7.5

0 1 0

b) Moment about y – axis = u_y . (r_OA x T) = │ 3.8 5.2 0 │ = - [(3.8)(7.5) ] = - 28.5 KN-m

0 -13.1 7.5

0 0 1

c) Moment about z – axis = u_z . (r_OA x T) = │ 3.8 5.2 0 │ = [(3.8)(-13.1) ] = - 49.8 KN-m

0 -13.1 7.5

_______________________________________________________________________________________________

Summary:

The vector representation of moment about a point is: M_A = r_A x F and the vector representation of the moment about the axis is: M_a-a = [u_a-a . (r_A x F)] u_a-a.

Lecture 15

Force System Resultant

Moment of a Couple (4.6)

Moment of a Couple:

It was pointed out earlier that a force or a system of forces acting on a rigid body produces two effects: It moves the body, and rotates it about an axis. When two forces that are equal in magnitude and opposite in direction act on a rigid body, they tend to rotate the body without moving it. The system of forces that produce rotation but not displacement is called a couple. The moment produced by the couple is called couple moment.

A simple example of a couple consists of two forces F and –F separated by a finite distance. The couple about point A is:

M = r_A x ( - F) + r_B x ( + F ) = (r_B – r_A) x F

The above equation is equivalent to: M = (r_B – r_A) x F where the quantity (r_B – r_A) is the vector directed from point of action of one force to the point of action of the other. The effect of the couple is, therefore a moment M that induces rotation but not translation. It is important to note that M depends on position vector r = (r_B – r_A) but not on vector r_A or vector r_B. This means that the couple has no fixed point of application and same result would have been achieved if the moment had been computed about a different point.

r = r_B – r_A

+ F M = r_A x (-F) + r_B x (+F) = (r_B – r_A) x F

B

- F

r_A Moment of couple about O₁ Ξ about O₂

Moment of couple is Independent of position

r_B

O₂

O₁

O₁

Lecture 16 and 17

Equivalent System

Resultant of a Force & a Couple (4.7, 4.8)

Equivalent System:

The characteristics of a couple stay the same if:

- The couple is moved to a position in its original plane (see a).

- The couple is moved to a plane parallel to its original plane (see b).

- The magnitude of the force and distance can be changed provided that the moment

Remain the same.

Equivalent System

A set of forces and couple moments acting on a body can be simplified and replaced by an equivalent single force and a single moment. Similarly, a single force can be replaced by an equal parallel force through a different point. To compensate for the move of the force into different location, it is necessary to add a couple.

Consider a force F acting on a rigid body at point A. It is desired that the force be moved to another point, say B. Since the concept of transmissibility allows the force to slide anywhere along its line of action without altering the external effects on the body, we can move a force equal to F to point B and attach equal but opposite force at the same point. As a result, a force F is now applied to point B along with a couple formed by –F and F. Thus “any force acting on a body can be replaced by the same force at another arbitrary point, provided that a couple of moment of F about A is added”. Note that since M is a free vector and can be applied anywhere.

Example 1:

Replace the force F=100 N acting on point A by a force acting on point B and a vertical couple acting on C and D as shown in the figure.

Resultant of a Force and Couple System:

Example 1: (From Hibbeler)

Two couples act on the frame. If d = 4 ft, determine the resultant couple moment. Compute the result by (a) resolving each force into x and y components and (b) summing the moments of all the force components about B.

a) F₁ = - 50 sin 30^O i - 50cos 30^O j F₂ = (4/5)(80) i + (3/5)(80) j

r₁ = 3i r₂ = - 4 j

Since: M_C = ∑ (r x F)

i j k i j k

M_B = │ 3 0 0 │+│ 0 -4 0 │

-50sin 30^O -50cos 30^O 0 (4/5)(80) (3/5)(80) 0

= (-130 +256)k

= 126 k lb-ft

Negative because it is opposite to +z Axis

b) M_B = F ^’_1y (2) – F_1y (5) – F ^’_2x (1) + F_2x (5)

M_B = 50 cos 30^O (2) – 50 cos 30^O (5) - (4/3) (80)(1) + (4/5) (80)(1) + (4/3) (80)(5)

= 126 k lb-f

Example 2:

A 100 m beam subjected to the forces shown. Reduce the system of forces to (a) equivalent force-couple system at A, (b) equivalent force-couple system at B

(-)

a) R = ∑ F = -100 j - 80 j + 150 j – 20 j = - 50 j (M_R)_A = (25i)x(-80j)+(53i)x(+150j)+(100i)x(-20j) = - (3950 N-m) k

b) Transferring the resultant R to B must be accompanied by a moment of R to B:

(+)

Moment at B due to R = (M_R)_B = (-100i) x (- 50 j) = (5000 N-M) k

(-)

Total moment at B = (5000 k) + (-3950k) = (1050 N-M) k

Example (From Meriam)

Determine the resultant of the four forces at point O and the couple acting on the plate shown

R_X = 40 + 80 cos 30^O – 60 cos 45^O = 66.9 N and: R_Y = 50 + 80 sin 30^O + 60 sin 45^O = 132.4 N

R = √ (R_X² + R_Y²) R = √ (66.9)² + (132.4)² = 148.3 N and: θ = tan ^-1 (R_X / R_Y ) = tan ^-1 (132.4 / 66.9) = 63.2^O

M_O = 140 – 50 (5 m) + 60 cos 45^O (4 m) – 60 sin 45^O (7) = - 237 N.m Located at O

Example:

Replace the forces, moment, distributed force and couple by a force-couple system at A.

Equivalent force: ½ (20)(8) = 80 N

20 N/m 8 m F = 60 N

A 2m M = 200 N.m

5 m 8 m 2m 5 m 40^O

F = 100 N

F = 60 N

ΣF_x = 100 cos 40 i = 76.6 i

ΣF_y = - 100 Sin 40 – (1/2) (8m) (20 N/m) = -144.28 j

F_R = SQRT (76.6² + 144.28² ) = 163.35 N @ θ = tan^-1 (-144.28 / 76.6) = -62.04^O

+

Σ M_R = 200 N.m – (60 N)[4m] – (100) sin 40 [5+8+8+5] – (80) [8/3 + 5]

= 200 – 240 – 1671.25 – 613.33 =

= 2324.58 N

The equivalent Forces and moments are shown below:

M_R = 2324.58 N

F_R = 163.35 N

Note that M_R and F_R act on A will produce the same external effects at the supports as those produced by the original force and moment system.

Lecture 18

Force System Resultant

Distributed Loading (4.10)

Distributed Loading:

Occurrence: Distributed loading occur in various structures such as: bridges, slabs, beams, fork lifts, Retaining walls, platforms and other structures subjected to body weight and/or loads.

If the distributed load along the x-axis can be represented by a single force:

F = Area under the curve = ∫_L w dx = ∫_A dA

Location of the force: x₀ = ∫ x w dx / ∫ dA

Example: (Hibbeler)

Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member AB, measured from A.

+ ← ∑ F_{R x} = ∑ F_x = (5) (200 N) = 1000 N and: + ↓ ∑ F_{R y} = ∑ F_y = [(200 + 100) / 2] (6) = 900 N

F_R = √ [(1000)² + (900)²] = 1345 N θ = tan^-1 (900/1000) = 42^O

M_{RA =} ∑M_A = 1345 cos 42^O (y) = 1000 (2.5) – 300 (2) – 600 (3) gives y = 0.1 m F_R = 900 N

F1 = 300N

F₂ = 600N

200 N/m 100 N/m

B 6 m

x^’

Example 2:

Find the position of the resultant d of the loading shown below.

F = ∫ w(x) dx = ∫_0→a (b/a²) x² dx = (1/3)(b/a²) a³ = 1/3 ba

M = (x) ( F ) = ∫_0→a x(b/a²) x² dx =(1/4) ba² Since d = M / F

Then: d = (1/4ba²) / (1/3 ba) = ¾ a

Lecture 19

Equilibrium of a Rigid Body

2-Dimensional Free Body Diagram (5.1, 5.2)

Conditions for Rigid Body:

The necessary conditions required to attain equilibrium in a rigid body are:

1) The summation of internal and external forces must be zero.

∑ f_i + ∑ F_i = 0

2) The summation of all moments about a point must be zero.

∑ M_fi + ∑ M_Fi = 0

Where M_fi and M_Fi are the moment due to internal and external forces respectively.

Internal Forces:

Internal forces are the forces that hold together the various particles forming the rigid body. They usually appear when a rigid body is dismembered and the free body diagram is drawn.

External Forces:

External forces are the forces that act on the rigid body as a whole. They include: Weight, tension and reaction forces.

Equilibrium in Two Dimensions:

Free-Body Diagrams:

When a free body diagram involves a set of unknown reactive forces acting at a point (For example: reactive forces acting on ball and socket, thrust bearing, hinges, etc.). To minimize the number of the unknowns, it is often convenient to take the moment about these forces.

Support Reactions:

Rule of Thumb: If a support prevents translation (movement) of an object, in a given direction, a force is developed on the body in that direction. Examples are shown below:

Equations of Equilibrium:

If the free-body diagram of the whole rigid body is taken, Newton’s law states that all internal forces occur in equal and opposite in direction, the above equations become:

∑ F_i = 0 and: ∑ M_O = 0

Which states that, if the rigid body is to be at equilibrium, the sum all external forces acting on the body and the summation of all the moments due external forces about a point is zero

Methods of solution:

- Establish x, y coordinate system.

- Outline the free-body diagram.

- Assume the directions of the unknown forces and moments.

- write the equations of equilibrium ∑ M_A = 0 by choosing point A so that

point o lies on line of action of two unknown forces. This is done to ensure

the direct solution of the third force (see figure).

- Write the equations of forces: ∑ F_x = 0 , ∑ F_y =0

- Orient x-y axis so as to minimize the number of the unknowns (see figure).

- if the solution yields negative values reverse the assumed directions of the unknown

Forces and couple moments.

F₁ A F₂ B A_x

Pin Ball A_y B_y

F₁ F₁

A

Beam fixed to wall

θ

Example:

+

∑ M_B = 0 then: (9) A_x + (7)100 = 0

A_x = -77.78 Then A_{X ←}

∑ F_x = 0 then B = 77.78 →

∑ F_y = 0 then: A_y = 100 N

Example: (Hibbeler)

The cord shown supports a force of 100 lb and wraps over a frictionless pulley. Determine the tension in the cord at C and the horizontal and vertical reaction at A.

Solution:

Equations of equilibrium

Σ M_A = 0 Then: 100 (0.5) – T (0.5) = 0 Then T = 100 lb

Σ F_x = 0 Then: A_x + 100 sin 30^o = 0 → A_x = - 50 Then: A_x = 50 lb ←

Σ F_y = 0 Then: A_y – 100 – 100 cos 30^o = 0 Then: A_y = 187 lb ↑

Example: (Hibbeler)

The box wrench is used to tighten the bolt at A. If the wrench does not turn when the load is applied at B and C, determine the moment applied at the bolt and the force at the wrench

Solution:

Σ F_x = 0 Then: A_x – 52 (5/13) + 30 cos 60^o = 0 Gives: A_x = 5 N

Σ F_y = 0 Then: A_y – 52 (12/13) + 30 sin 60^o = 0 Gives: A_y = 74 N

Σ M_A = 0 Then: M_A – 52 (12/13) (0.3) - 30 sin 60^o (0.7) = 0 Gives: M_A = 32.6 N-m

Note: We added the unknown moment M because it is free vector and must be included.

F_A = SQRT[(5²) +(74²)] = 74.1 N and θ = tan^-1 (74/5) = 86.1^o /_

Checking the calculation:

+

Σ M_C = 0 Then: 52 (12/13 m) (0.4) + 32.6 N-m – 74 (0.7 m) = 0 OK

Most Common Support Reactions:

Example: (Hibbeler)

Determine the reactions at the pin support A and ball support B

3000 N 750 N

We have three unknowns & three equations, ΣF_x = 0 A_x = 0

ΣF_y = 0 Then: A_y + B_y – (250) (12 ft) – (1/2) (250) (6 ft) = 0

Σ M_A = 0 Then: (6) B_y – (2+6)(750) – (0) (3000) = 0 Then: B_y = 1000 lb

Substituting into the first equation:

A_y = -1000 + (250) (12) + (1/2) (250) (6) Then: A_y = 2750 lb A_x = 0

Two and Three Force Members:

Two Force Members: If a body is held by two forces at equilibrium, the forces must be collinear, equal and opposite in magnitudes. This requires that the resultant moment and resultant forces are zero.

Three Force Members: When a body is held by three forces at equilibrium, the forces must be concurrent and coplanar, or they must be parallel.

Lectures 20-22

Equilibrium of a Rigid Body

3-Dimensional Free Body Diagram (5.3, 5.2, 5.3, 5.4, 5.5)

Equilibrium in three Dimensions:

a) Vector Equations of Equilibrium:

The conditions are: Σ F = 0 and: Σ M_o = 0 Where Σ M_o is the sum of couple moments and moments of all forces about point O.

b) Scalar Equations of Equilibrium:

Σ F = Σ F_x i+ Σ F_y j+ Σ F_z k = 0 Or: Σ F_x = 0, Σ F_y = 0 and: Σ F_z = 0.

Σ M_o = Σ M_x i + Σ M_y j + Σ M_z k = 0 Or: Σ M_x = 0, Σ M_y = 0 and: Σ M_z = 0

Methods of Solution:

a) Example Using Scalar Solution:

Example: (5.15, Hibbeler)

The homogeneous plate shown has a mass of 100 Kg is subjected to a force and a couple moment along its edges. If it is supported in the horizontal plane by means of a roller at A, a ball and socket at B, and a cord at C, determine the components of reaction at A, B & C.

Solution:

Unknowns: A_z, B_x, B_y, B_z and T_C ∑ F_x = 0 Then: B_x = 0 ∑ F_y = 0 Then: B_y = 0

∑ F_z = 0 Then: A_z + B_z + T_c – 300 N – 981 N = 0 (1)

Summing the moments of the forces on the free-body diagram, with positive along x and y axis yields:

+

∑ M_x = 0 T_C (2m) – 981 N (1m) + B_z (2m) = 0 (2)

+

∑ M_y = 0 300 N (1.5m) + 981 N (1.5 m) – B_z (3m) – A_z (3m) – 200 N.m = 0 (3)

Solving (1), (2), (3) Gives: A_z = 790 N B_z = - 217 N and T_C = 707 N

b) Example Using Vector Solution:

To determine the unknown forces take the summation of moment bout a line that passes as many unknowns as possible:

EXAMPLE: (5.17, Hibbeler)

Let F = - 1000 j (in the opposite y direction). Determine the tension in cables BC & BD and the reactions at the ball and-socket joint A.

F_A = A_x i + A_y j + A_z k ………………….(reaction at ball-and-socket , unknown)

T_BC = T_BC [ 6i - 6k) /SQRT (6²+6²)

T_BC = 0.707 T_BC i – 0.707 T_BC k ………………………………………(T_BC unknown)

T _BD = T _BD (- 3i + 6j – 6k) / SQRT (3² + 6² + 6²)

T _BD = - 0.333 T_BD i + 0.666 T_BD j - 0.666 T_BD k …………………….( T_BD unknown)

Σ F_x = 0 A_x + 0.707 T_BC – 0.333 T_BD = 0 (1)

Σ F_y = 0 A_y - 0.666 T_BD – 1000 = 0 (2)

Σ F_z = 0 A_z - 0.707 T_BC – 0.666 T_BD = 0 (3)

Summing moments about point A:

Σ M_A = o Or: r _BA X ( F + T_BC + T_BD ) = 0

(6 k) X (- 1000 j + 0.707 T_BC i – 0.707 T_BC k - 0.333 T_BD i + 0.666 T_BD j - 0.666 T_BD k = 0

Evaluating the cross products and equating components of i = 0, components of j = 0

(-4 T_BD + 6000) i + ( 4.24 T_BC – 2 T_BD ) j = 0 (4,5)

(Note that components of k =0), we have 5 equations and five unknowns: T_BC, T_BD, A_x, A_y and A_z. Then: T_BC = 707 N, T_BD = 1500 N, A_x = 0 N, A_y = 0 and A_z = 1500 N

EXAMPLE: (Hibbeler)

The rod is supported at A by a joint bearing and at D by a ball-and-socket joint and at B by a cable as shown. Find the tension in cable BC.

Solution:

There are six unknowns including three force components caused by the ball-and-socket at D (D_x, D_y & D_z), two caused by the journal bearing at A (Az & Ay), and one tension caused by the cable (T).

One way to solve the problem is to solve five equations ( 3 from Σ F_x = 0, Σ F_y = 0 and Σ F_z = 0 and 2 from Σ M_A = o and Σ M_D = o

As an alternative, the tension T at the cable can be obtained by taking the moment about the line passing through D and A which will eliminate unknown forces A_x, A_y, D_x, D_y and D_z

Σ M_DA = U_DA . (r_DB x T + r_DE x W) = 0

U_DA = r_DA / |r_DA| = - 1/√2 i – 1/√2 j

U_DA = - 0.707 i – 0.707 j

r_DE = - 0.5 j

W = - 981 k

r_DB = - j

T = T( 0.2/0.7) i – T(0.3/0.7) j + T(0.6/0.7) k or:

T = 0.286 T i – 0.429 T j + 0.857 T k

Σ M_DA = U_DA . (r_DB x T + r_DE x W) = 0

(- 0.707 i – 0.707 j) . [(- j) x [0.286 T i – 0.429 T j + 0.857 T k] + (- 0.5j) x (-981 k)] = 0

(- 0.707 i – 0.707 j) . j X [( - 0.286) Ti + (- 0.857) T k + ( - 0.5j) x (-981 k)] = 0

(- 0.707 i – 0.707 j) . (0.286 T k - 0.857 T i + 490.5 i ) = 0

(- 0.707 i – 0.707 j) . [(- 0.857 T + 490.5 i) i + 0.286 T k) = 0

(- 0.707i) . (-0.8571T + 490.5 ) i + 0 + 0

- 0.707 (-0.857 T + 490.5) Then: T = 572 N

To Find D_z take the moment about line AB ( D_Z = 490.5 N)

Similarly, to find A_Z take the moment about y axis (A_z = 0)

Lecture 23

Structural Analysis

Simple Trusses, the Method of Joints (6.1, 6.2)

Simple Trusses:

A truss is a framework composed of straight wooden or metal members joined together by riveted connections, large bolts or gusset plates to form a rigid structure known as truss.

Simple Trusses: Trusses made up of basic triangles put together to form a truss.

Planar Truss: Trusses that lie in a single plane. They are used to support roofs and bridges. They start from a single triangular unit and joined together by additional units.

Assumptions:

- All members are joined together by pins to prevent bending of members.

- All loading is applied at the joints and weight of individual members are

negligible.

-All truss members are two-force members which makes them either in tension

or in compression.

Method of Joints:

Method of joints consists of applying the condition of equilibrium of forces at each point of connecting pin in the truss. The method is summarized as follows:

- Draw the free body diagram of any joint where at least one known load exists and

no more than two unknowns are present. If this point lies on one of the supports,

it is necessary to find external reaction at this point by considering the entire truss

as a free-body-diagram and using the equations of equilibrium of rigid body to solve

for the external reactions of the truss.

.

- Assume the member acting on the joint’s free-body-diagram to be in tension. If the

answer yields negative number, the member is in compression the correct sense is

used for the analysis of the subsequent joint.

- Apply the two equilibrium equations ∑ F_x = 0 and ∑ F_y = 0 to solve for the unknown

forces.

Example :(6.3, Hibbeler):

Determine the force in each member of the truss shown. Indicate whether the members are in tension or in compression.

To determine F_DC, we can do one of the following: F_DB F_DC

a) We can retain the sense of F_DB and apply the equation:

∑ F_y = 0 - F_DC – (4/5) (-250N) = 0 Then: F_DC = 200 N (C)

450 N 600 N

b) We can correct the sense of F_DB and apply the equation:

From the free body diagram of the entire truss, the support reactions can be determined by applying the equations of equilibrium:

∑ F_x = 0 600 N – C_x = 0 then: C_x = 600 N

∑ M_C = 0 -A_y (6m) + 400 (3m) + 600N (4m) = 0 then: A_y = 600 N

∑F_y = 0 600N – 400 N – C_y = 0 then: C_y = 200 N

Joint A: (see fig. (c)):

Three forces acting on pin A: F_AB is compressive (in order to balance the upward force 600N)

F_AD is tensile (in order to balance the x coordinate of F_AB)

∑F_y = 0 Then: 600N – (4/5) F_AB = 0 Then: F_AB = 750 N (C)

∑ F_x = 0 Then: F_AD – (3/5) (750) = 0 Then: F_AD = 450 N (T)

Joint D: (see fig. (d)):

The force in AD is known and it is tensile (why?) so we only have two unknown forces in joint D. The two forces can be determined by summing forces in the horizontal direction:

∑ F_x = 0 Then: -450N + (3/5) F_DB + 600N = 0 Then: F_DB = -250 N which means it acts in the opposite sense to that shown in figure (d) Then: F_DB = 250 (T)

∑ F_y = 0 - F_DC – (4/5) (250N) = 0 Then: F_DC = -200 N (T) or: F_DC = 200 N (C)

Joint C: (see fig. (d)): ∑ F_x = 0 Then: F_CB – 600N = 0 Then: F_CB = 600 N (C)

Example: (Beer & Johnston)

Using the method of joints, determine the forces in each member of the truss shown and determine whether the member is in tension or in compression.

∑M_C = 0 Then: (2000 lb)(24 ft) + (1000 lb)(12 ft) – E_y (6 ft) = 0 Then: E_y = + 10,000 lb ↑

∑F_x = 0 Then: C_x = 0

∑F_y = 0 Then: (-20000 lb) - `1000 lb + 10,000 lb +C_y = 0 Then: C_y = -7000 lb = 7000 lb↓

C_x = 0

Joint A:

F_DA = 2500 lb F_DB

F_DB

F_DA D ₄ F_DB = F_DA Then: F_DB = 2500 (T)

3 F_DE = (3/5) F_DA + (3/5) F_DB

D F_DE F_DE Then: F_DE = 3000 (c)

Joint D:

1000 lb ∑F_Y = 0 -1000 –(4/5)(2500) + (4/5) F_BE =0

A B F_BC Then: F_BE = 3750 lb or: F_BE = 3750 (C)

1500 C

∑ F_x = 0 F_BC -1500 –(3/5) (2500) – (3/5) (3700) = 0

2500 E Then: F_BC = 5250 lb or: F_BC = 5250 lb (T)

D F_BE

Joint B:

Joint E:

Lecture 24

Structural Analysis

Zero Force Members (6.3)

Zero force members are members in the truss that carry no forces, they are constructed to provide the following:

- Increase stability and rigidity of trusses.

- Provide support for future loading conditions if necessary.

- Support the weight of the truss

As a general rule: (Rule 2) (Rule 1)

“If three members form a joint in a truss, two of which are collinear, and if there is no external load or reaction at the joint, then the third member must be zero force member (Rule 1). If two members form a joint in a truss and there is no external force or reaction applied to the joint, the members must be zero force members (Rule 2).”

Example: Determine the zero force members in the truss shown below.

F₁

F₁

F=0 F=0 F₁

F₂ F₂ ≠ 0

F₂ F₃

F₁ = F₂ = 0 (No external load

F₁ = F applied at the joint)

Rule 2 Then: F₁ = F₂ = F₃ = 0

Example:

Determine the zero force members in the truss shown below.

In the above structure:

- Joint I, M and O carry three members, two of which lie in the same line. Since

Joints I, M and O carry no external loading, members DI, FM and GO are zero force

members.

- Applying the same reasoning to joint G and D, we find that the two joints are in the

Same situation as members O, M and I and, therefore, members GN and DJ must be

zero force members.

- Forces IH = IJ, KI = KL, OP = ON, NO = NM, GP = GC, FC = FL, DA = DH, and

MN = ML.

- Since no external force is applied at joints M, N, and O, the forces in members ML,

MN, NM, NO, ON and OP are all equal.

Example:

Determine the zero force members in the truss shown.

In the above:

- Joint F is has three members, two of its members lie in the same line. Since the

joint has no external loading, member FJ is a zero force member

- With the elimination of member FJ, joint J is practically a three member joint,

therefore, member JE is a zero force member.

- Since member JE is a zero force member, member EB is a zero force member.

- With the elimination of member EB, joint B is practically a three member joint,

therefore, member BD is a zero force member.

- Since member BD is a zero force member, member DH is a zero force member.

- With the elimination of member DH, joint His practically a three member joint,

therefore, member HA is a zero force member.

- Joint G has a horizontal reaction force, therefore, member JA is a zero force

member.

Lecture 25

Structural Analysis

Simple Trusses, the Method of Sections (6.4)

Method of sections is based on the principle that “if a body is in equilibrium then any part of the body is also in equilibrium”. This method is most useful when only select member forces are to be determined.

Method of Solution:

1) Draw the Free-body diagram of the whole structure to find the external reactions

2) Cut the section so that it passes through the members which are to be analyzed.

In this example, the members to be analyzed are BC, BG and EG.

3) Since the isolated cutoff body is in equilibrium, the three equations of

equilibrium can be solved to determine the three forces in members BC, BG

and EG.

Note that , if only the force in member BC is desired, only one equation of

equilibrium is needed ( ∑M _E = 0 ) to yield the value of F_BC.

Example:

Determine the forces in members JI, DI and DE of the truss shown.

Writing the equilibrium equation for the whole structure: ∑ M_A = 0, then:

(R_G) (60) = (500)10) + (200) (20) + (100) (40) + (200) (50) Then: R_G = 383.33 N

200 N

Force in member ED: ∑M_I = 0 Then:

- F_ED (10 m) - 200 N (10 m) + 383.33 N (20 m) = 0 Then: F_ED = 566.66 N (T)

Force in member IJ: We can evaluate the force F_{I J} by taking the moment ∑ M_D = 0 Then:

F_{I J} cos (26.6^O) (15 m) + 383.33 N (30 m) – 100 N (10 m) – 200 N (20 m) = 0 Then F_IJ = -484.63 N =F_{IJ =} 484.63 (C)

Force in member ID: We can evaluate force F_ID by taking the moment about point G:

F_ID cos 45^O (30 m) + 200 N (10 m) + 100 N (20 m) = 0 Then: F_ID = - 188.56 N or: F_ID = 188.56 (C)

Lecture 26

Structural Analysis

Review of Simple Trusses

ILLUSTRATIVE EXAMPLE:

Req’d: Use the method of joints and the method of sections to find the force and direction in member CB.

Using Method of Joints:

100 N

F_CB (Sin 26.57^O) = 50 N

F_AC F_CB ( C) F_CB = 111.8 N (C)

50 N 26.57^O 63.43^O

A B

100 N

100 N F_AB (T) 100 N 50 N

50 N

Using the Method of Sections:

CUT

100 N E

F_CB

(10) Sin (90-26.57)

+

A F_CB (10 Sin 63.43^O ) + (100) (10) = 0

F_CB = - 111.8 N (T) or: 111.8 (C)

Lecture 27,28

Structural Analysis

Frames & Machines (6.6)

Frames:

Frames are structures composed of joined members that are loaded at any location along their axis. Unlike simple trusses, frame members are multi force members that carry three or more forces and they are not necessarily directed along the axis of the member.

Method of Solution:

1) Start by taking the whole frame as free-body diagram in order to solve for the external forces acting on the joints and supports.

2) Isolate each part of the frame (dismember the frame), show all forces acting on the member and treat each component as rigid body under equilibrium and apply the equations of equilibrium to each component. Do not forget to take advantage of the moment equation to eliminate unknowns.

3)Since it is not possible to assign proper sense to all forces, make an arbitrary assignment and reverse the sense if it yields negative value. It is very important, however, that the sense of the force be properly represented in the interacting bodies as shown in the figure below.

Example:

Determine the forces in member AC.

Taking the FBD of the whole frame: ∑M_A = 0 (E_Y) (12m) – (6000N.m) – (4000N)(3m) = 0, Then:

+

E_Y = 1500 N ↑

∑ F_y = 0 1500 – 4000N + A_Y = 0 Then: A_Y = 2500 N ↑

Isolating part AC:

∑ F_Y = 0 Then: C_Y + 2500 N – 4000 N = 0

2500

Then: C_Y = 1500 ↑

∑ M_C = 0 Then: 4000 (3m) + A_X (8 m) – A_Y (6 m) = 0

+

Then: A_X = 375 N →

Example: (Beer & Johnston)

In a small frame shown, members EBF and ABCD are connected by a pin at B and by cable at EC. A 75 lb load is supported by a second cable which passes over a pulley at F. Determine the tension in cable EC and the components of the pin reaction at B.

External reactions on the frame involve three unknown: D_X , D_y and A

∑ F_y = 0 Then: D_y = 75 lb ↑

+ ∑ M_D = 0 Then: Ax (15) – 75 (12) = 0 Then: A_x = 60 lb ←

∑ F_X = 0 Then: D_X = 60 lb →

∑ F_Y = ∑ F_Y = 0 in the FBD of the pulley. Then F_x = 75 lb → and: F_y = 75 lb ↑

Member EBF and ABCD are connected at B so the forces acting at point B are equal and opposite as shown. Similarly, The forces exerted at E and C are equal and opposite and their direction is known.

Member EBF:

+ ∑ M_E = 0 Then: (B_Y) (4 ft) – (75 lb) (14 ft) = 0 Then: B_Y = 263 lb

+ ∑ M_B = 0 Then: ( T cos 33.7^O) (4 ft) – (75 lb) (10ft) = 0 Then T = 225 lb

∑ F_X = 0 Then: ( T sin 33.7^O) – B_X – 75 lb = 0 Then: B_X = 50 lb

Since the values are found to be positive, the forces are directed as shown above. The direction of forces on member ABCD are , therefore, opposite in sense to that of member EBF.

Checking the calculation on member ABCD: ∑ F_X = 0 ? - 60 lb + 75 lb + 50 lb – (225 lb) sin 33.7 + 60 = OK

Example: (Hibbeler)

Determine the horizontal and the vertical component force which the pin at C exerts on member ABCD of the frame shown.

Identify all two-force members (FB is a two-force member WHY?)

Step 1: Draw FBD of the entire frame to find external reactions:

∑ M_A = 0 Then: - 981 N (2m) + D_X (2.8m) = 0 Then: D_X = 700.7 N

∑ F_X = 0 Then: A_X – 700.7 N = 0 Then: A_X = 700.7 N

∑ F_Y = 0 Then: A_Y – 981 N = 0 Then: A_Y = 981 N

Step 2: Isolate Member CF:

∑ M_C = 0 Then: - 981 N (2m) – F_B sin 45^O (1.6m) = 0 Then: F_B = - 1734.2 N Or: F_{B =} + 1734.2 acting upward

and to the right.

∑ F_X = 0 Then: - C_X + ( 1734.2 cos 45^O) = 0 Then: C_X = 1226 N to the left

∑ F_Y = 0 Then: C_Y + 1734.2 sin 45^O) – 981 N = 0 Then: C_Y = - 245 N Or: C_Y = + 245 acting downward ↓

EXAMPLE (From Hibbeler)

Disassembling a Frame: (What if the we cannot determine external forces by using the FBD of the entire structure?)

An Illustrative Example:

Given A Frame consisting of members and pulley as shown

Civil Engineering Dept.

CE 201 Engineering Statics

Dr. Rashid Allayla

Internal Forces

7.1 Internal Forces:

Internal forces are the forces that exist within the structure when an imaginary section is passed through the structure. When the internal forces are exposed, the forces become external forces.

7.2) Shear and Moment Equations:

Example:

Find internal forces and bending moment at a section 2 m from A

Example: (Hibbeler)

Determine the internal normal force, shear force and the moment acting at point B of the frame shown.

EXAMPLE: (Hibbeler)

The beam supports the loading shown. Determine the internal force, shear force and bending moment acting just to the left of the external force and just to the right of it.

External forces:

∑ M_D = 0 Then: 9 KN-m + (6 KN) – A_Y (9 m) = 0 Then: A_Y = 5 KN

Segment AB:

∑ F_X = 0 Then: N_B = 0

∑ F_Y = 0 5 KN – V_B = 0 Then: V_B = 5 KN ↓

∑ M_B = 0 - (5 KN) (3 m) + M_B = 0 Then M_B = 15 KN.m

+

Segment AC:

∑ F_X = 0 Then: N_C = 0

∑ F_Y = 0 5 KN – 6 KN + V_C = 0 Then: V_C = 1 KN ↑

∑ M_B = 0 - (5 KN) (3 m) + M_C = 0 Then M_C = 15 KN.m

+

EXAMPLE:

Find the internal forces and the moments at points B and C for the cantilever beam shown.

(50)(6/2)

Analysis of External Forces and Moment:

From FBD (b) ∑ M_A = 0 Then: - [(50)(6/2) N] (10m) + M_A = 0 Then: M_A = 1500 N-m

∑ F_X = 0 Then: A_X = 0 and: ∑ F_Y = 0 Then A_Y = (50) (6/2) = 150 N

From FBD (c): ∑ F_X = 0 Then: N_B = 0, and: ∑ F_Y = 0 Then: + 150 – V_B = 0, Then: V_B = 150 N

∑ M_B = 0 Then: M_A –A_y (6) + M_B = 0, Then: 1500 – (150)(6) - M_B = 0

M_B = 600 N.m

From FBD (d): ∑ F_Y = 0 Then: + 150 – (50/2) (3/2) – V_C = 0 Then: V_C = 112.5 N

∑ M_C= 0 Then: M_A–A_y(9)+(50/2)(3/2)(1) +M_C=0,Then: 1500-(150)(9) +37.5+M_C =0

M_C = 187.5 N.m

Check the right side:

ΣM_C = 0 Then: - (25)(3) (1.5) – (25)(3/2)(2) +M_C = 0 Then M_C = 187.5 N.m

7.2) Shear and Moment Diagrams:

Beam: Is a structural member designed to carry loading perpendicular to the axis.

Simply Supported Beam: A beam that is pinned at one end and roller-supported at the other.

Shear Diagram: A graphical representation of shear as a function of x-axis.

Moment Diagram: A graphical representation of moment as a function of x-axis.

Sign Convention:

Construction of the Shear and Moment Diagrams:

- Determine the external reactions by applying the equilibrium equations on the entire

FBD of the beam.

- Cut the beam into various segments and draw the FBD of each segment.

- Solve for internal forces at the cut sections by summing perpendicular forces acting

on each segment to determine shear force

- Determine the bending moment at the cut sections.

- Follow positive sign convention as shown earlier.

- If the computed values of V and M are positive, the values are plotted above x axis,

and if they are negative, they are plotted below x axis.

Note:

Internal normal forces are not considered since the applied loads are generally normal to the x-axis producing only shear forces and bending moments. Another reason for not considering the normal forces is that the beam’s resistance to shear is more important in design than the beam’s resistance to normal forces.

Example : (Hibbeler)

Draw the shear and bending moment diagrams of the beam shown below.

Example:

Draw the shear and bending moment diagrams the beam shown below.

Problem 2: (Meriam)

Example 3:

Draw the shear and moment diagrams for the beams shown below

(12) + (1/2) [ (1080 - 600) ] (12)

d

Summary:

To Construct Shear & Moment Diagrams:

- Solve for support reactions by taking the FBD of the whole beam and applying the

equations of equilibrium.

- Locate the discontinuities along the beam.

- Cut the beam into sections between discontinuities.

- Construct the FBD of the cut section and apply the equations of equilibrium.

- Plot V vs. M as function of x.

Civil Engineering Dept.

CE 201 Engineering Statics

Dr. Rashid Allayla

Friction

Friction:

Friction is the force that acts on a body when it comes into contact with another body. This force, known as the frictional force, acts tangent to the contacting surface and directed in the opposite sense to motion or possible motion of the body.

Types of Friction:

- Dry friction: Is the type of friction occurring between two unlubricated surfaces. The direction of the friction force always opposes the motion or the impending motion. This type of friction is called Coulomb Friction.

- Fluid Friction: Is the friction that exists between two fluid elements moving at

different velocities. When there is no relative velocity between layers of fluid,

friction will not develop within the fluid.

8.1) Characteristics of Dry Friction:

Analysis:

If the force P is increased, the friction force F also increase until it reaches a maximum value equals F_S. If P is further increased, the block starts to slide because the frictional force can no longer keep the block from sliding. When the block starts to move, the maximum frictional force drops slightly to lower value F_K (called kinetic frictional force) and stays approximately constant. The reason for the drop is that when F_→ F_S, the surface peaks shear off and shear force drops off slightly to F_K.

Static Friction:

The frictional force F that can develop between two surfaces without having the surfaces slide relative to each other is defined by the equation:

F_S = μ_S N

Kinetic friction:

The frictional force F that develops when two surfaces are sliding on each other is proportional to the normal force applied on the surfaces and is directed opposite to the relative motion of the surfaces. This is defined by the equation:

F_K = μ_K N

Where μ_S and μ_K are the coefficients of static and kinetic frictions.

Coefficient of static friction > Coefficient of Dynamic Friction

Relates to the magnitude of maximum Relates to the magnitude of friction force

Friction force that can be exerted by dry exerted by dry surfaces when sliding occur

Surfaces when slip is impending:

f = μ_K N where μ_K is the coefficient of kinetic

f = μ_S N where μ_S is the coefficient of friction.

static friction.

Note that:

- The frictional force F acts tangent to the surface of contact and directed opposite

to the relative motion.

- The static frictional force F_s, the coefficient of static friction μ_s and the coefficient

of the kinetic friction μ_s are independent of the area of the surface of contact.

- Both coefficients strongly depend on the nature of the contact surface (see table)

- When slipping is about to occur between two surfaces, the maximum static force

of friction F_s is proportional to the normal force so that: F_s = μ_s N.

- When slipping start to occur, the kinetic frictional force is proportional to the

normal force so that: F_k = μ_k N.

8.2) Solution of Problems Involving Dry Friction:

Group 1: Equilibrium Problems:

This type of problems require that the total number of unknowns must be less or equal to the independent equations of equilibrium. Since the condition of equilibrium requires that no slipping occur, the solution of Forces must satisfy the inequality F_S ≤ μ_S N.

Group 2: Impending Motion Problems:

This type of problems require that forces in contact with surfaces satisfy the equation

F_S = μ_S N. The total number of unknowns = number of equilibrium equations + Frictional equations.

Group 3: Impending Motion at Some Point:

This type of problems require that forces in contact with surfaces will satisfy one of the following conditions: Impending motion, motion or tipping.

Analysis of the Problem:

In the above problem, the number of unknowns are: N_A + F_A + N_C + F_C + B_X + B_Y + P = 7 unknowns

The number of available equations are:

Equilibrium equations: ∑ F_X = 0, ∑ F_Y = 0, ∑ M = 0

Left frame: Friction equation: F_{A =} μ_A N_A (one equation) and equilibrium equations: ∑F_X = 0, ∑F_Y = 0, ∑M = 0 (Three equations)

Right frame: Friction equation: F_{C =} μ_C N_C (one equation) and equilibrium equations :∑F_X = 0, ∑ F_Y = 0, ∑ M = 0 (Three equations)

Total: = 8 equations. Since impending motion for both frames occurring in the same time is highly unlikely, the choice of which frictional equation to be used depends on the kind of motion which actually occurs: As P increases, one of the following possibilities will occur:

a) Slipping at A and no slipping at C: Then F_{A =} μ_A N_A and F_C < μ_C N_C

b) Slipping at C and no slipping at A: Then F_{A =} μ_A N_A and F_A < μ_A N_A

And the actual situation can be determined by calculating P and choosing smaller value.

EXAMPLE:

Let W = 100 N, μ_A = 0.3 and μ_C = 0.35 Find the value of P that will cause slipping of the frame shown.

The unknowns: N_A, N_C, F_A, F_C, B_X, B_y and P = 7 unknowns

a) Assume Slipping at A and no slipping at C: Then F_{A =} μ_A N_A and F_C < μ_C N_C

F_A = (0.3) N_A (1)

Writing the equation of equilibrium for the right member:

∑F_X = 0 Then: P - B_X – F_C = 0 (2)

∑F_y = 0 Then: N_C – B_Y - 100 = 0 (3)

∑ M_C = 0 Then: + (B_Y) (5 m) + (B_X) (10 m) – P (5) = 0 (4)

Writing the equation of equilibrium for the left member:

∑F_X = 0 Then: B_X – F_A = 0 (5)

∑F_y = 0 Then: N_A – 100 + B_Y = 0 (6)

∑ M_A = 0 Then: + (B_Y) (5 m) - (B_X) (10 m) – (100) (2.5 m) = 0 (7)

Solving equations 1→7 yield P = 156.25 N

b) Assume Slipping at C and no slipping at A: Then F_{C =} μ_C N_C and F_A < μ_A N_A

F_C = (0.35) N_C (1)

Writing the equation of equilibrium for the right member:

∑F_X = 0 Then: P - B_X – F_C = 0 (2)

∑F_y = 0 Then: N_C – B_Y - 100 = 0 (3)

∑ M_C = 0 Then: + (B_Y) (5 m) + (B_X) (10 m) – P (5) = 0 (4)

Writing the equation of equilibrium for the left member:

∑F_X = 0 Then: B_X – F_A = 0 (5)

∑F_y = 0 Then: N_A – 100 + B_Y = 0 (6)

∑ M_A = 0 Then: + (B_Y) (5 m) - (B_X) (10 m) – (100) (2.5 m) = 0 (7)

Solving equations 1→7 yield: P = 129.3 N < 156.25 …. Then C will slip before A

Example: (Hibbeler)

The uniform crate has a weight of 196.2 N. If a force P = 80 N is applied to the crate, determine if it remains in equilibrium. The coefficient of static friction is μ = 0.3.

Example:

A 150 lb man starts to climb a 26 – ft ladder that weighs 50 lbs. The ladder is placed against smooth wall with its lower end is 20 ft from the wall. The coefficient of friction between the ladder and the floor is 0.3. Determine the distance x from the man to the wall when the ladder starts to slip.

Example:

Two blocks are in top of each other, the top block is subjected to a force P. If W_A = 80 N, W_B = 120 N, μ between block A and B is μ_s = 0.5 and between B and the floor is μ_s = 0.1. What will happen when P increases?

Example: (Hibbeler)

Beam AB is subjected to uniform load of 200 N/m and is supported at B by post BC. If μ_B=0.2 and μ_C = 0.5, determine the force P needed to pull the post out from under the beam

200 N/m

A B

4m 0.75 m

P

0.25 m

C

800 N

A_X F_B

2 m 2 m

A_Y N_B =400 N N_B 400

F_B B

0.75 P

0.25

F_C C

N_C

From equilibrium equations: ΣFx = 0 P - F_B – F_C = 0 ………………………………. (1)

ΣF_Y = 0 N_C – 400 = 0 ……………………………... (2)

ΣM_C = 0 - P (0.25) + F_B (1) = 0 ………………………. (3)

If Post slips at B: Then F_C ≤ μ_C N_C and: F_B = μ_B N_B = 0.2 (400 N) = 80 N

Using this result solving equations 1→ 3 yields: P = 320 N

F_C = 240 N

N_C = 400 N

Since F_C = 240 N > μ_C N_C = 0.5 (400N) = 200 N the other case must be investigated.

If Post slips at C: Then F_B ≤ μ_B N_B and: F_C = μ_C N_C = 0.5 (N_C) (4)

Solving equations 1 → 4 yields: P = 267 N

N_C = 400 N

F_C = 200 N

F_B = 66.7 N

Since, in this case P is smaller, then post slips at C first.

EXAMPLE:

A box 1 meter wide and 2 meters high weighs 100 N. The coefficient of static friction between the box and the floor is μ_s = 0.5 and the coefficient of kinetic friction is μ_K = 0.3. What is the minimum force that will cause impending motion of the box?

Example: (Higdon & Stiles)

A 150 lb man starts to climb a 26 ft 50 lb ladder that is placed against a wall and its lower end is 10 ft from the wall. The ladder is placed on a 100 lb box which is 2 ft high. The coefficient of friction between the ladder and the wall is 0.20, between the ladder and the box is 0.35 floor is 0.3 and between the box and the floor is 0.25. Determine how far from the wall can the man climb before the motion of the ladder impends

Civil Engineering Dept.

CE 201 Engineering Statics

Dr. Rashid Allayla

Center of Gravity & Centroid

9.1,9.2) Center of Gravity, Center of Mass & Centroid for a Body:

Since a body consists of infinite number of particles, the center of gravity of the body is defined as the single equivalent weight that can represent all particles in the body. If the body is homogeneous and gravity is constant in magnitude and direction, the center of gravity would coincide with the center of mass.

Since the rigid body consists of infinite number of dV, the center of gravity of the whole body is the integral of the individual center of gravies of dV’s. If x^’, y^’ and z^’ are the location of the COG of dV, then:

X_G ( Σ W_i ) = Σ ( x_I ) ( W_i )

= dW

∫ x^’ γ dV ∫ y^‘ γ dV ∫ y^’ γ dV

X_G = _________ Y_G = ________ Z_G = _________

∫γ dV ∫γ dV ∫γ dV

Where γ is the specific weight of the body (dW/dV). If the density of the body is defined as ρ = γ / g where g is the gravity. The center of mass is defined as:

∫ x^’ ρ dV ∫ y^’ ρ dV ∫ y^’ ρ dV

X_G = _________ Y_G = ________ Z_G = _________

∫ρ dV ∫ρ dV ∫ρ dV

Centroid of a Body:

Centroid is defined as the geometric center of the body. If the body is uniform and homogeneous in density, the term γ drops out of the equation. The equation of the centroid is determined by summing the moments of the elements about the coordinate axis.

Centroid of a volume:

∫ x^’ dV ∫ y ^‘ dV ∫ y^’ dV

X_V = _________ Y_V = ________ Z_V = _________

∫ dV ∫ dV ∫ dV

Where ∫ is taken over the volume V.

Centroid of an Area:

∫ x^’ dA ∫ y ^‘ dA ∫ y^’ dA

X_A = _________ Y_A = ________ Z_A = _________

∫ dA ∫ dA ∫ dA

Where ∫ is taken over the area A

Centroid of a Line

∫ x^’ dL ∫ y ^‘ dL ∫ y^’ dL

X_A = _________ Y_A = ________ Z_A = _________

∫ dL ∫ dL ∫ dL

Where ∫ is taken over the line L

Example on Center of Gravity of an Area:

Example : (Hibbeler) Locate the Centroid of the area shown.

Example: (Hibbeler)

Locate the X coordinate of the shaded area bounded by the two curves y = x & y = x².

Example on Center of Gravity of a Line:

Example: (Hibbeler)

Locate the Centroid of the rod bent into the shape of parabolic arc.

9.3) Composite Bodies:

Example:

Example: (Hibbeler)

Locate the Centroid of the wire shown:

EXAMPLE:

Determine the y & z locations of the center of gravity of the object shown.

…

Example on Center of Gravity of Volumes:

Example: (Hibbeler 9.6)

Locate the y axis of the centroid for the paraboloid of revolution, which is generated by revolving the shaded area shown: The equation of the curve is: z² = 100 y

Choosing an element having the shape of thin disk with thickness (dy) and perpendicular to the axis of revolution with radius r = z. The area of the element is πz² and its thickness is dy. The elemental volume, therefore is: dv = (πz²) (dy) and its centroid is located at y_’ = y

Integrating to find the centroid of the paraboloid:

∫ y’ dV ∫ _{y = 0 → 100} [y (πz² )] dy 100 π ∫ _{y = 0 → 100} y² dy

y_c = -------------- = ------------------------------- --- = ------------------------------------- = 66.7 mm

∫ dV ∫_{y = 0 → 100} [(πz² )] dy ∫ _{y = 0 → 100} y dy

Example:

The radius of the truncated cone of circular cross section is given as a function of x by the equation r = 1 +0.25 x. Determine the coordinate of its centroid.

Because of axial symmetry shape, the centroid lies on the x axis. We will determine the x coordinate of the centroid using the elemental volume:

r

x

dV = π r² dx = π [1 + 0.25 x ]² dx

dx

∫ x dV ∫ _{0 → 4} π x [1 + 0.25 x ]² dx

x_c = _______ = ___________________ 4 ft

∫ dV ∫ _{0 → 4} π [1 + 0.25 x ]² dx

x_c = ∫ _{0 → 4} [ x + 0.5 x² + 0.063 x³ ] dx / ∫ _{0 → 4} [1 + 0.5 x + 0.063 x² ] dx

= ( 0.5 x² + 0.167 x³ + 0.016 x⁴ ) / ( x + 0.25 x² + 0.021 x³)

= (8 + 10.688 + 4.096) / (4 + 4 + 1.344) = 2.438 ft

Summary:

To locate the center of gravity, center of mass, Centroid of volume, Centroid of area or Centroid of length of composite section:

- Calculate the respective centers of gravity or Centroid of individual elements of the

composite section.

- Find the components of center of gravity or the Centroid of individual elements of

composite section.

- Assign a negative sign to the area of the cut section.

- Apply the moment principle X_C N = x₁ N₁ + x_n N_n where N represents gravity, mass

volume, area or length of composite section.

- The X, Y and Z component C.O.G. or Centroid of the composite section calculated

using the above equations.

Civil Engineering Dept.

CE 201 Engineering Statics

Dr. Rashid Allayla

Moments of Inertia

10.1) Definition:

Given an area A, the moment of inertia of the differential area dA is defined as:

I_x = ∫_A y² dA or, in general: I_x = ∫_A y² f(x) dx

Where f(x) is a function to be determined in terms of x.

I_y = ∫_A x² dA or, in general: I_y = ∫_A x² f(y) dy

Where f(y) is a function to be determined in terms of y.

EXAMPLE:

Determine the moment of inertia I_x and I_y of the triangle shown:

To evaluate I_x determine the moment of inertia of the small strip about x axis:

(I_x)_element = ∫_element y² dA

_y

= ∫₀ (y²) [dydx]

_h

I_x = ∫₀ 1/3 y³ dx

_h h

I_x = ∫₀ 1/3 [(h/b) x]³ dx = 1/12 (h/b) x⁴ │= ¹/₁₂ b h³

0

To evaluate I_y determine the moment of inertia of the small strip about y axis:

I_y = ∫_A x² dA Since: dA = y dx and y/x = h/b or y = (h/b) x , then:

I_y = ∫_A x² (ydx)

_b b

I_y = ∫₀ x² [(h/b) x]dx = ¼ x⁴ (h/b) │= ¼ h b³

⁰

Example: Hibbeler

Determine the moment of inertia for the area bounded by the two functions y = x and y = x² using: 1) horizontal elemental area and 2) Vertical elemental area.

1) Using horizontal elemental area: dA = (x₁ – x₂ ) dy

1 1

I_x = ∫ y² dA = ∫ y² (x₁ – x₂ ) dy = ∫ y² (√y – y ) dy = 2/7 y^7/2 – ¼ y⁴ | = 0.0357 m⁴

9 0

2) Using vertical elemental area: dA = (y₂ – y₁ ) dx , use the parallel-axis theorem (SEE NEXT TOPIC):

dI_x = [ dI_x^’] + [ dA ( y’ )²]

where: dI_x^’ is the moment of inertia about the center of gravity of the elemental area and y ^‘ is the distance of the center of gravity of the elemental area from the x axis, then:

b

h³

dI_x = [1/12 dx ( y₂ – y₁ )³] + [( y₂ – y₁ ) dx] [( y₁ + ( y₂ – y₁ ) / 2 )²]

= 1/3 ( y₂³ – y₁³ ) dx

h

= 1/3 (x³ – x⁶ ) dx

Then: b

1

I_x = 1/3 ∫(x³ – x⁶ ) dx = 1/12 x⁴ – 1/21 x⁷ | = 0.0357 m⁴

9

10.2) Parallel Axis Theorem for an area:

The moment of inertia in terms of x coordinate system was found to be:

I_x = ∫_A y² dA and: I_y = ∫_A x² dA

If y = y^’ + d_y and x = x’ + dx where y^’ and x^’ is the coordinates of dA relative to the x^’y^’ axis (passing through the center of gravity of the area A), then:

I_y = ∫_A (x^’ + d_x)² dA = ∫_A (x^’) ² dA +2dx∫ x^’ dA + d_x² ∫_A dA

= 0 because ∫ x^’ dA = x_G ∫ dA = 0 (since x_G = 0)

I_x = ∫_A (y^’ + d_y)² dA = ∫_A (y^’) ² dA +2dy∫_A y^’ dA + d_y² ∫_A dA

Therefore:

I_x = I_x^’ + A d_y² and: I_y = I_y^’ + A d_x²

Which states that the moment of inertia for an area about a given axis is equal to the moment of inertia of the area about an axis passing through its own centroid plus the product of the area and the square of the distance between the two axis.

10.5) Moment of Inertia for Composite Areas:

Method:

1) Divide the composite area into parts of easily determinable moments of inertia.

2) Determine the moment of inertia of each member and use the method of parallel-axis theorem to determine the moment of inertia about a given coordinate system.

3) Sum the results.

Illustrative Example:

Determine the moment of inertia for the beam’s cross-sectional area shown about the x and y centroidal

axis.

Rectangle A A_A ^y2_CG

I_x = I’_x + A d’_y = 1/12 (100) (300)³ + [(100) (300)] (200)²

= 1.425 x 10⁹ mm⁴

I_y = I’_y + A d’_x = 1/12 (300) (100)³ + (100) (300) (250)²

= 1.90 x 10⁹ mm⁴

Rectangle B

I_x = 1/12 (600) (100)³ = 0.05 (10)⁹ mm⁴

I_y = 1/12 (100) (600)³ = 1.8 (10)⁹ mm⁴

Rectangle C

I_x = I’_x + A d’_y = 1/12 (100) (300)³ + (100) (300) (200)²

= 1.425 x 10⁹ mm⁴ 300-50

I_y = I’_y + A d’_x = 1/12 (300) (100)³ + (100) (300) (250)²

= 1.90 x 10⁹ mm⁴

Summation: I_x = 2.9 (10)⁹ mm⁴