2.1

Chapter 2: Statics

2.1 Characteristics of a Force

Force

Force is defined as the action of one body on another that affects the state of

motion or rest of the body.

In the late 17th century, Sir Isaac Newton summarized the effects of force in

three basic laws.

First Law: A body at rest will remain at rest and a body in motion will move in a

straight line at a constant speed unless acted upon by a force. (Equilibrium)

Second Law: If the resultant force acting on a particle is not zero, the particle

will have acceleration proportional to the magnitude of the resultant force and

in the direction of the resultant force.

(F = ma)

Third Law: For every force of action, there is a reaction that is equal in

magnitude, opposite in direction, and has the same line of action.

Characteristics of a Force

A force is characterized by the following.

• Point of application.

- The point of application defines the point where the force is applied.

• Magnitude.

- Magnitude refers to the quantity of force, a numerical measure of intensity.

- In English units, force is expressed in units of pound (lb) and kilo-pound

(kip).

- In metric (SI) units, force is expressed in units of Newton (N), or kilo-Newton (kN = 1,000 N).

• Direction.

- The direction of a force is defined by

its line-of-action and sense.

- The sense of the force is indicated by

an angle and an arrowhead.

- The angle θ is used to denote the line of action of a force in relation to a

reference axis (e.g. horizontal or vertical).

2.2

In the study of forces and force systems, the word particle is used when the size

and shape of the body under consideration does not affect the solution.

Rigid Bodies

In reality, any body under the action of forces undergoes some kind of

deformation (i.e. a change in shape).

• In statics, we consider that a body theoretically undergoes no deformation

because the changes are small and considered insignificant.

• The concept of a rigid body is purely theoretical (often called an idealization).

Principle of Transmissibility

Principle of transmissibility - The external effects on a body remain unchanged

when a force F1 acting at point A is replaced by a force F2 of equal magnitude

acting at point B, if both forces have the same sense (direction) and the same line

of action.

• The two forces have the same effect on the rigid body and are said to be

equivalent.

An example of the principle of transmissibility

The “principle of transmissibility” is valid only in terms of the external effects on

a body remaining the same, where internally this may not be true (e.g. a test

cylinder).

External effects are the same.

Internal effects (stress and deformation) are different.

2.3

External and Internal Forces

Forces acting on rigid bodies are separated into two groups:

• “External forces” represent the action of other bodies on the rigid body under

consideration.

• “Internal forces” are the forces that hold together the particles forming the

rigid body.

Types of Force Systems

Force systems are identified based on the arrangement of the forces acting on the

rigid body or particle.

• Collinear – All forces act along the same straight line (including 2-D and 3-D

systems).

• Coplanar – All forces act in the same plane (i.e. 2-D system).

• Coplanar/parallel – All forces are parallel and act in the same plane (i.e. 2-D

system).

• Coplanar/concurrent – All forces intersect at a common point and lie in the same

plane (i.e. 2-D system).

• Noncoplanar/Parallel – All forces are parallel to each other, but not all lie in the

same plane (i.e. 3-D system).

• Noncoplanar/concurrent – All forces intersect at a common point but do not all

lie in the same plane (i.e. 3-D system).

• Noncoplanar/nonconcurrent – General force system (3-D system).

2.2 Vector Addition

Characteristics of Vectors

If forces act in the same direction (or in opposite directions), the magnitude of

the resultant force can be determined by simply adding the magnitudes of the

given forces.

In general, vectors (including forces) are added according to the Parallelogram Law.

• Vectors, characterized by both magnitude and direction, require a special

procedure for adding them to determine the resultant vector.

Using the Parallelogram Law, we may add vectors graphically or by trigonometric

methods.

2.4

Graphical Solution

Consider the forces acting on a particle

(point).

• The force R is called the “resultant” of

the forces F and P.

- The force R may be obtained by

constructing, to scale, a parallelogram,

using F and P as the sides of the

parallelogram.

- The diagonal line represents the “resultant” force R and is the vector

addition of forces F and P.

The magnitude of the resultant force R is determined graphically by measuring the

length of the line segment and determining the magnitude based on the scale used

to draw forces F and P.

An angle, such as the angle θ, is measured from a reference axis (such as the

horizontal axis) to specify the direction of the resultant force.

Trigonometric Solution

The magnitude and direction of the

resultant force R can also be determined by

trigonometry using the Law of Sines and the

Law of Cosines.

Law of Sines: sin A = sin B

a b

Law of Cosines: b2 = a2 + c2 – 2 a c (cos B)

Addition of Two Vectors

Vectors (including forces) are added according to the Parallelogram Law.

• The magnitude of the resultant force R determined from the two forces (F and

P) is not equal to the algebraic summation of the magnitudes (F + P) of the

vectors F and P, except in the special case where F and P are collinear.

• When F and P are collinear, the Parallelogram Law reduces to an algebraic or

scalar addition: R = F + P.

2.5

• The parallelogram constructed from the vectors F and P does not depend upon

the order in which F and P are selected.

- The addition of two vectors is said to be “commutative,” that is

F + P = P + F

An alternative method for determining the sum of two vectors is referred to as

the “triangle rule.”

• Since the side of the parallelogram opposite to F is equal to F in magnitude and

direction, we need only draw half of the parallelogram.

• Arrange F and P in a “tip-to-tail” fashion by connecting the tail of F with the tip

of P or by connecting the tail of P with the tip of F since the addition of

vectors is commutative.

Addition of Three or More Vectors

The resultant of any number of forces (vectors) may be obtained

• Graphically by applying repeatedly the Parallelogram Law (or the “tip-to-tail

method”) to successive pairs of vectors until all of the given vectors are

replaced by a single resultant vector.

• By methods of trigonometry (using the Law of Sines and Law of Cosines) by

solving interim triangles.

- However, this method can be tedious.

Other, more convenient methods for adding three or more vectors will be

presented.

2.6

Example Problem - Vector Addition

Problem 2.1 (p. 27)

Given: Forces shown acting on the pin.

Find: The resultant force.

a) Solve graphically.

b) Solve by trigonometry.

Solution (graphical)

Scale: 1” = 50 lb

Solution (by trigonometry)

Using the Law of Cosines

R2 = (100)2 + (200)2 – 2 (100) (200) cos 60°

R2 = 30,000

R = 173.2 lb

Using the Law of Sines

sin 60° = sin α sin α = 200 sin 60°/173.2 = 1.00

R 200 α = 90°

θ = 180° - 40° - α = 180° – 40° – 90° = 50°

2.7

Example Problem - Graphical Addition of Three or More Vectors

Problem 2.4 (p. 28)

Given: Forces shown.

Find: Magnitudes of T2 and T3 for a

resultant force of 10,000 lb acting

vertically.

Solution (graphical)

Scale: 1” = 2,000 lb

2.8

2.3 Force Systems

Resolution of Forces into Rectangular Components

In many problems, it is desirable to resolve a

force into two components that are

perpendicular to each other.

• These components (Fx and Fy) are called

the rectangular components (i.e. the

parallelogram is a rectangle).

• The x and y-axes of a rectangular

coordinate system are often assumed to

be horizontal and vertical, respectively;

however, the axes may be chosen in any

two mutually perpendicular directions for

convenience.

A force F with a direction θ from the horizontal x-axis can be resolved into its

rectangular components Fx and Fy in terms of F and θ.

Fx = F cos θ Fy = F sin θ

(Note: The force triangle is geometrically similar to the dimensional triangle.)

The force components Fx and Fy form the legs of a parallelogram, with the diagonal

representing the original force F.

• Using the Pythagorean theorem for right triangles,

F = (Fx2 + Fy

2)1/2

tan θ = Fy / Fx or θ = tan-1 (Fy / Fx)

2.9

Example Problems - Resolution of Forces into Rectangular Components

Problem 2.6 (p. 32)

Given: Cantilever beam loaded

as shown.

Find: The x and y-components

of the force.

Solution

Fx = F (4/5) = 1000 (0.800)

Fx = 800 lb

Fy = - F (3/5) = - 1000 (0.600)

Fy = - 600 lb

Problem 2.8 (p. 32)

Given: Roof system shown.

Find: Components of the force that are

parallel (x-axis) and perpendicular

(y-axis) to the axis of the rafter.

Solution

Parallel component

Px = - P sin θ = - P (4/ 160 )

Px = - 300 (0.316) = - 94.9 lb

Perpendicular component

Py = - P cos θ = - P (12/ 160 )

Py = - 300 (0.949) = - 284.6 lb

2.10

Vector Addition by Component Method

Vectors may be added based on an analytical approach by using the two rectangular

components of each vector.

• Resolve each force into its

rectangular components.

• Add algebraically the horizontal

and vertical force components to

determine the horizontal and

vertical components of the

resultant force, respectively.

Sign convention

Force acting to the right or up is + (positive)

Force acting to the left or down is - (negative)

Horizontal components are summed algebraically.

Rx = - Ax + Bx – Cx or Rx = ∑Fx

Vertical components are summed algebraically.

Ry = + Ay + By – Cy or Ry = ∑Fy

The magnitude of the resultant, or the vector sum of Rx and Ry, is found using the

Pythagorean Theorem.

R = (Rx2 + Ry

2)1/2

tan θ = Ry / Rx = ∑Fy / ∑Fx

θ = tan-1 (Ry / Rx)

2.11

Example Problems - Vector Addition by the Component Method

Problem 2.9 (p. 40)

Given: Steel gusset plate shown.

Find: Resultant force by analytical

methods.

Solution

Rx = ∑Fx = + F1x - F2x + F3x

= + F1 sin 30° - F2 + (1/ 2 ) F3

= + 10 (0.500) - 12 + (0.707) 18

= + 5.00 – 12.00 + 12.73

Rx = + 5.73 k

Ry = ∑Fy = + F1y - F2y + F3y

= + F1 cos 30° - 0 - (1/ 2 ) F3

= + 10 (0.866) - 0 - (0.707) 18

= + 8.66 – 0 - 12.73

Ry = - 4.07 k

R = (Rx2 + Ry

2)1/2 = [(5.73)2 + (- 4.07)2]1/2 = 7.03 k

tan θ = - 4.07/5.73 = - 0.7103

θ = -35.4°

Thus, R = 7.03 k

2.12

Problem 2.12 (p. 41)

Given: Roof truss shown.

F1 and F2 ≤ 7 kN

Find: Vertical resultant force by

analytical methods.

Given the nature of the loads, Rx = 0 and Ry = 0.

• When the loads are “balanced” (i.e. the resultant force is zero), the condition is

called “equilibrium.”

• Equilibrium will be discussed later in this chapter.

Solution

Rx = ∑Fx = 0

0 = + F1x - F2x = F1 – F2 cos 25°

F1 = F2 cos 25° = F2 (0.906)

Let F2 = 7 kN (since F2 is the larger force.)

Then F1 = (7) (0.906) = 6.34 kN

Ry = ∑Fy = 0

0 = R + F1y - F2y = R + 0 – F2 sin 25°

R = F2 sin 25° = F2 (0.423) = 7 (0.423) = 2.96 kN

R = 2.96 k ↑

2.13

Moment of a Force

The tendency of a force to rotate a body

about an axis (or point) is called the moment of a force.

The moment M of force F about a point A is

defined as the product of the magnitude of

the force F and perpendicular distance d

(referred to as the moment arm) from point

A to the line of action of the force F.

MA = F x d

To increase the moment, increase the force F and/or the distance d.

If the force passes through the axis (i.e. d = 0), there is no moment.

The moment of a force is a vector quantity (i.e. moment has both magnitude and

direction).

Units used for moment are:

• English units

Pound-inch (lb-in), pound-foot (lb-ft), kip-inch (k-in), kip-foot (k-ft).

• Metric (SI) units

Newton-meter (N-m) or kilo-Newton-meter (kN-m).

In a coplanar (i.e. a two-dimensional) system, the direction of the moment is either

clockwise or counterclockwise.

• Sign convention

Counterclockwise is + (positive)

Clockwise is - (negative)

The sign convention (+ or -) corresponds with “right-hand rule.”

2.14

Example Problems - Moment of a Force

Problem 2.14 (p. 46)

Given: Box with the force shown.

Find: Moment about point A.

Does the box tip over?

Solution

MA = - F d + W d

= - 20 (5) + 25 (4) = - 100 + 100 = 0

MA = 0 lb-ft The box is on the verge of tipping over.

2.15

Problem 2.16 (p. 46)

Given: Wrench with force shown.

Find: Moment MA at center of pipe;

maximum moment from the 39 lb force.

Solution

Find the component of the force F

perpendicular to the wrench handle.

• The component of the force that is

parallel to the wrench handle passes

through the axis of rotation and does not

cause rotation.

Find the angle α between wrench handle and horizontal axis.

tan α = 15/8 α = 61.93°

Find the angle φ between force F and horizontal axis.

tan φ = 5/12 φ = 22.62°

Find the angle θ between the wrench handle and the force F.

θ = α – φ = 61.93° - 22.62° = 39.31°

Find the component of the force F perpendicular to the handle of the wrench

(i.e. Fy).

Fy = F sin θ = 39 sin 39.31° = 39 (0.634) = 24.71 lb

Find the moment MA at the center of the pipe.

MA = Fy d = 24.71 (17) = 420.0 lb-in (clockwise)

The maximum moment MA at the center of the pipe from the 39 lb force occurs

when the 39 lb force is applied perpendicular to the handle of the wrench.

(MA)max = 39 (17) = 663 lb-in

2.16

Alternate Solutions

• Determine the moment MA using the horizontal and vertical components of the

39 lb force.

Applicable equation: MA = - Fx dy + Fy dx

Fx = (12/13) 39 = 36 lb

Fy = (5/13) 39 = 15 lb

tan Δ = 8/15 = 0.533

Δ = 28.07°

dx = 17 sin Δ = 17 sin 28.07° = 8.0”

dy = 17 cos Δ = 17 cos 28.07° = 15.0”

MA = - 36 (15.0) + 15 (8.0)

= - 540.0 + 120.0 = -420.0

MA = 420.0 lb-in (clockwise)

• Determine the moment MA using the initial definition of moment, that is,

moment is equal to force times a perpendicular moment arm distance.

Applicable equation: MA = F d

d = 17 sin θ

θ = α – φ

tan α = 15/8 = 1.875

α = 61.93°

tan φ = 5/12 = 0.417

φ = 22.62°

θ = α – φ = 61.93° - 22.62° = 39.31°

MA = F d = 39 (17 sin θ)

= 39 (17 sin 39.31°) = 39 (17) 0.6335 = 420.0

MA = 420.0 lb-in (clockwise)

2.17

Varignon’s Theorem

Varignon’s Theorem – “The moment of a force about a point (axis) is equal to the

algebraic sum of the moments of the components of the force about the same

point (axis).”

Varignon’s Theorem can simplify the calculation of moments.

Consider the force acting on the bracket.

• Find the moment about point A.

Find the moment about point A using the scalar definition of moment (i.e. M = F d).

• Calculate the perpendicular distance d

using similar triangles.

6/15 = d/12

d = (6/15) 12 = 4.80”

MA = F d = 20 (4.80) = 96.0 lb-in

Find the moment about point A using Varignon’s Theorem.

• Calculate the moment using the

rectangular components of the force.

MA = - Fx (dy) + Fy (dx)

= - 12 (12) + 16 (15) = - 144 + 240

MA = 96.0 lb-in

2.18

Find the moment about point A using Varignon’s Theorem and the Principle of

Transmissibility.

• Calculate the moment using the vertical

component of the force.

- The horizontal component of the

force passes through point A.

MA = 16 (6) = 96.0 lb-in

Find the moment about point A using Varignon’s Theorem and the Principle of

Transmissibility.

• Calculate the moment using the

horizontal component of the force.

- The vertical component of the force

passes through point A.

MA = 12 (8) = 96.0 lb-in

2.19

Example Problems - Varignon’s Theorem

Problem 2.20 (p. 51)

Given: Truss loaded as shown.

Find: Moment about points B and C.

MB = - Fx (dy) – Fy (dx)

= - (12/13) 1300 (5) - (5/13) 1300 (0)

= - 6000 – 0 = - 6000

MB = 6,000 lb-ft (clockwise)

MC = - Fx (dy) + Fy (dx)

= - (12/13) 1300 (5) + (5/13) 1300 (12)

= -6000 + 6000 = 0

MC = 0 lb-ft (Force F passes through point C.)

2.20

Problem 2.23 (p. 51)

Given: Boom loaded as shown.

Find: Weight W for maximum force in

the cable of T = 2,000 lb

Solution

MC = 0 = - W d1 + Tx d2 – Ty (d3)

0 = - W (7 + 3) cos 60° + 2000 cos 30° (7 sin 60°)

– 2000 sin 30° (7 cos 60°)

0 = - W (10) (0.500) + 2000 (0.866) (6.062) – 2000 (0.500) (3.50)

0 = - W (5.00) + 10,499.4 – 3,500.0

5.00 W = 10,499.4 – 3,500.0 = 6,999.4

W = 1,400.0 lb

Alternate solution

• The components of the force in the cable may be taken parallel and

perpendicular to the boom AC.

• The parallel component passes through point C and causes no moment about

point C.

MC = 0 = - W (7 + 3) cos 60° + 2000 sin 30° (7)

0 = - W (10) (0.500) + 2000 (0.500) (7)

0 = - 5.0 W + 7000

5.0 W = 7000

W = 1400.0 lb

2.21

Couple and Moment of a Couple

Two forces having the same magnitude, parallel lines of action, and opposite sense

(direction) are said to form a “couple.”

• Couples cause moment (rotation) of a body only.

• The forces cause no translation of the body in the horizontal or vertical

directions - the sums of the horizontal and vertical components of the forces

are zero.

Consider the two equal, opposite, and parallel forces F.

MA = + F (x) – F (x + d),

MA = + F x – F x - F d

MA = - F d

The moment M is called the “moment of the couple.”

• The moment M is said to be a “free

vector” (i.e. the moment M will

have the same magnitude and same

direction regardless of the

location of point A).

MA = - F d1 – F d2 = - F (d1 + d2),

but d1 + d2 = d

MA = - F d

The moment M of a couple is constant.

• The magnitude of M is equal to the product (F) x (d) – the magnitude of either

force F times the perpendicular distance d between their lines of action.

• The sense (direction) of M (i.e. clockwise or counterclockwise) is determined by

observation.

- The sign convention (+ or -) corresponds with “right-hand rule.”

2.22

Example Problems - Couple and Moment of a Couple

Problem

Given: Triangular plate subject to the

three couples.

Find: Plate dimension “d” so that the

resultant couple is 350 Nm

(clockwise).

Solution

MO = - 350 = - F1 d – F2 (d sin θ) + F3 (d cos θ)

- 350 = - 100 d – 600 d sin 30° + 200 d cos 30°

- 350 = - 100 d – 300 d + 173.2 d

- 350 = - 226.8 d

d = -350/-226.8 = 1.543

d = 1.543 m

Note: The forces F1’, F2’ and F3’ all pass through point O and cause no moment

about point O.

2.23

Problem

Given: Two couples act on the beam as

shown.

Find: Magnitude of F so that the

resultant couple moment M is 300

lb-ft (counterclockwise).

Solution

MA = + 300 = F (4/5) 1.5 + F (3/5) 4 – 200 (1.5)

300 = F (1.20) + F (2.40) – (200) 1.5

300 = 3.60 F – 300

3.60 F = 600

F = 600/3.60 = 166.7

F = 166.7 lb

2.24

Resolution of a Force into a Force and Couple Acting at Another Point

In some problems, it may be useful to change the location of an applied force to a

different point of application on the rigid body.

In a previous section, we discussed the possibility of moving a force F along its line

of action (principle of transmissibility) without changing the external effects on

the body.

F1 = F2 (same line of action)

However, we cannot simply move a force away from the original line of action to a

parallel line of action without modifying the external effects on the rigid body.

• For example, if the applied force F

is moved from point A to point B on

the cantilever beam, differing

deflections at the free end will

result.

Consider a force F applied at point B on rigid body AB.

• Our goal is to move the force F to point A without changing the external

effects on the rigid body.

• Two forces, both with a magnitude

of F, are added at point A.

• The lines of action of the two

forces are collinear and parallel to

the line of action of the original

force at B.

The addition of the equal and opposite

forces at A does not change the external effects on the rigid body.

• The external effects are the same: the same translation and the same rotation.

• The new system consisting of the three forces is said to be equivalent to the

original system consisting of the single force.

2.25

The force F at point B and the force F at point A are equal and opposite forces

with parallel lines of action, thus forming a couple M.

• The moment due to the couple is equal to

F x d and is a constant value anywhere on

the rigid body.

• The couple M can be placed at any

convenient location (such as point A) with

the remaining force at point A.

The system of the force and couple at point A is said to be an “equivalent force-couple system.”

2.26

Example Problem - Resolution of a Force into a Force and Couple Acting at Another

Point

Problem 2.26 (p. 58)

Given: Column-beam system shown.

Find: Equivalent force-couple system

through the column centerline.

Solution

R = 90 kN

Mcouple = - 90 (180 + 125) = - 90 (305)

= - 27,450 kN-mm

= - 27.45 kN-m

Mcouple = 27.45 kN-m (clockwise)

2.27

Resultant of Two Parallel Forces

Suppose we wish to represent the two forces A and B shown on the girder with a

single resultant force R, which produces an equivalent effect as the original forces.

• The equivalent resultant force R must

produce the same translational

tendency as the forces A and B.

• The equivalent resultant force R must

produce the same rotational effect as

well.

The magnitude of the resultant R of the parallel forces A and B equals the

algebraic summation of A and B, that is

R = A + B

The location of the resultant force R is obtained by the principle of moments.

• The moment of the original forces

about a point must be equal to the

moment of the resultant force

about the same point.

∑MC = - A a – B b = - R x

R x = A a + B b

x = (A a + B b)

R

2.28

Example Problem - Resultant of Parallel Forces

Problem

Given: Building slab subjected

to four parallel column

loadings.

Find: The equivalent resultant

force and specify its

location (x, y) on the floor

slab.

Solution

R = ∑Fz = - 30 – 40 – 20 – 50

R = - 140 kN

R = 140 kN (down)

Cx = ∑Mx = - 30 (11) – 40 (13) – 50 (3)

= - 330 – 520 – 150

Cx = - 1,000 kN-m

Cy = ∑My = + 40 (10) + 20 (10) + 50 (4)

= + 400 + 200 + 200

Cy = + 800 kN-m

ix = |Cy/R| = |+ 800/- 140| = 5.71 m

iy = |Cx/R| = |- 1,000/- 140| = 7.14 m

Location: (5.71, 7.14) m

2.29

2.4 Equilibrium Equations: Two-Dimensional

Equilibrium

When the force and couple are both equal to zero, the external forces form a

force-couple system equivalent to zero and the rigid body is said to be in

equilibrium.

Two equations of equilibrium for a rigid body can be summarized as follows.

∑F = 0 Necessary and sufficient conditions for

∑M = 0 the equilibrium of a rigid body.

We may express the necessary and sufficient conditions for the equilibrium of a

rigid body as

∑Fx = 0 ∑Fy = 0 ∑Mz = 0

Collinear Force System

A collinear force system involves the action of forces along the same line of action

(e.g. tug-of-war).

Concurrent Force System

Equilibrium of a particle: When the resultant of all the forces acting on a particle

is zero, the particle is in equilibrium.

• This condition may be stated mathematically as

∑F = 0

This mathematical expression is not only a necessary condition for equilibrium, but

also it is a sufficient condition.

• Since the force system is concurrent, the moment equation (∑Mz = 0) is

automatically satisfied.

Graphically, a closed polygon represents equilibrium.

Analytically, the following equations of equilibrium are required in order to satisfy

the conditions of equilibrium.

∑Fx = 0 and ∑Fy = 0

2.30

Consider the coplanar, concurrent force system of a weight suspended by two

cables.

Analytically, we can write the following equations of equilibrium.

∑Fx = 0 = - FAC cos 45° + FBC cos 30°

∑Fy = 0 = + FAC sin 45° + FBC sin 30° - W

If W is known, then FAC and FBC can be determined.

Nonconcurrent/Coplanar Force System

Equilibrium of a rigid body: When the resultant of all the forces and moments

acting on a rigid body is zero, the rigid body is in equilibrium.

• This condition may be stated mathematically as

∑F = 0 ∑M = 0

Three equations of equilibrium for a rigid body can be summarized as follows.

∑Fx = 0 ∑Fy = 0 ∑Mi = 0 (where i is any point on the rigid body)

These three equations (called the “equations of statics”) allow solution for no more

than three unknowns.

The following alternate sets of equations are possible.

∑Fx = 0 ∑Mi = 0 ∑Mj = 0

or

∑Fy = 0 ∑Mi = 0 ∑Mj = 0

or

∑Mi = 0 ∑Mj = 0 ∑Mk = 0

2.31

Free-Body Diagrams

An essential step in solving equilibrium problems involves drawing free-body diagrams.

• A free-body diagram is a simplified representation of a particle or rigid body on

which all of the applied forces and reactions are shown.

A free-body diagram shows the following.

• Particle or rigid body (with significant dimensions).

• Externally applied forces and moments.

• Weight of the rigid body (if significant).

• Reactive forces and moments.

Free-Body Diagram of Particles

The free-body diagram of a particle is relatively simple because it only shows

concurrent forces acting at a point.

2.32

Example Problems - Equilibrium of a Particle

Problem 2.29 (p. 72)

Given: Frame loaded as shown.

Find: Forces in members AC and BC.

Solution (assuming tension in AC and BC)

• The free body diagram is shown below.

∑Fx = 0 = - 500 cos 20° - AC sin 10° + BC sin 30°

0 = - 469.8 – 0.1736 AC + 0.500 BC

469.8 = – 0.1736 AC + 0.500 BC (Divide by 0.1736)

Eq. 1 2,706.20 = - AC + 2.88 BC

∑Fy = 0 = - 500 sin 20° - AC cos 10° - BC cos 30°

0 = - 171.0 – 0.985 AC – 0.866 BC

171.0 = – 0.985 AC – 0.866 BC (Divide by - 0.985)

Eq. 2 - 173.60 = AC + 0.879 BC

Solve Eq. 1 and Eq. 2 simultaneously by adding the two equations.

Eq. 1 2,706.20 = - AC + 2.88 BC

Eq. 2 - 173.60 = + AC + 0.879 BC

2,532.60 = 0 + 3.759 BC

BC = 2,532.60/3.759 = + 673.7 N

BC = 673.7 N (tension)

AC = - 2,706.20 + 2.880 BC

= - 2,706.20 + 2.880 (673.7) = - 2,706.20 + 1,940.26 = - 765.94

AC = - 765.9 N

AC = 765.9 N (compression)

2.33

2.5 Free-Body Diagrams of Rigid Bodies

Free-body diagrams of rigid bodies include a system of forces that no longer have

a single point of concurrency.

• This system of actions may include the following.

- Applied forces and couples (moments) and

- Reactive forces and couples (moments).

“Reactions” are the forces through which the ground and other bodies oppose a

possible motion of the free body.

• Reactions are exerted at points where the free body is supported or connected

to other bodies.

Procedure for Drawing a Free-Body Diagram

1. Draw the isolated body. Include slopes and dimensions, as these may be needed

to compute moments.

2. Show the magnitudes and directions of the known external forces (including the

weight of the body).

3. Show the unknown external forces (support reactions or constraints).

Support Reactions

It is important to understand how to replace certain supports with the appropriate

restraining forces. In general,

• If a support prevents translation, then a force is developed on the body in that

direction.

• If rotation is prevented, then a couple moment is exerted on the rigid body.

Reactions exerted on two-dimensional structure may be divided into three groups,

corresponding to three types of supports or connections.

1. Reaction equivalent to a force with a known line of action (1 unknown).

• This reaction prevents translation of the free body in one direction, but it

cannot prevent the body from rotating about the connection.

• The most common examples include the following.

a. Roller support

Prevents translation in the

direction perpendicular to

the plane on which the roller

rests.

2.34

b. Cables (weightless).

The reaction acts in one

direction along the cable.

c. Short links.

The reaction acts in one

direction along the link.

For all of these there is only one unknown involved, that is, the magnitude of

the reaction.

• The line of action is known.

2. Reaction equivalent to a force with an unknown line of action (2 unknowns).

• This reaction prevents translation of the free body in all directions, but it

cannot prevent the body from rotating about the connection.

• The most common examples include the following.

a. Pin

Prevents vertical and

horizontal translation.

b. Ball and socket (2-D)

The magnitude and direction of the reactive force is unknown.

• Normally we work with the components, thus fixing the directions but not

knowing the magnitude of the components.

2.35

3. Reaction equivalent to a force and a couple (3 unknowns).

• This reaction is caused by “fixed supports” which oppose any motion of the

free body and thus constrain it completely, preventing both translation and

rotation.

• The most common example include the following.

a. Fixed end.

Prevents vertical and

horizontal translation,

and prevents rotation.

Reactions of this group involve three unknowns, consisting usually of the two

components of the force and the moment of the couple.

If a free-body diagram for the rigid body is drawn, only external forces are shown.

• Internal forces are not represented on a free-body diagram for the entire rigid

body.

The weight of the rigid body is usually insignificant compared to the applied

forces.

• When the weight of a body must be considered, a force representing the

weight is used with its point of application acting through the center of gravity

of the rigid body.

When the direction of an unknown force or couple is not known, no attempt should

be made to determine the correct direction.

• The sense of the force or couple can be arbitrarily assumed.

• The sign of the answer obtained will indicate whether or not the assumption is

correct.

2.36

Consider the loaded beam shown and

the steps in drawing the free-body

diagram.

Steps

1. Draw the isolated body.

2. Add the applied forces.

3. Show the reactive forces.

a. Does the support at A prevent

1) Horizontal translation?

2) Vertical translation?

3) Rotation?

b. Does the support at B prevent

1) Horizontal translation?

2) Vertical translation?

3) Rotation?

Idealized Models

In order to perform a correct force analysis of any object, it is important to

consider a corresponding analytical or idealized model that gives results that

approximate as closely as possible the actual situation.

• Example: An underground pump station. Is reinforced concrete slab simply

supported by the concrete block walls, or fixed-fixed?

2.37

Example Problem - Equilibrium of Rigid Bodies

Problem 2.34 (p. 84)

Given: Pole loaded as shown.

Find: Horizontal and vertical components

of the reactions at A and B.

∑MA = 0 = + Bx (10) – 300 (3) + 100 (4)

Bx (10) = 300 (3) – 100 (4) = 900 – 400 = 500

Bx = 50 lb ←

∑Fx = 0 = - Bx - 100 + Ax = - 50 – 100 + Ax

Ax = 150 lb →

∑Fy = 0 = Ay – 300

Ay = 300 lb ↑

2.38

Example

Given: Beam loaded as shown.

Find: Range of values for P for a

safe beam:

RA and RB ≤ 25 kips

(i.e. compression in the

columns at A and B)

∑Fx = 0 = Ax Ax = 0 kips

Let RA = 0 kips

∑MB = 0 = 6 P - 6 (2) – 6 (4)

6 P = 12 + 24 = 36

P = 6.0 kips

Check RB: ∑Fy = 0 = - 6 – 6 – 6 + RB

RB = 18.0 kips < 25.0 kips OK

Let RA = 25 kips

∑MB = 0 = - 25 (9) + 6 P - 6 (2) – 6 (4)

6 P = 225 + 12 + 24 = 261

P = 43.5 kips

Check RB: ∑Fy = 0 = 25 - 43.5 – 6 – 6 + RB

RB = 30.5 kips > 25.0 kips NG

Let RB = 25 kips

∑MA = 0 = - 3 P + 9 (25) – 6 (11) – 6 (13)

3 P = 225 – 66 – 78 = 78

P = 27.0 kips

Check RA: ∑Fy = 0 = - 27 + 25 – 6 – 6 + RA

RA = 14.0 kips < 25.0 kips OK

Therefore, 6.0 kips ≤ P ≤ 27.0 kips Answer

2.39

2.6 Statical Indeterminacy and Improper Constraints

In the analysis of a beam, truss, or framework, the first step usually involves the

drawing of a free-body diagram. From the FBD, we can determine

• Whether the necessary and available equations of equilibrium are sufficient to

satisfy the given loads conditions and

• The unknown support forces and couples.

In the case where the support restraints adequately resist translation (x and y)

and rotation to satisfy the conditions of equilibrium, the reactions are said to be

statically determinate, and the rigid body is said to be completely constrained.

When the number of unknowns exceeds the number of equations of equilibrium, the

rigid body is said to be statically indeterminate.

• The degree of indeterminacy is the difference between the number of

unknowns and the number of equations of equilibrium.

Determinate

Determinate

Determinate

SI1

(Degree of Redundancy)

2.40

SI3

SI1

Mobile structure

(Improperly constrained)

Improper Constraints

Although the number of unknowns is equal to the number of equations of

equilibrium, no support exists that restrains the horizontal translation and the

equilibrium equation (e.g. ∑Fx = 0) cannot be satisfied.

• These constraints are improperly arranged; this condition is referred to as

improperly constrained.

If the reactions involve less than three unknowns, there are more equations than

unknowns, and some of the equations of equilibrium cannot be satisfied under a

general loading condition.

• The rigid body is only “partially constrained.”