6.1

Chapter 6: Structural Analysis

Chapter Objectives

• To show how to determine the forces in the members of a truss using the method

of joints and the method of sections.

• To analyze the forces acting on the members of frames and machines composed of

pin-connected members.

In this chapter, we shall consider problems dealing with the equilibrium of structures

made of several connected parts.

• These problems call for the determination of the following forces.

1. External forces acting on the structure (reactions).

2. “Internal forces” – forces that hold together the various parts of the

structure.

Newton’s Third Law states, “The forces of action and reaction between bodies in

contact have the same magnitude, same line of action, and opposite sense.”

In this chapter, we shall consider three broad categories of engineering structures.

1. Trusses.

• Trusses consist of straight members connected at joints located at the ends of

the members.

• Members of a truss are “two-force” members.

2. Frames.

• Frames always contain at least one “multi-force member.”

3. Machines.

• Machines are designed to transmit and modify forces, and are structures

containing moving parts.

• Machines, like frames, always contain at least one “multi-force member.”

6.1 Simple Truss

A truss is one major type of engineering structures and is used in bridges and

buildings.

• A truss consists of straight members connected at joints.

• Truss members are connected at their ends only.

• No member is continuous through a joint.

6.2

Truss Frame

The members of a truss are slender and can support little lateral load.

• All loads must be applied at the joints and not along the members themselves.

• In the case of bridge trusses, the dead

loads and traffic loads from the deck are

first carried by “stringers” which in turn

transmit the loads to “floor beams”, and

then the loads are finally transmitted to

the joints of the supporting side trusses

(ref. Fig. 6-2 in textbook).

Assumptions for Design

1. All loadings are applied at the joints.

2. The members are joined together by smooth pins.

Because of these two assumptions, each truss member acts as a two-force

member.

• The member is either in tension or compression.

Although the members are actually joined together by means of riveted, bolted, or

welded connections, the members are considered to be pinned together.

• The forces acting at each end of the member reduce to a single force and no

couple.

• Each member is treated as a two-force member.

Actual connection (Gusset Plate)

The moments that are created at

the ends of the members are small

and considered insignificant.

6.3

Simple Truss

The triangle ABC is considered the “basic truss.”

• The truss is said to be a “rigid truss” meaning

that the truss is stable and will not collapse.

• The only possible deformation involves small

changes in the length of its members.

A large truss may be obtained by successively adding two members, attaching them

to separate existing joints, and connecting them at a new joint.

• A truss constructed in this manner is called a “simple truss.”

• In a simple truss, the total number of members is related to the total number

of joints by the following equation.

m = 2 n – 3

where

n = the total number of joints

m = the total number of members

6.2 The Method of Joints

A truss may be considered as a group of pins and two-force members.

• We can dismember a truss and draw a free-body diagram for each pin.

• Since the entire truss is in equilibrium, each pin must be in equilibrium.

Consider the truss shown below.

6.4

Each pin represents a concurrent force system.

• At each pin we can write only two equations of equilibrium (a.k.a. equations of

statics).

∑ Fx = 0 and ∑ Fy = 0

In the case of a simple truss, the number of members and the number of pins

(joints) are related by the following equation.

m = 2 n – 3

where

n = the number of pins (joints)

m = number of members

and the number of unknowns that can be determined by the pins is

m + 3 (i.e. 2 n = m + 3).

• Thus, the forces in all the members plus the two components of the reaction at

A (i.e. Ax and Ay) and the vertical reaction at C (i.e. Cy) may be found using the

free-body diagrams of the pins.

• Typically, the entire truss is treated as a rigid body to determine the reactions

at the supports.

The “Method of Joints” solution is good if the entire truss is to be analyzed (i.e. all

the forces in each member is required).

• The solution must begin where there are only two unknown forces, usually at one

of the supports.

6.5

Example – Method of Joints

Given: The truss shown.

Find: Force in each member.

First, find the reactions at the supports.

∑ Fx = 0 = Cx Cx = 0

∑ MC = 0 = 1,000 (12) + 2,000 (24) – 6 Ey

6Ey = 1,000 (12) + 2,000 (24) = 12,000 + 48,000 = 60,000

Ey = + 10,000 lb The direction that was assumed is correct.

Ey = 10,000 lb ↑

∑ Fy = 0 = - 2000 – 1000 + Cy + Ey

Cy = 3000 – Ey = 3,000 – 10,000

Cy = - 7,000 lb The direction that was assumed is not correct.

Cy = 7,000 lb ↓

Next, draw a free-body diagram for each of the joints and solve for the forces.

• Assume tension (pulling on the joint) – tension is “good.”

For Joint A:

∑ Fy = 0 = - 2000 – (4/5) AD

AD = - (5/4) 2000 = - 2500 lb

AD = 2500 lb (C)

∑ Fx = 0 = AB + (3/5) AD

AB = - (3/5) AD = - (3/5) (- 2500)

AB = + 1500 lb

AB = 1500 lb (T)

6.6

For Joint D:

∑ Fy = 0 = (4/5) BD – (4/5) 2500

BD = + 2500 lb

BD = 2500 lb (T)

∑ Fx = 0 = (3/5) 2500 + (3/5) BD + DE

DE = - (3/5) 2500 - (3/5) BD

DE = - (3/5) 2500 - (3/5) (+2500) = - 3000 lb

DE = 3000 lb (C)

For Joint B:

∑ Fy = 0 = - 1000 - (4/5) 2500 – (4/5) BE

(4/5) BE = - 1000 - 2000

BE = - 3750 lb

BE = 3750 lb (C)

∑ Fx = 0 = - 1500 + BC - (3/5) 2500 + (3/5) BE

BC = + 1500 + (3/5) 2500 – (3/5) (- 3750)

BC = + 5250 lb

BC = 5250 lb (T)

For Joint E:

∑ Fy = 0 = - (4/5) 3750 + (4/5) CE + 10,000

(4/5) CE = (4/5) 3750 – 10,000 = - 7,000

CE = - 8,750 lb

CE = 8,750 lb (C)

Check (using Joint E):

∑ Fx = 0 = (3/5) 3750 + 3000 + (3/5) CE

0 = (3/5) 3750 + 3000 + (3/5) (- 8750)

0 = 2250 + 3000 – 5250 = 0 OK

Also, check Joint C:

∑ Fx = 0 = - 5250 + ( 3/5 ) 8750 = 0 OK

∑ Fy = 0 = - 7000 + ( 4/5 ) 8750 = 0 OK

6.7

6.3 Zero-Force Members

Using Joint C:

∑ Fy = 0 = (4/5) BC and BC = 0

In order to determine zero force members in a plane truss:

1. Examine the unloaded joints.

2. If a joint is unloaded, determine if there are more than three members framing

into the joint.

3. For three members framing into one joint, if two of the three members are

collinear, then the force in the third member is zero.

Zero force members are not useless.

1. These members may carry loads when the loading conditions change.

2. These members are needed to support the weight of the truss.

3. These members help to maintain the truss in the desired shape.

6.8

6.4 The Method of Sections

If the force in only one member or if the forces in only a few members are

desired, the “Method of Sections” is a more efficient method of solution.

• In practice, the portion of the truss to be analyzed is obtained by “passing a

section” through three members of the truss, at least one of which is the

desired member.

• The “section” is a line that is drawn which divides the truss into two completely

separate parts, but does not intersect more than three members if possible.

• Then, if the entire truss is in equilibrium, then any portion of the truss is in

equilibrium also.

Consider the truss shown at the right.

• To find the internal forces in

members BD, BE, and CE, make a cut

through these members.

Assuming tension in these members:

Use ∑ ME = 0 to find BD.

Use ∑ Fy = 0 to find BE.

Use ∑ MB = 0 to find CE.

6.9

Example – Method of Sections

Given: The truss shown.

Find: Force in members BC, BK, LK.

Find the reaction at the left support.

∑ MG = 0 = - Ay (180) + 2 (150) + 4 (120) + 4 (90) + 4 (60) + 2 (30)

180 Ay = 1440.0 Ay = 8.0 k ↑

For the portion of the truss left of the cut, find the forces in members BC, BK,

and LK.

∑ MB = 0 = - 8 (30) + 40 LK

LK = + 6.0 k LK = 6.0 k (T)

∑ MK = 0 = - (3/ 10 ) BC (40) – (1/ 10 ) BC (30) + 2 (30) – 8 (60)

= - 37.95 BC – 9.49 BC – 420

47.44 BC = - 420

BC = - 8.85 BC = 8.85 k (C)

∑ MO = 0 = - (3/5) BK (40) – (4/5) BK (120) - 2 (120) + 8 (90)

= - 24 BK – 96 BK + 480

120 BK = 480

BK = + 4.0 k BK = 4.0 k (T)

Check: ∑ Fx = 0 = (3/ 10 ) BC + (3/5) BK + LK

= (3/ 10 ) (- 8.85) + (3/5) (4.0) + 6.0

= - 8.40 + 2.40 + 6.0 = 0 OK

6.10

Example – Method of Sections (K truss)

Given: The truss shown.

Find: Force in upper chord EJ, lower chord GH, and diagonals FJ & FH.

Find the reactions at the supports.

∑ MV = 0 = - 80 Ay + 8 (60) + 8 (50)

Ay = + 11.0 k

Ay = 11.0 k ↑ (as assumed)

Certain trusses will require more than a single cut.

• For the K-truss, a vertical cut crosses four members; thus, there are four

unknowns.

• But using a different cut, forces in a couple of the required members may be

found.

For Cut 1:

∑ MG = 0 = - 11 (20) – EJ (20)

EJ = - 11.0 k

EJ = 11.0 k (C)

∑ ME = 0 = GH (20) - 11 (20)

GH = + 11.0 k

GH = 11.0 k (T)

6.11

For Cut 2:

∑ Fy = 0 = 11.0 – 8.0 + ( 1/ 2 ) FJ – ( 1/ 2 ) FH

0 = 3.0 + ( 1/ 2 ) FJ - ( 1/ 2 ) FH

∑ Fx = 0 = ( 1/ 2 ) FJ + ( 1/ 2 ) FH – 11.0 + 11.0

0 = ( 1/ 2 ) FJ + ( 1/ 2 ) FH FJ = - FH

Using the first equation, substitute “- FH” for “FJ” and solve for FH.

∑ Fy = 0 = 11.0 – 8.0 + ( 1/ 2 ) FJ – ( 1/ 2 ) FH

0 = 11.0 – 8.0 + ( 1/ 2 ) (- FH) – ( 1/ 2 ) FH

0 = 3.0 – (2/ 2 ) FH

FH = 3.0 ( 2 /2) = + 2.12 FH = 2.12 k (T)

FJ = - FH = - 2.12 k FJ = 2.12 k (C)

6.12

6.5 Space Trusses

A “space truss” consists of straight members joined together at their extremities

to form a three-dimensional (3D) configuration.

• The most elementary (or “basic”) rigid

space truss consists of six members to

form a tetrahedron.

• Starting from the “basic space truss,” a

“simple space truss” is obtained by adding

three new members, all joined at one end

at a new joint, and joined to three

existing joints.

In a simple space truss, the number of members and the number of joints are

related by the following equation.

m = 3 n – 6

where

m = the number of members

n = the number of joints

Assumptions for Design

The members of a space truss may be treated as two-force members provided the

external loading is applied at the joints and the joints consist of ball-and-socket

connections.

The conditions of equilibrium for each joint are expressed by the following three

equations.

∑ Fx = 0

∑ Fy = 0

∑ Fz = 0

6.13

6.6 Frames and Machines

A frame is a structure that contains at least one multi-force member; that is, a

member acted upon by three or more forces.

• Note the contrast to truss members that consist of all two-force members.

Analyzing a Frame

Consider the crane shown.

Using the three equations of equilibrium,

we can determine the tension T in the

cable, and the components of the

reaction at A.

∑ MA = 0 yields T

∑ Fx = 0 yields Ax

∑ Fy = 0 yields Ay

Free-Body Diagrams

In order to determine the internal forces holding the various parts of the frame

together, we must “dismember” the frame and draw a free-body diagram for each

part.

Points B, C, and E are pin connections and, therefore, are replaced by horizontal

and vertical restraining forces.

• As we assign directions to the components of the reactions, Newton’s third law

(equal and opposite forces) must be satisfied.

6.14

Check for determinacy.

• There are 3 equations for each free-body diagram.

• There are 4 free-body diagrams; thus, we can solve for as many as 12 unknowns.

In this problem there are only 9 unknowns.

• The two components for each of the forces at pins B, C, and E.

• The two components of the reaction at A.

• The tension in the cable.

Equations of Equilibrium

Using the equations of equilibrium, the solution follows as outlined below.

• Using the entire structure as a free-body diagram, solve for the components of

the reaction at A (that is, Ax and Ay) and the tension T in the cable.

• Then using a free-body diagram of member ABCD:

∑ MB = 0 yields Cx

∑ MC = 0 yields Bx

• Using a free-body diagram of member CEF:

∑ MC = 0 yields Ey

∑ ME = 0 yields Cy

• Using a free-body diagram of member BE:

∑ Fx = 0 yields Ex

∑ Fy = 0 yields By

Frames Which Cease to be Rigid When Detached from Their Supports

The reactions cannot be completely determined from the free-body diagram of the

entire frame.

• Thus, we must dismember the frame even to find the external forces.

• Equilibrium equations are said to be “necessary, but not sufficient” for a non-

rigid structure.

6.15

Example – Frame Analysis

Given: Frame shown.

Find: Components of the forces acting

on each member of the frame.

Find the reactions at the supports.

∑ MA = 0 = - 360 (15) – 240 (33) + 12 Ex

12 Ex = 5400 + 7920 = 13,320 Ex = 1110 lb →

∑ Fx = 0 = Ax + Ex

Ax = - Ex = - 1110 lb Ax = 1110 lb ←

∑ Fy = 0 = Ay – 360 - 240

Ay = + 600 Ay = 600 lb ↑

Find the components of the forces acting on each member.

FBD 1

∑ MB = 0 = - 600 (6) + 18 Dy

18 Dy = + 3600

Dy = + 200

Dy = 200 lb ↑ on ABD

∑ MD = 0 = - 600 (24) - 18 By

18 By = - 14,400

By = - 800

By = 800 lb ↓ on ABD

6.16

FBD 2

∑ MC = 0 = - 12 Bx + 18 By + 360 (9) – 240 (9)

12 Bx = 18 (- 800) + 360 (9) – 240 (9)

= - 14,400 + 3240 – 2160 = - 13,320

Bx = - 1110 Bx = 1110 lb ← on ABD

∑ Fx = 0 = - Bx + Cx

Cx = Bx = - 1110 Cx = 1110 lb ← on BC

∑ Fy = 0 = - By – 360 + Cy - 240

Cy = By + 360 + 240

= - 800 + 360 + 240 = - 200 Cy = 200 lb ↓ on BC

FBD 3

∑ MC = 0 = 1110 (24) – 12 Dx Dx = 2220 lb ← on CDE

6.17

Machines

Machines are structures designed to transmit or modify forces.

• The solution involving forces on a machine is similar to that for a frame.

• The solution will generally involve the use of one or more free-body diagrams.

• The free-body diagrams should be chosen to include the input forces and the

reactions to the input forces.