In the civil engineering field, most analysis of forces are done on structures. These analyses translate to different kinds of concepts and assumptions.

Here are the general types of structures:

1. Beam – a simple member subjected to transverse loadings and should be properly connected.

P

W

2. Truss – composed of members/

bars assumed to be connected

by frictionless pins; loads are

applied at nodes; members form

triangular divisions.

3. Frames – composed of members

that are connected by rigid joints;

passing a cutting plane through

any point exposes internal shear,

axial and moment at that section.

Trusses are used for two general purposes: (1) as roof trusses and (2) bridge trusses.

For truss analysis, the following assumptions govern:

pin-connected members

Joint/nodal loads only

These assumptions simplify the analysis for the truss. Each truss member may also be called a two-force member because it only resists vertical and horizontal forces, not moments.

A simple truss is a planar truss which begins with a triangular element and can be expanded by adding two members and a joint. For these trusses, the number of members (M) and the number of joints (J) are related by the equation M = 2J -3

Member Forces in trusses maybe characterized into two depending on their direction.

If the member force is towards the joint, the member is under compression.

If the member force is away from the joint, the member is under tension.

Remember that the body we are analyzing is the node or joint.

There are three ways to determine the member forces in trusses, namely:

Method of Joints

Method of Sections

Graphical Method or Maxwell Diagram

Method of Joints

STEPS FOR ANALYSIS

1. If the truss’s support reactions are not given, draw a FBD of

the entire truss and determine the support reactions (typically using scalar equations of equilibrium).

2. Draw the free-body diagram of a joint with one or two

unknowns. Assume that all unknown member forces act in

tension (pulling the pin) or compression (pushing the pin)

3. Apply the scalar equations of equilibrium, Fx = O and

Fy = O, to determine the unknown(s). If the answer is

positive, then the assumed direction is correct,

otherwise it is in the opposite direction.

4. Repeat steps 2 and 3 at each joint in succession until all the

required forces are determined.

If a joint has only two non collinear

members and there is no external load

or support reaction at that joint, then

those two members are zero-force

members. In this example members DE,

DC, AF, and AB are zero force members.

You can easily prove these results by

applying the equations of equilibrium to

joints D and A.

Zero-force members can be removed (as

shown in the figure) when analyzing the

truss.

ZERO-FORCE MEMBERS

If three members form a truss joint for which two of the members are collinear and there is no external load or reaction at that joint, then the third non-collinear member is a zero force member.

* Please note that zero-force members are used to increase stability and rigidity of the truss, and to provide support for various different loading conditions.

ZERO-FORCE MEMBERS

Example 1: Determine the forces in the members of the roof

truss shown in the figure below.

30○

30○ 30○

B

A C

D

100 lb

100 lb

Example 2: The cantilever truss shown in the figure below is

hinged at D and E. Find the force in each member.

A

B

C

D

E

30○ 60○ 60○

10 ft 5 ft

1000 lb

1000 lb 1000 lb

Example 3: Determine the each force member in the figure below

using method of joints

A

B

C D 10ft

10ft

2000lb

30ft

5ft

30○

Example 4: Compute the force in each member of the Warren

truss shown in the figure below.

60○ 60○

B

A C

E

4000 lb

D

60○ 60○

10 ft 10 ft

3000 lb 2000 lb

Example 5: Determine the force in each member of the truss

shown in the figure below using method of joints.

B

A C

300 lb

D

9 ft 9 ft

900 lb 300 lb

E

F

G

12 ft

9 ft H

9 ft 9 ft

9 ft

Example 6: Determine the force in each member of the crane

shown in the figure below using method of joints

A

C

B

E

D

40○

12 ft

4200 lb

9 ft

Example 7: In the cantilever truss shown in the figure below,

compute the force in members AB, BE and DE using method of joints.

(Hint: Use rotation of axis)

A

B

C E H

D

G

30○ 30○

30○

30○

60○

1000 lb

1000 lb

1000 lb

1000 lb

F

Example 8: Determine the member forces using method of joints.

10KN

10KN

10KN

10KN

10KN

1.5m 1.5m 1.5m 1.5m 1.5m 1.5m

1.5m

A

B

C

D

E

F

G

H

I

J

K L

Example 9: The bicycle shown below is loaded with weight =

600N, horizontal load=200N at the seat and 100N at the pedals. Solve for the member forces of the truss

forming the bike.

Assume a hinge

support at A

and roller

at C.

Example 10: Three bars, hinged at A and D and pinned at B and C

as shown, form a four-link mechanism. Determine the value of P that will prevent motion.

P 200 lb

A

B

C

D

75○

45○ 45○

60○

Method of Sections

In the method of sections, a truss is divided into two parts by taking an imaginary “cut” (shown here as a-a) through the truss. Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut members will also be either tensile or compressive with the same magnitude. This result is based on the equilibrium principle and Newton’s third law.

1. Decide how you need to “cut” the truss. This is based on: a) where you need to determine forces, and, b) where the total number of unknowns does not exceed three (in general).

2. Decide which side of the cut truss will be easier to work with (minimize the number of reactions you have to find).

3. If required, determine any necessary support reactions by drawing the FBD of the entire truss and applying the E-of-E.

4. Draw the FBD of the selected part of the cut truss. We need to indicate the unknown forces at the cut members. Initially we may assume all the members are in tension, as we did when using the method of joints. Upon solving, if the answer is positive, the member is in tension as per our assumption. If the answer is negative, the member must be in compression. (Please note that you can also assume forces to be either tension or compression by inspection as was done in the figures above.)

5. Apply the scalar equations of equilibrium (E-of-E) to the

selected cut section of the truss to solve for the unknown

member forces. Please note, in most cases it is possible to write one equation to solve for one unknown directly. So look for it and take advantage of such a shortcut!

Example 11: Determine the force member DG of the Howe truss

shown in the figure below using method of sections.

B

A C

2000 lb

D

12ft 12 ft

1200 lb 800 lb

E

F

G 12 ft 12 ft H

9 ft 9 ft

Example 12: Use the method of sections to determine the forces

in member BD, CD, and CE of the Warren Truss.

60○ 60○

B

A C

E

4000 lb

D

60○ 60○

10 ft 10 ft

3000 lb 2000 lb

Example 13: Determine the force in members AB, BD, BE and

DE of the Howe roof truss shown in the figure below using method of joints and method of

sections.

A

B

C

D

F

H E G

30○ 30○ 30○ 30○

600 lb 1000 lb 400 lb

10 ft 10 ft 10 ft 10 ft

Example 14: The Warren truss loaded as shown in the figure is

supported by a roller at C and a hinge at G. By the method of sections, compute the force in the members BC, DF and CE.

A

20 ft

20 ft

G C

B D

E

20 ft 20 ft

1000 lb 400 lb

600 lb F 800 lb

Example 15: For the truss shown in the figure, determine the force BF

by the method of joints and

then check this result using

method of sections.

(Hint: To apply method

of sections, first

obtain the value

of BE by

inspection.)

12 ft

9 ft

12 ft

9 ft

1200 lb 2400 lb

1200 lb

A

B

C

D

E

F

Example 16: In the fink truss shown in the figure, the web

members BC and EF are perpendicular to the inclined members at their midpoints. Use method of sections to determine the force in members DF, DE and CE.

A

B

C

D

E

F

G

1000 lb 1000 lb

2000 lb

2000 lb

2000 lb

10 ft

40 ft

Example 17: Determine the member forces DF, DG, EG, BD, BE and

CE using method of sections.

10KN

10KN

10KN

10KN

10KN

1.5m 1.5m 1.5m 1.5m 1.5m 1.5m

1.5m

A

B

C

D

E

F

G

H

I

J

K L

Example 18: A beam carrying the loads shown in the figure below is

composed of three segments. It is supported by a four vertical reactions and joined by two frictionless hinges. Determine the values of the reactions.

400 lb 200 lb/ft 100 lb/ft

R1 R2 R3 R4

6 ft 4 ft 4 ft 10 ft 3 ft 3 ft

Hinge Hinge

Example 19: For the frame loaded as shown

in the figure below, determine

the horizontal and vertical

components of the pin

pressure at B. Specify

directions (up or down;

left or right) of the force

as it acts upon members.

A

B

D

C

2ft

4ft

300 lb

200 lb

2ft 2ft

Example 20: A beam carrying the loads shown in the figure below is

composed of three segments. It is supported by a four vertical reactions and joined by two frictionless hinges. Determine the values of the reactions.

100N

300N/m

150N/m

R1 R2 R3 R4

3 m 2 m 2 m 5 m 2 m 2 m

Hinge Hinge

Example 21: A billboard BC weighing

1000lb is subjected to a wind pressure of 300 lb per ft as shown in the figure. Neglecting the weights of the supporting members, determine the components of the hinge forces at A and F.

A

B

C

D

E

F

6 ft 6 ft

4 ft

4 ft

4 ft

2 ft

300

lb

/ft

Example 22: The frame shown in

the figure is supported by a

hinged at E and a roller at

D. Compute the horizontal

and vertical components

of the hinge force at C as

it acts upon BD.

A

E

B

C

D

240 lb

4ft 2ft

8ft

45○ 45○

Example 23: For the frame shown in the figure is supported by a

hinge at A and a roller at E.

Compute the horizontal and

vertical components of the

hinge forces at B and C as they act upon

member AC.

A E

B

C

6ft

4ft

240 lb

2ft 2ft

D

6ft 1ft

Example 24: The bridge shown in the figure consists of two end

sections, each weighing 200 tons with center gravity at G, hinged to a uniform center span weighing 120 tons. Compute the reactions at A,B,E and F.

30 ft 30 ft 20 ft 20 ft 20 ft 20 ft 60 ft

40 ft

40 ft 20 ft 60 tons

G G

A B

C D

E F

Example 25: The frame shown in the figure is hinged at E and

roller supported at A. Determine horizontal and vertical components of the hinge forces at B,C, and D. Neglect the weights of the members

Example 26: Determine the member forces using graphical

method or Maxwell diagram.

10KN

10KN

10KN

10KN

10KN

1.5m 1.5m 1.5m 1.5m 1.5m 1.5m

1.5m

A

B

C

D

E

F

G

H

I

J

K L