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Pin-Jointed Frame Structures

(Frameworks)

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Pin Jointed Frame Structures (Frameworks)

A pin-jointed frame is a structure constructed from a number of straight members connected together

at their ends by frictionless pinned joints. Since the joints are free to rotate they cannot transmit torque

from one to another, they are therefore only subjected to axial loading, which may be tensile or

compressive. A member in tension is called a tie, while one in compression is a strut.

Tie in Tension Strut in Compression

The arrow head within the members, represent the equal and opposite forces acting in the member.

An example of a typical framework is given below:

Such a frame is said to be statically determinate, that is, capable of being solved using equations of

equilibrium. When this is not the case it would be referred to as statically indeterminate. We shall only

be considering the former.

Problem of this type can be solved either graphically or analytically. We will consider both methods.

Graphical Method

Using this method of solution, we make use of Bow’s notation which is use to describe the framework

and the forces within each member. We first place capital letter between all of the external forces and

then internally between the member which make up the framework. You need to work around the frame

in a clockwise direction and from left to right across it, similar to that shown below. Where you start is

unimportant.

A

D B

C

The load on the top right-hand corner of the framework would be referred to as force AB and its sense

which is downwards would be ab as we go clockwise from A to B.

Commencing at a joint at which there are no more than two unknown forces, we construct a force

polygon for all internal and external forces acting at that joint. Proceed systematically though the frame,

constructing a force polygon at each joint, to produce one composite force diagram. The magnitude and

direction of each unknown force can then be scaled of the diagram.

Typical framework with

loading and supports

E

F

G

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This method has largely been superseded by modern computer methods but is still useful with complex

frames. (See later for a worked example.)

Analytical Methods

We will consider two analytical methods, namely; Method of Resolution at Joint and Method of Section.

Method of Resolution at Joint

This method is best suited when forces in all members are required. At any joint in a plane frame at

which there are no more than two unknown forces, write down two equations of equilibrium by

resolving forces at the joint in two mutually perpendicular directions. Solve these equations for the two

unknown forces. Proceed systematically through the frame until all forces are known. External support

reactions are usually determined first by the use of moment equilibrium. (See worked example later.)

Method of Section This method is useful if you only need to determine the forces in one or a few members of the

framework. Cut the frame, as shown in the example below by a section through the member under

consideration and no more than two other members in which the member forces are unknown. Both

parts of the structure can then be treated as structures in equilibrium and either part can then be solved

by resolving forces or by taking moments of force about a suitable point. The choice as to whether to

resolve or to take moments will depend upon the geometry of the frame and will be illustrated in a

worked example that follows shortly.

Frame cut for solution using Method of Section

Let’s use the following example to illustrate all of the above three methods. Note that Bow’s notation

has been used to label the frame.

50kN A 30 kN

D B

C

R1 R2

Graphical Solution

1. Calculate the support reactions

Taking moments about R1 and assuming clockwise moments to be positive

∑M = 0 = (50 x 1.5) + (30 x 4.5) – 6R2

E

F

G All angles are 60o and

sides are 3m

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This gives R2 = 35 kN

Equating vertical forces will give R1 (assume forces up to be positive)

∑FV = 0 = R1 +35 - 50 – 30

This gives R1 = 45 kN

2. Draw the force diagram to scale. Start with the external forces which should all be in a straight

line.

d

e g c

f a

Force Diagram (NTS)

b

3. Scale off the magnitude of all unknown forces.

GB = -40

FG = +5.7

EF = -5.7

DE = -52

CE = +26

CG = +20

AF = -23

4. Determine whether forces are compressive or tensile using Bow’s notation – will be explained

in class.

50kN A 30 kN

D B

C

R1 R2

Accuracy of answers depends upon scale of diagram and

how well you draw it. Note the + and – signs have come

from the next stage.

E

F

G Diagram shows nature of forces in

each member.

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Method of Resolution at Joint

We already have values for the support reactions otherwise calculate them at this point.

Next consider each joint in turn starting with one where at least one force is known.

We will begin by considering joint BCG

BG

CG

R2

Resolving vertical forces

∑FV = 0 = R2 + BGCos30 (We are assuming forces up are positive)

Therefore 0 = 35 +BGCos30

This gives BG = -40.41 kN

Resolving Horizontal forces (We are assuming that forces left to right are positive)

∑FH = 0 = -CG – BGSin30

Therefore 0 = -CG – (-40.41Sin30)

Hence CG = 40.41Sin30 = 20.21 kN

Consider next joint ABGF

AB

AF

FG

BG

Resolving vertical forces

∑FV = 0 = -AB – BGCos30 – FGSin60

0 = -30 – (-40.41Cos30) –FGSin60

This gives FG = 5.77 kN

Resolving horizontal forces

∑FH = 0 = -AF –FGCos60 + BGSin30

Notice that we are assuming that all

internal forces are tensile. Actual values

are then substituted. A negative result

implies a force is compressive.

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0 = -AF – 5.77Cos60 + (-40.41Sin30)

This gives AF = -23.09 kN

The remainder of the problem is carried out in a similar fashion. You can complete the rest as a tutorial

problem.

Method of Section

1. Calculate the support reactions – already done.

2. Let’s assume we require the forces in members AF, FG, and CG. We therefore cut the frame as

shown below.

A

D B

C

3. Draw the FBD

A 30 kN

F

B

G

C

35 kN

4. Calculate the forces

It should be obvious from the free body diagram that we can determine FG by equating vertical

forces.

∑FV = 0 = -30 + 35 – FGSin60 (taking forces up as positive)

This gives FG = 5.77 kN

Again by careful examination of the FBD we see that by taking moments about AB we should be

able to determine CG.

∑MAB = 0 = (-35 x 1.5) + CG x 3Cos30 (taking clockwise moments as positive)

This gives CG = 20.21 kN

Next by equating horizontal forces we should be able to determine AF

∑FH = 0 = -AF – FGCos60 – CG

0 = -AF – 5.77Cos60 – 20.21

E

F

G

Again the cut members are assumed

to be in tension.

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This gives AF = - 23.09 kN

Summary of the Procedure of Method of Section

1. Calculate the reactions of the frame using a FBD of the whole frame.

2. Isolate a portion of the frame by passing a cutting plane through the frame, cutting no more

than three members in which the forces are unknown (unless all but one of the lines of action

of the cut member intersect at a point).

3. Apply the three equations of equilibrium (∑M = 0, ∑FH = 0, and ∑FV = 0) to the isolated portion

of the frame and solve for the unknowns.

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Tutorial Problems

1. For the framework given in below determine the magnitude of the force in each member, using

the graphical method, and state whether each is a tie or a strut. (Note that it is not usual to label

the framework for you.)

2kN 3kN

A

All members are 1m

E B

D C

R1 1kN R2

[R1= 2.75kN; R2= 3.25kN - All of the following have been found by measurement - members:

hb = 3.75kN (strut), ch = 1.9kN (tie), fd = 1.6kN (tie), ef = 3.2kN (strut), ag = 2kN

(strut), fg = 0.85kN (tie), gh = 0.3kN (tie)]

2. Use the method of resolution at joints to verify the forces in the members that make up the top

right hand joint of the framework shown on question1 above.

(hb = -3.75kN hg = 0.286 kN ga = -2.018 kN)

3. For the pin jointed frame shown below solve all the forces and reactions in the frame stating

which members are tie and which are struts. The length of all vertical and horizontal members

are 3m.

A

D C B

15kN 30kN

[Reactions: R1 = 20kN; R2 = 25kN; members: ha = 35.35kN (strut), bh = 25kN (tie), ed = 20kN

(tie), ea = 29kN (strut), cg = 25kN (tie), gf = 7.07kN (strut), fa = 20kN (strut), ef = 20kN (tie),

gh = 30kN (tie)]

4. The Warren truss shown below is composed of similar members all of which are 3 m long.

Determine the forces in all members due to a vertical load of 90 kN at G. Use the method of

resolution at joints to solve this problem.

E

F

G H

F

G

H

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(Reactions: RE = 30; RA = 60 kN; members: DE = -34.64; FE = 17.32; DF = 34.64; CD = -

34.64; CF = -34.64; FG = 51.96; CG = 34.64; BC = -69.28; BG = 69.28; AG = 34.64; AB = -

69.28 kN)

5. Structure shown, is a truss which is pinned to the floor at point A, and supported by a roller at

point D. Determine the forces in all members of the truss. It is suggested that you use the method

of joints to solve the problem.

Note: Start by calculating reaction at D which you should find is 12 kN.

(FAB = 8.73 kN tensile, FAG = 21.82 kN compressive, FBC = 15.71 kN tensile, FBF = 8.73 kN

compressive, FBG = 8.73 kN tensile, FCD = 5.24 kN tensile, FCE =13.09 kN tensile, FCF = 13.09

kN compressive, FDE = 13.9 kN compressive, FEF = 0.48 kN compressive, FFG = 12.22 kN

compressive)

6. Using the method of section, determine the forces in members BD, CD, and CE of the roof truss

shown below.

(BD = -160 kN; CD = -200 kN; CE = 320 kN)

Note: You need to determine reaction at A which you should fine is 120 kN

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7. Determine the forces, and state whether tensile of compressive, in all members of the

framework shown below. It is suggested you use Method of Joints.

(FAB = 5.56 kN tensile, FAE = 75.56 kN tensile, FBC = 4.45 kN tensile, FBE = 3.34 kN

compressive, FCD = 88.87 kN compressive, FCE = 5.57 kN tensile, FCF = 50 kN tensile, FDF =

71.11 kN tensile, FFE = 71.11 kN tensile)

8. The framework shown below is supported by a roller at C and hinge at G. By the method of

section determine the forces in BC, DF, and CE.

(FBC=44.72 kN compressive, FCE=10 kN tensile, FDF=80 kN compressive)

9. Determine the force in members DF, DG, and EG of the Howe truss shown below.

(DF = -280 kN, DG = -150 kN, EG = 400 kN)

10. For the framework shown below determine the forces in members BC, CF and FE, stating

whether they are tensile or compressive.

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G F E

1.2 m

A B C D 1.2 m 1.2 m 1.2 m

2.7 kN 2.7 kN 3.6 kN

[FBC = 9.9 kN (C), FFC =8.91 kN (T), FFE = 3.6 kN (T)]

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