# Analysis of Pin-Jointed Plane Trusses

A structure made up of several bars (or members) riveted or welded together is known as truss is composed of such members which are just sufficient to keep the truss in equilibrium, when the truss is supporting an external load, then the frame is known as perfect truss. Though in actual practice the members are welded or riveted together at their joints, yet for calculation purposes the joints are assumed to be hinged or pin-joined. A plane truss is a truss which carries applied loads in a plane. The basic plane truss consists of three straight members which are connected by pins at their ends to form a triangle.

2.12.1. Types of Frames or Trusses. The different types of frames or trusses are:

i.            Perfect frame, and
ii.            Imperfect frame.

Imperfect frame may be a deficient frame or a redundant frame.

1.     Perfect Frame. The frame which is composed of such members, which are just sufficient to keep the frame in equilibrium, when the frame is supporting an external load, is known as perfect frame, The simplest perfect frame is a triangle as shown in Fig. 2.55 which consists three members and three joints. The three members are: AB, BC and AC whereas the three joints are A, B and C. This frame can be easily analysed by the condition of equilibrium.

Let the two members CD and BD and a joint D are added to the triangular frame ABC. Now, we get a frame ABCD as shown in Fig.2.56 (a). This frame can also be analysed by the conditions of equilibrium. This frame is also known as perfect frame.

* The couple at C is 4 kN m clockwise. Hence its sense is positive.

** The couple at D is 8 kN m anti-clockwise. Hence its sense is negative. The couple is also moment.

Suppose we add a set of two members and a joint again, we get a perfect frame as shown in Fig.2.56 (b). Hence for a perfect frame, the number of joints and number of members are given by,

n = 2j – 3

where n = Number of members, and

j = Number of joits.

Fig. 2.56 (a)

Fig. 2.56 (b)

2.     Imperfect Frame. A frame in which number of members and number of joints are not given by

n = 2j – 3

is known, an imperfect frame. This means that number of members in an imperfect frame will be either more or less than (2j – 3).

(i)  If the number of members in a frame are less than (2j – 3), then the frame is known deficient frame.

(ii)  If the number of members in a frame are more than (2j – 3), then the frame is known as redundant frame.

2.12.2. Assumptions Made in Finding Out the Forces in a Frame. The assumptions made in finding out the forces in a frame are:

i.  The frame is a perfect frame
ii. The frame carries load at the joints
iii. All the members are pin-joined.

2.12.3. Reactions of Supports of a Frame. The frames are generally supported

i.on a roller support or
ii.on a hinged support.

If the frame is supported on a roller support, then the line of action of the reaction will be at right angles to the roller base as shown in Figs. 2.57 and 2.58.

If the frame is supported on a hinged support, then the line of action of the reaction will depend upon the load system on the frame.

The reaction at the support of a frame are determined by the conditions of equilibrium frame and the reactions at the supports must form a system of equilibrium.

1.12.4.   Analysis of a Frame or Truss. Analysis of a frame consists of:

(i)  Determinations of the reactions at the supports and

(ii) Determinations of the reactions at the supports and

The reactions are determined by the condition that the applied load system and the
induced reactions at the supports form a system in equilibrium.

The forces in the members of the frame are determined by the condition that every joint
should be in equilibrium and so, the forces acting at every joint should form a system in
equilibrium

A frame is analysed by the following methods:

(i)  Method of joints,

(ii  Method of sections, and

(iii) Graphical method.

1.12.5.    Method of Joints. In this method, after determining the reactions at the supports, the equilibrium of every joint is considered. This means the sum of all the vertical forces as well as the horizontal forces acting on a joint is equated to zero. The joint should be selected in such a way that at any time there are only two members, in which the forces are unknown. The force in the member will be compressive if the member pushes the joint to which it is connected whereas the force in the member will be tensile if the member pulls the joint to which it is connected.

Problem 2.31. Find the forces in the members AB, AC and BC of the truss shown in Fig. 2.59

Sol. First determine the reactions RB and RC. The line of action of load of 20 kN acting at A is vertical. This load is at a distance of AB × cos 60° from the point B. Now let us find the distance AB.

The triangle ABC is a right-angled triangle with angle BAC = 90°. Hence AB will be equal to BC × cos 60°.

AB = 5 × cos 60° = 5 × 1/2 = 2.5 m

Now the distance of line of action of 20 kN from B is

AB × cos 60° or 2.5 × 1/2 = 1.25 m.

Resolving the forces acting on the joint B, vertically

F1 sin 60° = 15

Taking the moments about B, we get

RC × 5 = 20 × 1.25 = 25

As F1 is pushing the joint B, hence this force will be compressive. Now resolving the forces horizontally, we get

F2 = F1 cos 60° = 1.7.32 × 1/2 = 8.66 kN (tensile)

As F2 is pulling the joint B, hence this force will be tensile.

Joint C

Let F3 = Force in the member AC
F2 = Force in the member BC

The force F2 has already been calculated in magnitude and direction. We have seen that force F2 is tensile and hence it will pull the joint C. Hence it must act away from the joint C as shown in Fig. 2.59 (b).

and RC = Total load -RC = 20 – 5 = 15 kN
Now let us consider the equilibrium of the various joints.

Joint B

Let F1 = Force in member AB

F2 = Force in member BC

Let the force F1 is acting towards the joint B and the force F2 is acting away* from the joint B as shown in Fig. 2.59 (a). (The reaction RB is acting vertically up. The force F2 is horizontal. The reaction RB will be balanced by the vertical component of F1. The vertical component of F1 must act downwards to balance RB. Hence F1 must act towards the joint B so that its vertical component is downward. Now the horizontal component of F1 towards the joint B. Hence force F2 must act away from the joint to balance the horizontal component of F1):

Resolving forces vertically, we get

F3 sin 30° = 5 kN

As the force F3 is pushing the joint C, hence it will be compressive.   Ans.

*The direction of F2 can also be taken towards the joint B. Actually when we consider the equilibrium of the joint B, if the magnitude of F1 andF2 comes out to be positive then the assumed direction of F1 and F2 are correct. But if anyone of them is having a negative magnitude then the assumed direction of that force is wrong. Correct direction then will be the reverse of the assumed direction.

Problem 2.32. A truss of span 7.5 m carries a point load of 1 kN at joint D as shown in Fig.2.60. Find the reactions and forces in the members of the truss.

Sol. Let us first determine the reactions RA and RB.

Taking moments about A, we get RB × 7.5 = 5 × 1

RA = Total load – RB

∴ = 1 – 0.667 = 0.333 kN

Now consider the equilibrium of the various joints.

Joint A

Let F1 = Force in member AC
F2 = Force in member AD.

Let the force F1 is acting towards the joint A and F2 is acting away from the joint A as shown in Fig. 2.60 (a).

Resolving the forces vertically, we get

F1 sin 30° = RA

= 0.666 kN (Compressive)

Resolving the forces horizontally, we get

F2 = F1 × cos 30°

= 0.666 x 0.866 = 0.5767 kN (Tensile)

Joint B

Let F4 = Force in member BC
F
5 = Force in member BD

Let the direction of F4 and F5 are assumed as shown in Fig. 2.60 (b).

Resolving the forces vertically, we get

F4 sin 30° = RB = 0.667

Resolving the forces horizontally, we get

F5 = F4 cos 30° = 1.334 × 0.866 = 1.155 kN (Tensile)

Joint D

Let F3 = Force in member CD. The forces F2 and F5 have been already calculated in magnitude and direction. The forces F2 and F5 are tensile and hence they will be pulling the joint D as shown in Fig. 2.60 (c). Let the direction* of F3 is assumed as shown in Fig. 2.60 (c).

Resolving the forces vertically, we get
F3  sin 60° = 1

= 1.1547 kN (Tensile)

Hence the forces in the members are:

F1 = 0.666 kN (Compressive)
F2 = 0.5767 kN (Tensile)

F3 = 1.1547 kN (Tensile)

F4 = 1.334 kN (Compressive)

F5 = 1.155 kN,(Tensile). Ans

Problem 2.33. A truss of span 5 m is loaded as shown in Fig. 2.61. Find the reactions and forces in the members of the truss.

Sol. Let us first determine the reactions RA and RB

Triangle ABD is a right-angled triangle having angle ADB = 90°.

AD = AB cos 60° = 5 × 0.5 = 2.5 m

The distance of the line of action of the vertical load 10 kN from point A will be AD
cos 60°

or 2.5 × 0.5 = 1.25 m.

From triangle ACD, we have AC =AD = 2.5 m

∴ BC = 5 – 2.5 = 2.5 m

In right-angled triangle CEB, we have

*The horizontal force F5 is more than F2 Hence the horizontal component of F3 must be in the direction of F2. This is only possible if F3 is acting away from D.

The distance of the line of action of the load of 12 kN from point A will be (BE × cos )

The distance of the line of action of the load of 12 kN from point A will be (5-1.875)=3.125 m.

Now taking the moments about A, we get

RB × 5 = 10 × 1.25 + 12 × 3.125 = 50

RA = Total load -RB = (10 + 12) -10 = 12 kN

Now consider the equilibrium of the various joints.

Joint A

F1 = Force in member AD, and
F2 = Force in member AC

Let the directions of F1 and F2 are assumed as shown in Fig. 2.61 (a)

Resolving the forces vertically,

F1 sin 60° = 12

= 13.856 kN (Compressive)

Resolving the forces horizontally,

F2 = F1 cos 60° = 13.856 × 0.5
= 6.928 kN (Tensile)

Now consider the joint B .

Joint B

Let F3 = Force in member BE, and
F4 = Force in member BC

Let the directions of F3 and F4 are assumed as shown in Fig-. 2.61 (b).

Resolving the forces vertically, we get

F3 sin 30° = 10

Now resolving the forces horizontally, we get

F4 = F3 cos 30o = 20 × 0.866

= 17.32 kN (Tensile)
Now consider the joint C.

Joint C

Let F5 = Force in member CE
F6 = Force in member CD

Let the directions of F5 and F6 are assumed as shown in Fig.2.61 (c).

The forces F2 and F4 are already known in magnitude and directions. They are tensile and hence will be pulling the joint C as shown in Fig. 2.61 (c).

Resolving forces vertically, we get

F6 sin 60° + F5 sin 60° = 0

or F6 = -F5
Resolving forces, horizontally, we get

F2 F6 cos 60° = F4 F5 cos 60°

The magnitude of F6 is -ve, hence the assumed direction of F 6 is wrong. The correct direction F6 will be as shown in Fig. 2.61 (d).

F5 = 10.392    (Compressive)

F6 = 10.392    (Tensile)

Now consider the joint E.

Joint E

Let F7 = Force in member ED

Let F7 is acting as shown in Fig. 2.61 (e).

The forces F3 and F5 are known in magnitude and directions. They are compressive hence they will be pushing the joint E as shown in Fig. 2.61 (e).

Resolving the forces along BED, we get

F7 + 12 cos 60° = F3

F7 = F3 – 12 × 0.5

= 20 – 6 = 14 kN (Compressive)

As F7 is positive hence the assumed direction of F7 is correct. Ans.

Problem 2.34. A truss of span 9 m is loaded as shown in Fig. 2.62. Find the reaction and forces in the member of the truss.

Sol. Let us first calculate the reactions RA and RB.

Taking moments about A, we get

RA × 9 = 9 × 3 + 12 × 6 = 27 + 72 = 99

and RA = Total load – RB = (9 + 12) -11 = 10 kN

In this problem, there are some members in which force is zero.

These members are obtained directly as given below:

“If three forces act at a joint and two of them are along the same straight line, then for the equilibrium the joint, the third force should be equal to zero.”

1. Three forces are acting at the point A (i.e., RA, FAC and FAG), two of which (i.e., RA, RAC) the same straight line. Hence the third force (i.e., RAG) is zero.

2. Similarly, three forces are acting at the joint B (i.e., RB, FBF and FBH), two of which (i.e., RB and FBH) are along the same straight line. Hence the third force FEH should be zero.

3. At the joint E also, three forces (i.e., FED, FEF and FEH) are acting, two of which (i.e., FED and FEF) are along the same straight line. Hence the third force FEH must be zero.

Now the equilibrium of various joints can be considered.

FAG = Force in member AG = 0
FAG = Force in member AC

= RA = 10 kN (Compressive)
Now consider the equilibrium of joint C.

Joint C [See Fig.2.62 (b)]

Let FCD = Force in member CD
FCG = Force in member CG
FAC = 10 kN (Compressive)

Let the directions of FCG and FCD are assumed as shown in Fig. 2.62 (b).

Resolving forces horizontally, we get

FCG cos θ = 10

Now consider the equilibrium of joint G.

Joint G

The force in member CG is 12.5 kN (Tensile).

Hence at the joint G, this force will be pulling the joint G as shown in Fig. 2.62 (c).

Resolving the forces vertically, we get

12.5 cos θ +FGD = 9

FGD = 9 – 12.5 cos θ

As the magnitude of FGD is negative, hence its assumed direction is wrong. The correct direction will be as shown in Fig. 2.62 (d).

Then, FGD = 1 kN (Compressive)

Resolving the forces horizontally, we get

12.5 sin θ = FGH

= 7.5 kN (Tensile)

Now consider the equilibrium of joint D.

Joint D

The forces in the members CD and GD have been already calculated. They are 7.5 kN and 1 kN respectively. Both are compressive.

Let FDH = Force in member DH, and

FDE = Force in member DE

Resolving the forces vertically, we get

FDH    θ = 1 kN

Resolving the forces horizontally, we get

7.5 + FDH sin θ = FDE

= 7.5 + 0.75 = 8.25 kN (Compressive)

Now consider the equilibrium of joint E.

Joint E

As shown in Fig. 2.62 (f), a joint E three forces are acting. The forces i.e., FDE and FEF are in the same straight line.

Hence force FEH must be zero.

Force in EF, i.e., FEF = FDE

= 8.25 kN (Compressive)

Now consider the joint H.

Joint H

It is already shown that forces in the members EH and BH are zero.

Also the forces in the member GH is 7.5 kN tensile and in the member DH is 1.25 kN tensile.

Let FHF is the force in the member HF.

Resolving forces vertically, we get

1.25 cos θ + FHF cos θ = 12

Now consider the joint B.

Joint B

See Fig. 2.62 (h)

The force in member BF = 11 kN (Compressive)
Now the forces in each member are known.

They are shown in Fig. 2.63. Also these forces are shown below.

 Member Force in member AC 10 kN (Comp.) AG 0 CG 12.5 kN (Tens.) CD 7.5 kN (Comp.) DG 10 kN (Comp.) DE 8.25 kN (Comp.) DH 1.25 kN (Tens.) GH 7.5 kN (Tens.) EH 0 EF 8.25 kN (Comp.) HB 0 HF 13.75 kN (Tens.) BF 11 kN (Comp.)

Problem 2.35. A plane truss is loaded and supported as shown in Fig. 2.64. Determine the nature and magnitude of the forces in the members 1, 2 and 3.

Sol. First calculate the reactions RA and RB

Taking moments about A, we get

RB × 4 = 1 × 1000

Consider the equilibrium of joint A.

Joint A [See Fig. 2.64 (a)]

Resolving the forces vertically,

Resolving the forces horizontally, we get

FAE = FAD cos θ = 1250 × 0.8

= 1000 N (Tensile)

Now consider joint E.

Joint E

Three forces, i.e., FAE, FEF and FED are acting at the joint E. Two of the forces, i.e., FAE and FEF are in the same straight line. Hence the third force, i.e., FED should be zero.

and FEF = FAE = 1000 N (Tensile)

Now consider the equilibrium of joint D.

Joint D

Let F1 = Force in member DG

FDF = Force in member DF

Let us assume their directions as shown in Fig. 2.64 (b).

The forces in the member AD and DE are 1250 N (Compressive) and 0 respectively.

Resolving forces vertically, we get

1250 sin θ + F1 sin θ + FDF sin θ = 1000

or 1250 × 0.6 + F1 × 0.6 + FDF × 0.6 = 1000                                           ∵ sin θ = 0.6)

or F1 + FDF = 1666.66 – 1250 = 416.66                              … (i)

Resolving the forces horizontally, we get

1250 cos θ + F1 cos θ = FDF × cos θ

or 1250 + F1 = FDF or F1 – FDF = -1250                                      … (ii)

Adding (i) and (ii), we get

2F1 = 416.66 – 1250 = – 833.34

Substituting the value of F1 in equation (i), we get

– 416.67 + FDF = 416.66

or FDF = 416.66 + 416.67

= 833.33 N (Comp.)

The magnitude of F1 is negative. Hence its assumed direction is wrong. The correct direction of F1 is shown in Fig. 2.64 (c).

∴ F1 = 416.67 N (Compressive). Ans.

To find the forces F2 and F3 consider the joint F.

Joint F

The forces in the members DF and EF are already known. They are:

FDF = 833.33 N (Compressive)

FEF = 1000 N (Tensile).

These forces are acting at the joint F as shown in

F2 = Force in member FG, and

F3 = Force in member FH

Resolving forces vertically, we get

833.33 sin θ = F2

F2 = 833.33 × 0.6

= 499.998 N = 500 N (Tensile). Ans.

Resolving forces horizontally, we get

F3 + 833.33 cos θ = 1000

F3 = 1000 – 833.33 × 0.8

= 333.336 N (Tensile). Ans.

Method of Joints Applied to Cantilever Trusses. In case of cantilever trusses, it is not necessary to determine the support reactions. The forces in the members of cantilever truss can be obtained by starting the calculations from the free end of the cantilever.

Problem 2.36. Determine the forces in all the members of a cantilever truss shown in Fig. 2.65.

Sol. Here the calculations can be started from end C. Hence consider the equilibrium of the joint C.

Joint C

Let FCD = Force in member CD, and

FCA = Force in member CA.

Their assumed directions are shown in Fig. 2.65,

Resolving the forces vertically, we get

FCD × sin 60o = 1000

Resolving the forces horizontally, we get

FCA = FCD × cos 60o

= 1154.7 × 0.5

=577.35 N (Compressive)

Now consider the equilibrium of the point D.

Joint D

[See Fig. 2.65 (a)]

The force FCD = 1154.7 N (tensile) is already calculated.

FBD = Force in member BD, and

Their assumed directions are shown in Fig.2.65 (a)

Resolving the forces vertically, we get

FAD cos 30o = 1154.7 cos 30o

Resolving the forces horizontally, we get

FBD = FAD sin 30° + FDC sin 30°

= 1154.7 × 0.5 + 1154.7 × 0.5 = 1154.7 N (Tensile)

Now the forces are shown in a tabular form below :

 Member Force in the member Nature of force AC 577.35 N Compressive CD 1154.7 N Tensile AD 1154.7 N Compressive BD 1154.7 N Tensile

Problem 2.37. Determine the forces in all the members of a cantilever truss shown in Fig. 2.66.

Sol. Start the calculations from joint C.

From triangle ACE, we have

Joint C

The direction of forces at the joint C are shown in Fig. 2.66.

Resolving the forces vertically, we get

FCD sin θ = 1000

Resolving the forces horizontally, we get

FCB = FCD × cos θ = 1666.66 × 0.8 = 1333.33 N (Tensile)

Now consider the equilibrium of joint B.

Joint B

Resolving vertically, we get

FBD = 1000 N (Compressive)
FBA = FCB = 1333.33 (Tensile)

Now consider the joint D.

Joint D

The forces in member CD and BD have already been calculated. They are 1666.66 N and 1000 N respectively as shown in Fig. 2.66 (a).

Let FDA = Force in member DA, and

FDE = Force in member DE
Resolving forces vertically, we get

1000 + 1666.66 sin θ = FAD sin θ + FED sin θ

or 1000 + 1666.66 × 0.6 = FAD × 0.6 + FED × 0.6

Resolving forces horizontally, we get

1666.66 cos θ + FAD cos θ = FED cos θ

Adding equations (i) and (ii), we get

2FED = 3333.32 + 1666.66 = 4999.98

Substituting this value in equation (i), we get

∴ FAD = 3333.32 – 2500 = 833.32 N (Tensile)

Now the forces are shown in a tabular form below:

 Member Force in the member Nature of force AB BC CD DE AD BD 1333.33 N 1333.33 N 1666.66 N 2500 N 833.32 N 1000 N Tensile Tensile Compressive Compressive Tensile Compressive

Method of Joints Applied to Trusses Carrying Horizontal Loads. If a truss carries horizontal loads (with or without vertical loads), hinged at one end and supported on rollers at the other end, then the support reaction at the roller supported end will be normal, whereas the support reactions at the hinged end will consists of: (i) horizontal reaction and (ii) vertical reaction.

The horizontal reaction will be obtained by adding algebraically all the horizontal loads; whereas the vertical reaction will be obtained by subtracting the roller support reaction from the total vertical loads. Now the forces in the members of the truss can be determined.

Problem 2.38. Determine the forces in the truss shown in Fig. 2.67 which carries horizontal load of 12 kN and a vertical load of 18 kN.

Sol. The truss is supported on rollers at B and hence the reaction at B must be normal to the roller base i.e., the reaction at B, in this case, should be vertical,

At the end A, the truss is hinged and hence the support reactions at the hinged end A will consists of a horizontal reaction HA and a vertical reaction RA.

Taking moments of all forces at A, we get

RB × 4 = 18 × 2 + 12 × 1.5 = 36 + 18 = 54

Let us first consider the equilibrium of joint A.

Joint A

The reactions RA and HA are known in magnitude and direction. Let the directions of the forces in the members AC and AD are as shown in Fig. 2.67(a).

Resolving the forces vertically, we get

Resolving the forces horizontally, we get

= 12 + 7.5 × 0.8 = 18 kN (Tensile)

Now consider the joint C.

Joint C

At the joint C, the force 0 in member CA and vertical load 18 kN are known in magnitude and directions. For equilibrium of the joint C.

FBC = FCA = 18 kN (Tensile)
FCD = 18 kN (Tensile)

Now consider the joint B.

Joint B

At the joint B, RB and force FBC are known in magnitude and direction.

Let FBD is the force in member BD.
Resolving the forces vertically, we get
FBD × sin θ = RB

Now the forces are shown in a tabular form below :

 Member Force in the member Nature of force AC AD CD CB BD 18 kN 7.5 kN 18 kN 18 kN 2.5 kN Tensile Compressive Tensile Tensile Compressive

Problem 2.39. Determine the forces in the truss shown in. Fig. 2.68 is subjected to horizontal and vertical loads, Mention the nature of forces in each case.

Sol. The truss is supported on rollers at B and hence RB will be vertical. The truss is
hinged at A and hence the support reactions at A will consists of a horizontal reaction HA and a vertical reaction RA.

Taking moment about A, we get

RB × 12 = 8 × 1.5 + 3 × 4 + 6 × 8

= 72

and RA = Total vertical loads – RB
= (3 + 6) – 6

= 3 kN (↑)

and HA = Sum of all horizontal loads

= 8 kN(←)

Now the forces in the members can be calculated. Consider the joint A.

Joint A

The reactions RA and HA are known in magnitude and direction. Let the directions of the forces FCA and FFA are as shown in Fig.2.68 (b).

Resolving the forces vertically, we get

FCA × sin θ = 3 kN

= 5 × 0.8 + 8 = 12 kN   (Tensile)

Now consider joint C

Joint C

The force FCA is known in magnitude and direction. The assumed direction of the forces FCD and FCF are shown in Fig. 2.68 (c).

Resolving forces vertically, we get

FCA sin θ = FCF sin θ

Resolving forces horizontally, we get

FCD = FCA cos θ + FCF cos θ

= 5 × 0.8 + 5 × 0.8 = 8 kN    (Compressive)

Now consider the joint F.

Joint F

The forces FFA and FFC are known in magnitude and directions. The assumed directions of the forces FDF and FGF are shown in Fig. 2.68 (d).

Resolving the forces vertically, we get

5 × sin a + FDF sin θ = 3

Resolving the forces horizontally, we get

12 + 5 cos θ = FGF + FDF cos θ

or 12 + 5 × 0.8 = FGF + 0 or 12 + 4 = FGF

FGF = 12 + 4 = 16 kN (Tensile)

Now consider the joint D.

Joint D

The forces FDC and FFD are known in magnitude and direction. The assumed directions of FDG and FDE are shown in Fig. 2 68 (e).

Resolving vertically, we get

FDG sin θ = FDF × sin θ = 0

∴ FDG = 0

Resolving forces horizontally, we get

FDE = FCD = 8 kN

FDE = 8 kN (Compressive)

Now consider the joint G.

Joint G

The forces FDG and FFG are known in magnitude and direction. The assumed directions of FGE and FGB are shown in Fig. 2.68 (f).

Resolving the forces vertically, we get

FGE sin θ = FDG sin θ + 6 = 6

Resolving forces horizontally, we get

FGB = 16 – FGE cos θ

= 16 – 10 × 0.8 = 8 kN (Tensile)

Now consider the joint E.

Joint E

The forces FGE and FDE are known in magnitude and directions. Let FBE is acting in a direction as shown in Fig. 2.68 (g).

Resolving forces vertically, we get

FGE sin θ = FBE sin θ

FBE = FGE = 10

FBE = 10 kN (Compressive)

If we have calculated the forces in member BE and BG, considering joint B, we would have got the same results.

Now the forces in each member are shown in Fig. 2.68 (h).

Method of Joints Applied to Trusses Carrying Inclined Loads. If a truss carries
inclined loads, hinged at one end and supported on rollers at the other end, then the support reaction at the roller supported end will be normal, whereas the support reactions at the hinged end will consists of:

(i)Horizontal reaction and

(ii)Vertical reaction.

The inclined loads are resolved into horizontal and vertical components.

The horizontal reaction will be obtained by adding algebraically all the horizontal components of the inclined loads; whereas the vertical reaction will be obtained by subtracting the roller support reaction from the total vertical components of the inclined loads. Now the forces in the members of truss can be determined.

Problem 2.40. Determine the forces in the truss shown in Fig. 2.69 which is subjected to inclined loads.

Sol. The truss is supported on roller at B and hence RB will be vertical.

The truss is hinged at A and hence the support reactions at A will consists of a horizontal reaction HA and a vertical reaction RA.

Now length AC = 4 × cos 30 = 4 × 0.866 = 3.464 m

and length AD = 2 × AC = 2 × 3.464 = 6.928 m

Now taking moments about A, we get

RB × 12 = 2 × AC + 1 × AD + 1 × AE

= 2 × 3.464 + 1 × 6.928 + 1 × 4 = 17.856

Total vertical components of inclined loads

= (1 + 2 + 1) × sin 60° + 1.0

= 4 × 0.866 + 1.0 = 4.464 kN

Total horizontal components of inclined loads

= (1 + 2 + 1) cos 60° = 4 × 0.5 = 2 kN

Now RA = Vertical components of inclined loads – RB

= 4.464 – 1.49 = 2.974 kN (↑)

HA = Sum of all horizontal components = 2 kN

Now the forces in the members can be calculated.

Consider the equilibrium of joint A.

Joint A

FAE = Force in member AE

FAC = Force in member AC

Their directions are assumed as shown in Fig. 2.69 (a).

Resolving the forces vertically, we get

FAC × sin 30° + 1 × sin 60° = 2.974

FAC × 0.5 + 0.866 = 2.974

= 4.216 kN (Compressive)

Resolving the forces horizontally, we get

FAE = 2 + FAC cos 30° – 1 × 60°

= 2 + 4.216 × 0.866 – 0.5 = 5.15 kN (Tensile)

Now consider the joint C.

Joint C

From Fig. 2.69 (b), we have

FCD = FAC = 4.216 (Compressive)

and FCE = 2 kN (Compressive)

Now consider joint E.

Joint E

Joint E [See Fig. 2.69 (c)]

Resolving forces vertically, we get

1 + 2 × sin 60° = FED × sin 60°

Resolving forces horizontally, we get

5.15 – 2 × cos 60° – FED cos 600 – FEF = 0

FEF = 5.15 – 1 – 1.57 = 2.58 kN  (Tensile)

At the joint G, two forces, i.e., FBG and FDG are in the same straight line and hence the third force, i.e., FGF should be zero.

FGF = a

Now consider the joint F.

Joint F [See Fig. 2.69 (d)]

Resolving forces vertically, we get

FDF × sin 60° = 0

FDF = 0

Resolving horizontally, we get

FFB = FEF = 2.58 kN

FFB = 2.58 kN (Compressive)

Now consider the joint B.

Joint B

Resolving vertically, we get

FBG × sin 60° = 1.49

The forces are shown in a tabular form as

 Member Force in the member Nature of force AC AE CE CD ED EF DF DG GB FB FG 4.216    kN 5.15    kN 2    kN 4.216    kN 3.155    kN 2.58    kN 0 2.98    kN 2.98    kN 2.58    kN 0 Compressive Tensile Compressive Compressive Tensile Tensile Nil Compressive Compressive Compressive Nil

2.12.6. Method of Sections. When the forces in a few members of a truss are to be determined, then the method of section is mostly used. This method is very quick as it does not involve the solution of other joints of the truss.

In this method, a section line is passed through the members, in which forces are to be determined as shown in Fig. 2.70. The section line should be drawn in such a way that it does not cut more than three members in which the forces are unknown. The part of the truss, on anyone side of the section line, is treated as a free body in equilibrium under the action of external forces on that part and forces in the members cut by the section line. The unknown forces in the members are then determined by using equations of equilibrium as

ƩFx = 0 ƩFy = 0 and ƩM = O.

Fig. 2.70

If the magnitude of the forces, in the members cut by a section line, is positive then the assumed direction is correct. If magnitude of a force is negative, then reverse the direction of that force.

Problem 2.41. Find the forces in the members AB and AC of the truss shown in Fig. 2.71 using method of section.

Sol. First determine the reaction RB and RC

The distance of line of action of 20 kN from point B is

Now draw a section line (1-1), cutting the members AB and BC in which forces are to be determined. Now consider the equilibrium of the left part of the truss. This part is shown in Fig. 2.71 (a).

Let the directions of FBA and FBC are assumed as shown in Fig. 2.71 (a).

Now taking the moments of all the forces acting on the left part about point C, we get

15 × 5 + (FBA × AC)* = 0

The perpendicular distance between the line of action of FBA and point C is equal to AC)

or 75 + FBA × 5 × cos 30° = 0

The negative sign shows that FBA is acting in the opposite direction (i.e., towards points B). Hence force FBA will be a compressive force.

∴ FBA = 17.32 kN (Compressive). Ans.

Again taking the moments of all the forces acting on the left part about point A, we get 15 × Perpendicular distance between the line of action of

15 kN and point C = FBC × Perpendicular distance between FBC and point A.

15 × 2.5 × cos 60° = FBC × 2.5 × sin 60°

= 8.66 kN (Tensile). Ans.

These forces are same as obtained in Problem 2.31.

* The moment of the force FBA about point C, is also taken by resolving the force FBA into vertical and horizontal components at point B. The moment of the horizontal component about C is zero, whereas the moment of vertical component will be (FBA × sin 60°) × 5 = FBA × 5 × sin 60° or FBA × 5 × cos 30°.

Problem 2.42. A truss of span 5 m is loaded as shown in Fig. 2.72. Find the reactions and forces in the members marked 4, 5 and 7 using method of section.

Sol. Let us first determine the reactions RA and RB

Triangle ABD is a right-angled triangle having angle

AD == AB cos 60° = 5 × 0.5 = 2.5 m

The distance of line of action the vertical load 10 kN from point A will be AD cos 60° or 2.5 × 0.5 = 1.25 m.

From triangle ACD, we have

BC = 5 – 2.5 = 2.5 m

In right-angled triangle CEB, we have

Now draw a section line (1-1), cutting the members 4, 5 and 7 in which forces are to be determined. Consider the equilibrium of the right part of the truss (because it is smaller than the left part).

This part is shown in Fig. 2.72 (a). Let F4, F5 and F7 are the forces in members 4, 5 and 7. Let their directions are assumed as shown in Fig. 2.72 (a).

Now taking the moments of all the forces acting on the right part about point E, we get

RB × BE cos 30° = F4 × (BE × sin 30°)

12 × BE cos 30° + F5 × BE = 0

or 12 × cos 30° + F5 = 0

F5 = – 12 × cos 30° = – 10.392 kN

F5 = 10.392 kN (Compressive). Ans.

Now taking the moments about point C of all the forces acting on the right parts, we get

12 × (2.5 -BE cos 30°) = F7 × CE + RB × BC

Negative sign shows that F7 is compressive.

F7 = 14 kN (Compressive). Ans.

These forces are same as obtained in Problem 2.33.

Problem 2.43. A truss of span 9 m is loaded as shown in Fig. 2.73. Find the reactions and forces in the members marked 1, 2 and 3.

Sol. Let us first calculate the reactions RA and RB.

Taking moments about A, we get

RB × 9 = 9 × 3 + 12 × 6 = 27 + 72 = 99

Now draw a section (1.1), cutting the members 1,2 and 3 in which forces are to be determined. Consider the equilibrium of the left part of the truss (because it is smaller than the right part). This part is shown in Fig.2.73 (a). Let F1, F2 and F3 are the forces members 1,2, and 3 respectively. Let their directions are assumed as shown in Fig. 2.73 (a)

Taking moments of all the forces acting on the left part about point D, we get

10 × 3 = F2 × 4

Now taking the moments of all the forces acting on the left part about point G, we get

10 × 3 + F1 × 4 = 0

Negative sign shows that force F1 is compressive.

F1 = 7.5 kN (Compressive) Ans.

Now taking the moments about the point C, we get

F2 × 3 – 9 × 3 + F3 × 4 = 0

or F2 × 3 – 27 + 7.5 × 4 = 0

Negative sign shows that force F2 is compressive.

F2 = 1.0 kN (Compressive). Ans.

Problem 2.44. A truss of 12 m span is loaded as shown in Fig. 2.74. Determine the forces in the members DG, DF and EF, using method of section.

Sol. The truss is supported on rollers at B and hence RB will be vertical. The truss is
hinged at A and hence the support sections at A will consists of a horizontal section HA and a vertical section RA.

In triangle AEC, AC = AE × cos 30°

= 4 × 0.866 = 3.464 m

Now length   AD = 2 × AC = 2 × 3.464 = 6.928 m

Now taking the moments about A, we get

RB × 12 = 2 × AC + 1 × AD + 1 × AE

= 2 × 3.464 + 1 × 6.928 + 1 × 4 = 17.856

Now draw the section line (1-1), passing through members DG, DF and EF in which the forces are to be determined. Consider the equilibrium of the right part of the truss. This part is shown in Fig. 2.74 (a). Let FDG, FFD and FEF are the forces in members DG, FD and EF respectively. Let their directions are assumed as shown in Fig. 2.74 (a). Taking moments of all forces acting on right part about point F, we get

RB × 4 + FDG × FG = 0

or 1.49 × 4 + FDG × (4 × sin 30°) = 0

-ve sign shows that the force FDG is compressive.

FDG = 2.98 kN (Compressive). Ans.

Now taking the moments about point D, we get

RB × BD cos 30° = FFE × BD × sin 30°

RB × cos 30° = FFE × sin 30°

= 2.58 kN (Tensile). Ans.

Now taking the moments of all forces acting on the right part about B, we get

FFD × distance between FFD and B = 0

FFD = 0. Ans.  (distance between FFD and B is not zero)

2.12.7. Graphical Method. The force in a perfect frame can also be determined by a graphical method. The analytical methods (such as method of joints and method of sections) give absolutely correct results, but sometimes it is not possible to get the results from analytical method. Then a graphical method can be used conveniently to get the results. The graphical method also provides reasonable accurate results.

The naming of the various members of a frame are done according to Bow’s notations. According to this notation of force is designated by two capital letters which are written on either side of the line of action of the force. A force with letters A and B on either side of the line of action is shown in Fig. 2.75. This force will be called AB.

The following steps are necessary for obtaining a graphical solution of a frame.

(i)  Making a space diagram

(ii) Constructing a vector diagram

(iii) Preparing a force table.

1.  Making a space diagram. The given truss or frame is drawn accurately according to some linear scale. The loads and support reactions in magnitude and directions are also shown in the frame. Then the various members of the frame arc named according to Bow’s notation. Fig. 2.76 (a) shows a given truss and the forces in the members AB, BC and AC are to be determined. Fig. 2.76 (b) shows the space diagram to same linear scale. The member AB is named as PS and so on.

Fig. 2.76 (a) Given Diagram

Fig. 2.76 (b) Space Diagram

Fig. 2.76 (c) Vector Diagram

2.   Constructing a vector diagram. Fig. 2.76 (c) shows a vector diagram, which is drawn as given below:

i. Take any point p and draw pq parallel to PQ vertically downwards. Cut pq = 4 kN to same scale.

ii. Now from q draw qr parallel to QR vertically upwards and cut qr = 2 kN to the same scale.

iii. From r draw rp parallel to RP vertically upwards and cut rp = 2 kN to the same scale.

iv.  Now from p, draw a line ps parallel to PS and from r, draw a line rs parallel to RS, meeting the first line at s. This is vector diagram for joint (A). Similarly the vector diagrams for joint (B) and (C) can be drawn.

3.   Preparing a force table. The magnitude of a force in a member is known by the length of the vector diagram for the corresponding member, i.e. the length ps of the vector diagram will give the magnitude of force in the member PS of the frame.

Nature of the force (i.e., tensile or compressive) is determined according to the following procedure:

1. In the space diagram, consider any joint. Move round that joint in a clockwise direction. Note the order of two capital letters by which the members are named. For example, the members at the joint (A) in space diagram Fig; 2.76 (b) are named as PS, SR and RP.
2.  Now consider the vector diagram. Move on the vector diagram in the order of the letters (i.e., ps, sr and rp).
3. Now mark the arrows on the members of the space diagram of that joint (here joint A).
4. Similarly, all the joints can be considered and arrows can be marked.
5. If the arrow is pointing towards the joint, then the force in the member will be compressive whereas if the arrow is away from the joint, then the force in the member will be tensile

Problem.2.45. Find the forces in the members AB, AC and BC of the truss shown in Fig. 2.77.

Sol. First determine the reactions RB and RC

Now draw the space diagram for the truss along with load of 20 kN and the reaction RB and RC equal to 15 kN and 5 kN respectively as shown in Fig. 2.77 (b). Name the members AB, AC and BC according to Bow’s notations as PR, QR and RS respectively. Now construct the vector diagram as shown in Fig. 2.77 (c) and as explained below:

i.  Take any point p and draw a vertical line pq downward equal to 20 kN to some suitable scale. From q draw a vertical line qs upward equal to 5 kN to the same scale to represent the reaction at C. Then sp will represent the reaction RB to the scale.

Fig. 2.77 (c) Vector Diagram

ii.  Now draw the vector diagram for the joint (B) From p, draw a line pr parallel to PR and from s draw a line sr parallel to SR, meeting the first line at r. Now prs is the vector diagram for the joint (B). Now mark the arrows on the joint B. The arrow in member PR will be towards to joint B, whereas the arrow in the member RS will be away from the joint. B as shown in Fig. 2.77 (b).

iii.Similarly draw the vector diagrams for joint A and C. Mark the arrows on these joints in space diagram.

Now measure the various sides of the vector diagram. The forces are obtained by multiplying the scale factor. The forces in the members are given a tabular form as.

 Member Force in member Nature of force According to given truss According to Bow’s notation AB AC BC PR QR RS 17.3 kN10.0 kN 8.7 kN CompressiveCompressive Tensile

Problem 2.46. A truss of span 7.5 m carries a point load of 1000 N at joint D as shown

in Fig. 2.78. Find the reactions and forces in the member of the truss.

Sol. First determine the reactions RA and RB

Taking moments about A, we get

RB × 7.5 = 5 × 1000

Now draw the space diagram for the truss along with load of 1000 N and reactions RA and RB equal to 333 N and 667 N respectively as shown in Fig. 2.78 (b). Name the members AC, CB, AD, CD and DB according to Bow’s notations as PR, PQ, RT, QR and QS respectively. Now construct the vector diagram as shown in Fig. 2.78 (c) and as explained below:

i.                    Take any point s and draw .a vertical line st downward equal to load 1000 N to some suitable scale. From t draw a vertical line tp upward equal to 333 N to the same scale to represent the reaction at A. The ps will represent the reaction RB to the scale.

ii.                 Now draw the vector diagram for the joint A. From p, draw a line pr parallel to PR and from t draw a line tr parallel to RT, meeting the first line at r. Now prt is the vector diagram for the joint A. Now mark the arrows on the joint A. The arrow in the member PR will be towards the joint A, whereas the arrow in the member RT will be away from the joint A as shown in Fig. 2.78 (b).

iii.               Similarly draw the vector diagrams for the joint C, Band D. Mark the arrows on these joints as shown in Fig. 2.78 (b).

Now measure the various sides of the vector diagrams. The forces in the members are obtained by multiplying the scale factor to the corresponding sides of the vector diagram. The forces in members are given in a tabular form as :

 Member Force in member Nature of force According to given truss According to Bow’s notation AC AD CB CD DB PR RT PQ QR QS 666 N 576.7 N 1333 N 1155 N 1555 N Compressive Tensile Compressive Tensile Tensile

Problem 2.47. Determine the forces in all the members of a cantilever truss shown in Fig. 2.79.

Sol. In this case the vector diagram can be drawn without knowing the reactions. First of all the space diagram for the truss along with loads of 1000 N of joints B and C. Name the members AB, BC, CD, DE, AD and BD according to Bow’s notation as PT, QS, SR, RV, VT and ST respectively. Now construct the vector diagram as shown in Fig. 2.79 (c) and as explained below:

1. The vector diagram will be started from joint C where forces in two members are unknown. Take any point q and draw a vertical line qr downward equal to load 1000 N to some suitable scale. From r, draw a line rs parallel to RS and from q draw a line qs parallel QS, meeting the first line at s. Now qrs is the vector diagram for the joint C. Now mark the arrows on the joint C. The arrow in the member RS will be towards the joint C, whereas the arrow in the member SQ will be away from the joint C as shown in Fig. 2.79 (b).

Fig.2.79 (c) Vector Diagram

ii.Now draw the vector diagram for the joints B and D similarly.

Mark the arrows on these joints as shown in Fig. 2.79 (b).

Now measure the various sides of the vector diagram. The forces in the members are given in a tabular form as:

 Member Force in member Nature of force According to given truss According to Bow’s notation AB BC CD DE AD BD PT QS SR RV VT ST 1333 N 1333 N 1666 N 2500 N 833 N 1000 N Tensile Tensile Compressive Compressive Tensile Compressive

From the vector diagram, the reactions R1 and R2 at A and E can be determined in magnitude and directions.

Reaction R2 = rv = 2500 N. This will be towards point E.

Reaction R1 = vp = 2000 N. This will be away from the point A as shown in Fig. 2.79 (b).  The reaction R1 is parallel to vp.